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Week 5Stewart §2.9, 3.1, 3.2154 min full read
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05Linear approximation, differentials, extreme values, and the Mean Value Theorem

No calculator, and the paper wants a decimal for $\sqrt{26}$. You know $\sqrt{25}=5$ without thinking, and $26$ sits one step to the right of $25$. The whole question is what that one step is worth, and you already own the tool that prices it.

By the end of this section you can turn $\sqrt{25}=5$ into a hand estimate for $\sqrt{26}$ that is within $0.001$ of the truth, say which way it errs, and find the largest and smallest value any continuous function takes on a closed interval.

In 60 seconds

Close to a point, a differentiable function and its tangent line are interchangeable; over a closed interval, the largest and smallest values of a continuous function hide only at critical numbers and at the two endpoints.

at a
$$L(x)=f(a)+f'(a)(x-a)$$

you need $f$ at an awkward point that sits near a friendly one

$$dy=f'(x)\,dx$$

a small input change $dx$ is given and you want the output change

Closed Interval Method
$$\max_{[a,b]}f=\max\{f(c_1),\dots,f(c_k),f(a),f(b)\}$$

the interval is closed, $f$ is continuous, and the question says absolute

Mean Value Theorem
$$f'(c)=\frac{f(b)-f(a)}{b-a}\quad\text{for some }c\in(a,b)$$

you know something about $f'$ everywhere and want a conclusion about $f$

Three most common mistakes

  1. Writing the tangent line as $L(x)=f'(a)(x-a)$ and losing the $f(a)$ term, so the estimate misses by the whole height of the base point.

  2. Reporting the location $c$ when the question asks for the maximum value $f(c)$, or handing back $f'(c)$ instead of $f(c)$.

  3. Running the closed interval search on critical numbers only and never evaluating the two endpoints, which are candidates precisely because no calculus points at them.

Midterm 1, Midterm 2 and the final carry 28% each, quizzes 10% and homework 6%. A midterm total under 40 points out of 200 means an FZ and no final exam, so this material, which belongs to the first midterm window, is expensive to skip.

How much time do you have?
10 minutes

You walk in knowing the tangent line formula, the four step search for absolute extrema, and the three mistakes that cost the most marks.

card, c-linearization, c-closed-interval, formula card
45 minutes

You can also tell $dy$ from $\Delta y$, find critical numbers where the derivative fails to exist, and use the Mean Value Theorem to bound a value.

card, c-linearization, c-differentials, c-critical-numbers, c-closed-interval, c-mvt, scaffolding ladder, practice C
full read

Everything above plus the two existence theorems with their failure cases, the worked examples, and the mixed practice that decides whether you can pick the right method without being told which one it is.

everything in order, practice A, practice B, practice C, practice D, self check
By the end of this section

  1. Compute the linearization of a function at a chosen base point and use it to estimate a nearby value by hand.

  2. Distinguish the differential $dy$ from the true change $\Delta y$ and use $dy$ to push a measurement tolerance through to a computed quantity.

  3. Classify a point as a local or an absolute extremum and report the value and the location separately.

  4. State the two hypotheses of the Extreme Value Theorem and produce a function that loses its extreme values when one of them fails.

  5. Find every critical number of a function, including those where the derivative fails to exist, and reject points outside the domain.

  6. Apply the Closed Interval Method to report the and minimum of a continuous function on a closed interval.

  7. Use the Mean Value Theorem to convert a bound on $f'$ into a bound on the values of $f$.

Syllabus coverage
2.9

Linear approximations and differentials

The tangent line as an estimator comes first; the same line rewritten in terms of changes, which is where measurement error lives, is the block after it.

covered
3.1

Maximum and minimum values

Four blocks share this token: the definitions, the Extreme Value Theorem, critical numbers with Fermat's theorem, and the search method they combine into.

covered
3.2

The Mean Value Theorem

Rolle's theorem appears inside this block as the special case with equal endpoint values, since the proof of the general statement runs through it.

covered
derivatives of exponential functions

Not part of this week's line

Differentiating $e^x$ and $a^x$ needs the exponential and logarithm functions themselves, which this course meets in the transcendental functions weeks late in the term. Nothing in this section uses them, and no problem here requires them.

deferred
Recall first
Tangent line at a point

$y=f(a)+f'(a)(x-a)$: the line through $(a,f(a))$ with slope $f'(a)$.

The first half of this section is this one line, read as a formula for approximate values instead of as a picture.

Chain rule

$\dfrac{d}{dx}f(g(x))=f'(g(x))\,g'(x)$.

Every derivative taken here goes through it, and the mixed practice at the end leans on it heavily.

Related rates

Differentiating one relation with respect to $t$ turns a relation between quantities into a relation between rates: from $x^{2}+y^{2}=L^{2}$ we get $2x\frac{dx}{dt}+2y\frac{dy}{dt}=0$.

The differential $dy=f'(x)\,dx$ is the same move with time removed, and one mixed question deliberately puts the two versions side by side.

Continuity at a point

$f$ is continuous at $a$ when $\lim_{x\to a}f(x)=f(a)$, which already presumes that $a$ belongs to the domain of $f$.

It is a hypothesis of both existence theorems in this section, so checking it is part of every answer.

The second derivative

$f''$ is the derivative of $f'$, so it measures how fast the slope itself is changing.

It is used once, in a check on how large a linear estimate's error ought to be; no problem in this section requires computing it.

Derivatives of sine and cosine

$(\sin x)'=\cos x$ and $(\cos x)'=-\sin x$, with $x$ measured in radians.

Two of the harder searches here are trigonometric, and the radian condition decides whether the derivative formula is even true.

Try it yourself first (2 questions)
1§05.0 — the tangent line, written down correctly●●○○○

Everything in the first half of this section is a tangent line used as a formula, so it is worth being sure the line itself comes out right.

Given

  • $f(x)=\sqrt{x}$

  • base point $a=4$, so $f(4)=2$

  • $f'(x)=\dfrac{1}{2\sqrt{x}}$

Find

  1. (a) Which equation is the tangent line to $f$ at $a=4$?

Hint 1/4

You need two numbers from the base point: the height there and the slope there.

Hint 2/4

The tangent at $a$ is $y=f(a)+f'(a)(x-a)$: height first, then slope times the step.

Hint 3/4

Here $f(4)=\sqrt4=2$ and $f'(4)=\dfrac{1}{2\sqrt4}=\dfrac14$, and the step is $(x-4)$.

Hint 4/4

So the line is $y=2+\tfrac14(x-4)$.

Show solution
Collect the two numbers the line needs
$$f(4)=\sqrt{4}=2$$

the height of the point the line has to pass through

$$f'(x)=\tfrac12x^{-1/2}=\frac{1}{2\sqrt{x}}\ \Rightarrow\ f'(4)=\tfrac14$$

the slope must be evaluated at the base point; leaving it as a function of $x$ would give a curve, not a line

Assemble
$$y=2+\tfrac14(x-4)$$

point plus slope times step, in that order

Answer $$\boxed{y=2+\tfrac14(x-4)}$$
Check

At $x=4$ the line returns $2$, which is $f(4)$, and its slope $\tfrac14$ matches $f'(4)$; a line that fails either test is not tangent there.

2§05.0 — a trigonometric equation with two answers●●○○○

Searching for critical numbers of a trigonometric function ends in an equation like this one, and the usual loss of marks is not the calculus but stopping at the first solution.

Given

  • $1-2\sin x=0$

  • $x$ ranges over the closed interval $[0,\pi]$

Find

  1. (a) Find every $x$ in $[0,\pi]$ that satisfies the equation.

Hint 1/4

Isolate $\sin x$ first, then ask how many angles in the given range have that sine.

Hint 2/4

On $[0,\pi]$ the sine takes each value in $(0,1)$ twice: once in the first quadrant, once in the second.

Hint 3/4

From $1-2\sin x=0$ we get $\sin x=\tfrac12$; on $[0,\pi]$ that happens at $x=\pi/6$ and at $x=\pi-\pi/6$.

Hint 4/4

So $x=\pi/6$ and $x=5\pi/6$, two solutions, not one.

Show solution
Isolate the sine
$$\sin x=\tfrac12$$

one algebraic step, and now the question is purely about angles

Count the angles, do not just name one
$$x=\tfrac{\pi}{6}$$

the first quadrant solution, the one the reflex answer gives

$$x=\pi-\tfrac{\pi}{6}=\tfrac{5\pi}{6}$$

sine is symmetric about $\pi/2$, so the second quadrant carries a twin; it lies inside $[0,\pi]$ and therefore counts

Answer $$\boxed{x=\tfrac{\pi}{6},\ \tfrac{5\pi}{6}}$$
Check

Substitute back: $\sin(\pi/6)=0.5$ and $\sin(5\pi/6)=0.5$, so $1-2(0.5)=0$ holds at both.

Whenever a search ends in $\sin x = k$ or $\cos x = k$, count the solutions in the interval before moving on. A dropped solution is a dropped candidate, and the dropped candidate is often the winner.

Notation
symbolreads asmeanswatch out
$L(x)$

the linearization of f at a

the tangent line to $f$ at $a$, used as a formula for approximate values of $f$ near $a$

$L$ carries its base point inside it; change $a$ and every number in $L$ changes.

$dx,\ dy$

the differential of x, the differential of y

$dx$ is any input change you choose; $dy=f'(x)\,dx$ is the change the tangent line predicts

$dx$ is an ordinary finite number here, not a limit and not an infinitely small quantity.

$\Delta y$

delta y

the true change $f(x+dx)-f(x)$

$\Delta y$ is what the function does, $dy$ is what the tangent says it does; they agree exactly only when $f$ is linear.

$c$

a critical number of f

a point of the domain where $f'(c)=0$ or $f'(c)$ fails to exist

A point outside the domain is never a critical number, however badly $f'$ behaves near it. The name also varies: and critical number mean the same thing here, and both are values of $x$.

$[a,b]\ \text{versus}\ (a,b)$

closed interval, open interval

the square bracket includes the endpoint, the round bracket excludes it

The Extreme Value Theorem needs a closed bounded interval; the Mean Value Theorem needs continuity on the closed one and differentiability only on the open one.

$f(c)\ \text{versus}\ c$

the value versus the location

the maximum value is a number on the $y$ axis, the point where it happens is a number on the $x$ axis

Find the maximum asks for the value; where asks for the location. Give both and you cannot be marked down for either.

Conventions used here
Domain comes first

Continuity, extrema and critical numbers are only ever discussed at points that belong to the domain of $f$. A point where $f$ is undefined is not a point where $f$ is discontinuous, and it is never a critical number.

Local means open interval

$f$ has a at $c$ when $f(c)\ge f(x)$ for every $x$ in some open interval around $c$, so an endpoint is never a local extremum here, although it can certainly be an absolute one. No answer in this section depends on that choice, because the Closed Interval Method tests endpoints separately.

Value versus location

The absolute maximum is the number $f(c)$; $c$ is where it occurs. A complete answer names both.

Radian convention

Every derivative formula for $\sin$ and $\cos$ used here assumes it, so $31^{\circ}$ becomes $31\pi/180$ before any differentiation.

Approximately equal is not equal

The symbol $\approx$ marks an estimate whose error we are willing to bound, not an equality.

Unbounded means no maximum

If $f$ grows without bound on an interval it has no maximum there; we say the maximum does not exist rather than that it is infinite.

Critical point and critical number

We use the two names for the same object: a number $c$ in the domain of $f$ with $f'(c)=0$ or $f'(c)$ undefined. It is a location on the $x$ axis, so a count of critical points is a count of $x$ values, and an answer that lists pairs $(c,f(c))$ is answering a different question from the one that was asked.

5.1Trading a curve for its tangent line

Replace $f$ near a friendly base point $a$ by its tangent line, so awkward values like $8.06^{1/3}$ cost one multiplication.

Where we stand: we can differentiate almost anything, and we still cannot put a decimal on $\sqrt{26}$. The derivative we already own answers exactly that.

Solvable with what we have

  • differentiate $f(x)=\sqrt{x}$ and get $f'(x)=\dfrac{1}{2\sqrt{x}}$

  • evaluate $f$ at a friendly point: $\sqrt{25}=5$, exactly

  • write the tangent line to $f$ at $(25,5)$

Not solvable yet

  • produce a decimal for $\sqrt{26}$ by hand

  • produce a decimal for $\sin(0.1)$ by hand

  • say how far off any hand estimate of ours is

Round the input: $26$ is nearly $25$, so $\sqrt{26}$ is nearly $\sqrt{25}=5$.

Why it fails

That answer is the value of $f$ at the wrong point, and it throws away the one thing we know about the neighbourhood: how fast $\sqrt{x}$ is climbing there. The true value is $5.09902$, so rounding is off by $0.099$ and the first decimal is already wrong. Notice what the rounding used and what it did not: it used $f(25)$, never $f'(25)$.

RuleRule 5.1: Linear approximation
Conditions

  • $f$ is differentiable at $a$

  • $x$ is near $a$, and how near decides how good the estimate is

  • The error $f(x)-L(x)$ is a graded object in its own right, not a footnote to the estimate. Its sign is fixed by the concavity of $f$ between $a$ and $x$, so concave up means the line runs below the curve; and for a concrete $f$ the error can be written in closed form over a common denominator, after which an inequality such as $\lvert f(x)-L(x)\rvert<\varepsilon$ is solved for $x$ and answered with an interval rather than a number.

$$\boxed{\ f(x)\approx L(x)=f(a)+f'(a)(x-a)\quad\text{for }x\text{ near }a\ }$$

Start at the height of the friendly point, then add the slope there times the horizontal step you took.

Why the tangent line is the right line to use

Differentiability at $a$ says the difference quotient $\frac{f(x)-f(a)}{x-a}$ tends to $f'(a)$, so near $a$ it equals $f'(a)+\varepsilon$ with $\varepsilon$ small. Multiplying by $(x-a)$ gives $f(x)=f(a)+f'(a)(x-a)+\varepsilon(x-a)$, and the error term is small times small. That product is the whole story: it explains why the line is excellent near $a$ and why it decays once $x$ moves away, since then only one factor is small.

Looks like this, but is not

$L$ is a formula for $f$, so $L(x)=5+0.1(x-25)$ can be used wherever I like.

At $x=100$ it returns $12.5$ while $\sqrt{100}=10$: an error of $2.5$, a quarter of the true value. Watch the error grow with the distance from the base point: $0.00098$ at $x=26$, $0.1$ at $x=36$, $2.5$ at $x=100$. A linearization is a local instrument, and the base point is part of it.

Estimating √26 by hand

Use a linearization to estimate $\sqrt{26}$ without a calculator, and decide whether the estimate is too big or too small.

Given

  • $f(x)=\sqrt{x}$

  • the target input is $x=26$

Find

a decimal estimate of $\sqrt{26}$ and the direction of its error

Solution
Choose the base point
$$a=25$$

the nearest input where both $f(a)$ and $f'(a)$ are exact; $a=16$ is also exact but ten units away, and the error grows with the step

$$f(25)=5,\qquad f'(x)=\frac{1}{2\sqrt{x}}\ \Rightarrow\ f'(25)=\frac{1}{10}$$

the two numbers the line is made of

Build the line and walk one step
$$L(x)=5+\tfrac{1}{10}(x-25)$$

height at the base point plus slope times step

$$L(26)=5+\tfrac{1}{10}(1)=5.1$$

the step is $26-25=1$, so the whole estimate costs one multiplication

Say which way it errs
$$5.1^{2}=26.01>26$$

squaring the estimate overshoots the target, so $5.1$ is above the true root; the tangent of a curve that bends downward lies above it

Answer $$\boxed{\sqrt{26}\approx 5.1\ \text{(slightly too big)}}$$
Check

Square the answer instead of taking a root: $5.1^{2}=26.01$, only $0.01$ past $26$. An estimate whose square lands that close cannot be far from the root. The true value is $5.09902$, so we are high by $0.00098$.

One derivative, one multiplication, one addition.

The base point is the only real decision in this method: take the nearest input where the function and its derivative are both exact.

Why sin x ≈ x, and where that stops being usable

Linearize $f(x)=\sin x$ at $a=0$ and test the result at $x=0.1$ and at $x=1$.

Given

  • $f(x)=\sin x$ with $x$ in radians

  • base point $a=0$

Find

the linearization and its error at two different distances

Solution
Build the line at zero
$$f(0)=0,\qquad f'(x)=\cos x\ \Rightarrow\ f'(0)=1$$

both numbers are exact at $0$, which is what makes $0$ the natural base point for sine

$$L(x)=0+1\cdot(x-0)=x$$

the famous small angle rule is nothing but a tangent line

Test it near and far
$$\sin(0.1)\approx 0.1$$

true value $0.0998334$, so the error is $0.00017$: three good decimals

$$\sin(1)\approx 1$$

true value $0.841471$, so the error is $0.159$, almost a fifth of the answer

Answer $$\boxed{\sin x\approx x\ \text{near }0;\ \text{error }0.00017\text{ at }x=0.1,\ 0.159\text{ at }x=1}$$
Check

No calculator needed for the far case: $\lvert\sin x\rvert\le 1$ for every $x$, while $L(x)=x$ passes $1$ at $x=1$ and keeps climbing. The approximation is guaranteed to fail eventually, and the numbers show it starting to.

Multiplying the step by ten multiplied the error by roughly a thousand here. Distance from the base point is not a detail, it is the whole error budget.

Checkpoint
§05.1 — choosing the base point●●○○○

Choosing $a$ is the only decision in a linearization, and it is where the marks go. You are asked to estimate $(1.98)^{4}$ by hand.

Given

  • the quantity to estimate is $(1.98)^{4}$

  • $f(x)=x^{4}$

Find

  1. (a) Which base point $a$ should the linearization use?

Hint 1/4

Ask what the base point has to give you: exact values of both $f$ and $f'$ there, as close to the target as possible.

Hint 2/4

The estimate is $L(x)=f(a)+f'(a)(x-a)$, so a good $a$ makes $f(a)$ and $f'(a)$ cheap and makes $x-a$ small.

Hint 3/4

The target is $x=1.98$. At $a=2$ we get $f(2)=16$ and $f'(2)=32$ with no effort, and the step is $x-a=-0.02$.

Hint 4/4

So take $a=2$, giving $L(1.98)=16+32(-0.02)=15.36$.

Show solution
Test the candidates against the two requirements
$$a=2:\ f(2)=16,\ f'(2)=4(2)^{3}=32,\ x-a=-0.02$$

exact and close, which is everything the method wants

$$L(1.98)=16+32(-0.02)=15.36$$

the estimate; the true value is $15.3695$, so we are low by about $0.0095$

Answer $$\boxed{a=2,\quad (1.98)^{4}\approx 15.36}$$
Check

Bound the answer without computing it: $1.98<2$ and the fourth power is increasing, so the true value is under $16$, and it should be close since $1.98$ is close to $2$. Our $15.36$ fits both.

⚠ Dropping the height and keeping only the slope

the slope feels like the calculus part, so $f(a)$ gets treated as scenery

wrong$$L(x)=f'(a)(x-a)$$
right$$L(x)=f(a)+f'(a)(x-a)$$
⚠ Leaving the derivative as a function of x

the formula is copied before $f'$ has been evaluated anywhere

wrong$$L(x)=f(a)+f'(x)(x-a)$$
right$$L(x)=f(a)+f'(a)(x-a)$$

5.2dy and Δy: the predicted change and the real one

$dy=f'(x)dx$ is the tangent's predicted output change for a small input step; reach for it when a tolerance is given.

The tangent line answered where is $f$ near $a$. Rewrite the same line in terms of changes and it answers how much does $f$ move when $x$ moves, which is the form every measurement question arrives in.

DefinitionDefinition 5.2: The differential
Conditions

  • $f$ is differentiable at $x$

  • $dx$ is any number you choose; it does not have to be small, though the usefulness of what follows does depend on its being small

$$\boxed{\ dy=f'(x)\,dx\qquad\text{against}\qquad \Delta y=f(x+dx)-f(x)\ }$$

The differential is the rise of the tangent line over a step of $dx$; delta y is the rise of the function itself over the same step.

Looks like this, but is not

$dy$ and $\Delta y$ are never equal, so the differential always lies a little.

For $f(x)=3x+1$ they agree exactly, for every $dx$: $\Delta y=3(x+dx)+1-(3x+1)=3\,dx$ and $dy=3\,dx$. A straight graph is its own tangent. The gap $\Delta y-dy$ is not a defect of the method, it is a measurement of how far the graph is from being straight.

step $dx$predicted $dy$true $\Delta y$$\Delta y-dy$

$1$

$0.1$

$0.099020$

$-0.00098$

$0.5$

$0.05$

$0.049752$

$-0.00025$

$0.1$

$0.01$

$0.009990$

$-0.0000100$

$0.01$

$0.001$

$0.0009999$

$-0.0000001$

Divide the step by ten and the gap divides by about a hundred: from $0.00098$ to $0.0000100$ to $0.0000001$. The error falls like the square of the step, which is the entire reason a straight line is worth using on a curved graph. Over a short step the mistake is not merely small, it is small compared with the step itself.

A cube measured to ±0.05 cm: how wrong is the volume?

A cube is measured and its side comes out as $10$ cm with a tolerance of $\pm0.05$ cm. Estimate the resulting uncertainty in the computed volume, in absolute and in relative terms.

Given

  • $V=s^{3}$

  • $s=10$ cm

  • $ds=0.05$ cm

Find

$dV$ in cm³, and $dV/V$ as a percentage

Solution
Differentiate the relation, then feed the tolerance in
$$dV=3s^{2}\,ds$$

the tolerance is a change in the input, so the differential is the natural language; nothing about time appears here

$$dV=3(10)^{2}(0.05)=15\ \text{cm}^{3}$$

substitute only after differentiating, exactly as with rates

Convert to relative error, which travels better
$$\frac{dV}{V}=\frac{3s^{2}\,ds}{s^{3}}=3\,\frac{ds}{s}$$

the algebra cancels the size of the cube, so the answer no longer depends on the units

$$\frac{dV}{V}=3\left(\frac{0.05}{10}\right)=0.015=1.5\%$$

half a percent in the side becomes one and a half percent in the volume

Answer $$\boxed{dV=15\ \text{cm}^{3}\quad\text{and}\quad dV/V=1.5\%}$$
Check

Compute the true change for comparison: $10.05^{3}-1000=15.075$ cm³. The differential said $15$, so it missed the change by $0.075$ cm³, which is half a percent of the change and $0.0075\%$ of the volume.

One differentiation and one substitution; the relative form took one extra line and removed the units from the answer.

The exponent is the amplifier. A in the side shows up three times over in a volume, twice over in an area, and half as much in a square root.

Checkpoint
§05.2 — when the differential is exact●●○○○

The differential is usually an estimate. There is one family of functions where it is not, and recognising it is a quick way to check whether you have understood what $dy$ measures.

Given

  • $f(x)=3x+1$

  • any base point $x$ and any step $dx$

Find

  1. (a) True or false: $dy=\Delta y$ for every choice of $dx$.

Hint 1/4

Do not test numbers yet. Ask what $dy$ and $\Delta y$ mean geometrically and how the graph of this particular $f$ relates to its tangent.

Hint 2/4

$dy=f'(x)\,dx$ is the rise along the tangent; $\Delta y=f(x+dx)-f(x)$ is the rise along the graph.

Hint 3/4

For $f(x)=3x+1$: $f'(x)=3$, so $dy=3\,dx$, while $\Delta y=\left(3(x+dx)+1\right)-(3x+1)=3\,dx$.

Hint 4/4

The two expressions are identical, so the statement is true.

Show solution
Compute both sides in general, not at one sample point
$$dy=f'(x)\,dx=3\,dx$$

the derivative is the constant $3$, so no base point enters

$$\Delta y=\bigl(3(x+dx)+1\bigr)-(3x+1)=3\,dx$$

the $3x$ and the $+1$ cancel, which is exactly what a straight graph does

Answer $$\boxed{dy=\Delta y=3\,dx\ \text{for every }dx}$$
Check

Consistency check against the error rule: the gap $\Delta y-dy$ is controlled by the bending of the graph, and a line does not bend, so a gap of zero is what the general theory predicts.

A non zero gap is a measurement of curvature. No curvature, no gap.

⚠ Reporting the quantity when the question asks for its change

the formula for $V$ is more familiar than the formula for $dV$

wrong$$dV=s^{3}=1000\ \text{cm}^{3}$$
right$$dV=3s^{2}\,ds=15\ \text{cm}^{3}$$
⚠ Substituting the numbers before differentiating

the same haste that flattens a related rates problem into a constant

wrong$$V=10^{3}=1000\ \Rightarrow\ dV=0$$
right$$dV=3s^{2}\,ds\ \Rightarrow\ dV=15$$

5.3Largest where, and largest among what

Distinguishes absolute extrema, which beat every value on the domain, from local ones, and the value $f(c)$ from its location $c$.

Second half of the section, and the question changes: not what is $f$ near here, but what is the biggest value $f$ ever takes here.

DefinitionDefinition 5.3: Absolute and local extrema
Conditions

  • $c$ belongs to the domain $D$ of $f$

  • for the local version, $f$ must be defined on an open interval around $c$

$$\boxed{\ \text{absolute max at }c:\ f(c)\ge f(x)\ \text{for all }x\in D;\qquad \text{local max at }c:\ f(c)\ge f(x)\ \text{for all }x\text{ in some open interval around }c\ }$$

Absolute means the value beats every other value the function takes; local means it only has to beat its immediate neighbours. Minima are the same sentences with the inequality turned round.

Looks like this, but is not

The tallest peak of the graph is the absolute maximum, so the deepest dip must be the absolute minimum.

In the figure the deepest interior dip sits at $x=3$ with value $1$, but the smallest value on $[0,6]$ is $0.5$, taken at the right endpoint, where the graph is still falling and simply runs out of interval. Dips compete with endpoints, and endpoints often win.

The same formula, three domains, three different answers

For $f(x)=x^{2}$, name the absolute extrema on each of these domains: all real numbers, the closed interval $[1,3]$, and the open interval $(1,3)$.

Given

  • $f(x)=x^{2}$

  • domains: $\mathbb{R}$, then $[1,3]$, then $(1,3)$

Find

the absolute maximum and minimum on each domain, if they exist

Solution
All real numbers
$$f(x)=x^{2}\ge 0=f(0)$$

a square is never negative, so $0$ is beaten by nobody and the minimum is attained at $x=0$

$$\text{no absolute maximum}$$

for any candidate $M$ the input $\sqrt{M}+1$ produces something larger, so no value can be the largest

The closed interval [1, 3]
$$f(1)=1,\qquad f(3)=9$$

on $[1,3]$ the function only increases, so the ends carry the extremes

$$\min=1\text{ at }x=1,\qquad \max=9\text{ at }x=3$$

both are endpoints; neither is a local extremum in the open interval sense, and both are perfectly good absolute ones

The open interval (1, 3)
$$\text{neither exists}$$

the values get arbitrarily close to $1$ and to $9$ without ever reaching them: for any $x_{0}\in(1,3)$ the midpoint of $x_{0}$ and $3$ is larger, and the midpoint of $1$ and $x_{0}$ is smaller

Answer $$\boxed{\mathbb{R}:\ \min 0,\ \text{no max};\quad [1,3]:\ \min 1,\ \max 9;\quad (1,3):\ \text{neither}}$$
Check

The construction is the check: on $(1,3)$ every proposed winner is beaten by an explicit competitor built from it, so the failure is not an oversight in the search but a fact about the domain.

Extrema belong to the pair, function and domain. Changing the domain without touching the formula changes the answer, and losing a bracket in the question changes the domain.

Checkpoint
§05.3 — the value, and where it happens●●●○○

A continuous function on a closed interval is described only by a few of its values. Nothing else about it is known, and nothing else is needed.

Given

  • $f$ is continuous on $[-2,5]$

  • $f(-2)=4$

  • a local maximum $f(1)=7$

  • a local minimum $f(3)=2$

  • $f(5)=9$

Find

  1. (a) What is the absolute maximum value of $f$ on $[-2,5]$, and where is it attained?

Hint 1/4

Separate the two words in the question: one asks for a number on the vertical axis, the other for a number on the horizontal axis.

Hint 2/4

The absolute maximum is the largest value taken anywhere on the interval, and endpoints are part of the interval; local maxima are only local claims.

Hint 3/4

The listed values are $f(-2)=4$, $f(1)=7$, $f(3)=2$ and $f(5)=9$, so the largest listed value is $9$, taken at the right endpoint.

Hint 4/4

The absolute maximum value is $9$, attained at $x=5$.

Show solution
List every candidate, endpoints included
$$f(-2)=4,\ f(1)=7,\ f(3)=2,\ f(5)=9$$

a local maximum is a candidate, not a verdict, and the two endpoints are candidates too

$$\max\{4,7,2,9\}=9\ \text{at}\ x=5$$

the comparison is between heights, and the winner is an endpoint even though nothing about $f$ turns around there

Answer $$\boxed{\text{value }9,\ \text{at }x=5}$$
Check

Cross check the wording: $9$ is a value of $f$ while $5$ is an input, and swapping them would claim $f$ reaches $5$ at the point $9$, which is outside the interval entirely.

⚠ Answering with the location when the value was asked for

the search produces $c$ last, so $c$ is the number still in your hand

wrong$$\max_{[-2,5]}f=5$$
right$$\max_{[-2,5]}f=9\ \text{at}\ x=5$$
⚠ Assuming the absolute maximum must be one of the local maxima

the interior of the graph is where the interesting shape is, so the ends stop being looked at

wrong$$\max=\max\{\text{local maxima}\}$$
right$$\max=\max\{\text{local maxima},\,f(a),\,f(b)\}$$

5.4The Extreme Value Theorem: when a largest value is guaranteed to exist

Guarantees a continuous function on a closed bounded interval attains a maximum and a minimum; drop either hypothesis and it fails.

Before hunting for the largest value, it pays to know whether there is one to find. Two hypotheses settle it.

TheoremTheorem 5.4: Extreme Value Theorem
Conditions

  • $f$ is continuous at every point of $[a,b]$

  • the interval is closed and bounded: both endpoints included, both finite

  • The theorem is a guarantee, not a test. On an interval that is not closed or not bounded the extreme values may perfectly well still exist, so a failed hypothesis licenses no conclusion at all, and the maximum and the minimum have to be settled one at a time by hand: typically one of them survives on a half open interval while the other is only approached at the excluded end.

$$\boxed{\ f\ \text{continuous on }[a,b]\ \Longrightarrow\ \exists\,c,d\in[a,b]\ \text{with}\ f(c)\le f(x)\le f(d)\ \text{for all }x\in[a,b]\ }$$

A continuous function on a closed bounded interval actually reaches a highest value and a lowest value somewhere on that interval.

Why we take this one on trust

The proof rests on the completeness of the real numbers rather than on anything about derivatives, and it belongs to a later analysis course. Nothing in this section depends on the argument; what you have to be able to do is check the two hypotheses, and notice when a problem quietly breaks one of them.

Looks like this, but is not

The theorem is about continuity, so a continuous function always has a maximum.

$f(x)=x$ on the open interval $(0,1)$ is continuous at every point of its domain and has no maximum: given any $x_{0}<1$, the midpoint $(x_{0}+1)/2$ is larger and still inside. Closed and bounded is a separate hypothesis, and both must hold at once.

A closed interval is not enough either

Let $f$ be defined on all of $[0,1]$ by $f(x)=1/x$ for $0<x\le 1$ and $f(0)=0$. Does $f$ attain an absolute maximum on $[0,1]$? An absolute minimum?

Given

  • $f(0)=0$

  • $f(x)=1/x$ for $0<x\le 1$

  • the interval $[0,1]$ is closed and bounded

Find

whether the two absolute extrema exist

Solution
Check the hypothesis that is actually in question
$$\lim_{x\to 0^{+}}\frac1x=+\infty\neq 0=f(0)$$

$f$ is defined at $0$, so continuity there is a fair question to ask, and the answer is no; the interval being closed does not repair that

$$\text{EVT does not apply}$$

one hypothesis fails, so the theorem promises nothing; it does not promise failure either

Settle each extremum by hand
$$\text{no maximum: given any }M,\ x=\frac{1}{M+1}\in(0,1]\ \text{gives}\ f(x)=M+1>M$$

an explicit competitor for every candidate, which is stronger than saying the function looks unbounded

$$\min=0\ \text{at}\ x=0$$

because $f(x)=1/x\ge 1>0$ on $(0,1]$ while $f(0)=0$, so the minimum survives

Answer $$\boxed{\text{no absolute maximum};\quad \text{absolute minimum }0\ \text{at}\ x=0}$$
Check

Test the competitor rule on a number: propose $M=1000$; then $x=1/1001$ lies in the interval and $f(x)=1001>1000$. The recipe defeats any candidate you name, which is what having no maximum means.

When a hypothesis fails, the theorem goes silent, not negative. Here the maximum genuinely disappears while the minimum is untouched, so each extremum has to be argued separately.

Checkpoint
§05.4 — which hypothesis actually buys you the guarantee●●●○○

Four descriptions of a function $f$. Exactly one of them forces $f$ to attain a largest value.

Given

  • in each option $f$ is a real valued function

  • no formula for $f$ is given, and none is needed

Find

  1. (a) Which description guarantees that $f$ attains an absolute maximum?

Hint 1/4

Do not look for the option that sounds strongest. Look for the one that matches both hypotheses of the guarantee word for word.

Hint 2/4

The Extreme Value Theorem needs continuity on an interval that is closed and bounded; drop either half and the conclusion can fail.

Hint 3/4

Continuity on $[0,1]$ has both: continuity, and a closed bounded interval with the endpoints included.

Hint 4/4

So the guarantee comes from continuity on the closed interval $[0,1]$.

Show solution
Match each description against the two hypotheses
$$\text{continuous on }[0,1]:\ \checkmark$$

continuity plus a closed bounded interval is exactly the theorem

$$\text{continuous on }(0,1]:\ 1/x\ \text{is a counterexample}$$

the interval is not closed, and we have already met a function that escapes

$$\text{bounded on }[0,1]:\ \text{a ceiling need not be reached}$$

boundedness says the values stay under some number, not that any value equals it

Answer $$\boxed{f\ \text{continuous on }[0,1]}$$
Check

Counterexamples confirm the three rejects: $1/x$ on $(0,1]$ is continuous with no maximum, and $f(x)=x$ for $x<1$ with $f(1)=0$ is bounded on $[0,1]$ with no maximum.

⚠ Reading the theorem backwards

an implication read as an equivalence, which is easy to do at speed

wrong$$f\ \text{not continuous}\ \Rightarrow\ \text{no maximum}$$
right$$f\ \text{not continuous}\ \Rightarrow\ \text{the theorem says nothing}$$
⚠ Applying the theorem on an interval that is not closed

the bracket at the end of the interval is small and gets copied wrongly

wrong$$\max_{(0,1]}\frac1x\ \text{exists}$$
right$$\frac1x\ \text{has no maximum on}\ (0,1]$$

5.5Critical numbers: the only interior places an extremum can hide

Narrows the interior search to points where $f'$ is zero or undefined, the only places a local extremum can sit.

The existence theorem told us the largest value is somewhere. Fermat tells us where it is allowed to be, which turns an infinite search into a short list.

TheoremTheorem 5.5: Fermat's theorem, and what a critical number is
Conditions

  • $c$ lies in the domain of $f$

  • $c$ is interior, not an endpoint: Fermat needs room on both sides

  • $f'(c)$ exists — that is the hypothesis doing the work

$$\boxed{\ f\ \text{has a local extremum at }c\ \text{and}\ f'(c)\ \text{exists}\ \Longrightarrow\ f'(c)=0\ }$$

If the graph turns around at a point where it has a tangent, that tangent has to be horizontal. A critical number is any point of the domain where the derivative is zero or fails to exist, so the two kinds together cover every way an interior extremum can occur. The three lines above are Fermat's hypotheses, not that definition: being interior is what the theorem needs, not what makes a point critical.

Where the zero comes from

Say $f$ has a local maximum at $c$ and $f'(c)$ exists. Approaching from the right, $\frac{f(c+h)-f(c)}{h}$ has a numerator that is at most zero and a positive denominator, so the limit is $\le 0$. Approaching from the left the denominator flips sign, so the limit is $\ge 0$. Since $f'(c)$ exists, both one sided limits equal it, and a number that is at once $\le 0$ and $\ge 0$ must be $0$. Notice which hypothesis each half used: the extremum controls the numerator, the differentiability lets the two sides be compared at all.

Looks like this, but is not

$f'(c)=0$, so $c$ is a local maximum or a local minimum.

$f(x)=x^{3}$ has $f'(0)=0$ and no extremum anywhere: it is negative to the left of $0$ and positive to the right, so $f(0)=0$ beats nothing. Fermat runs one way only, from extremum to zero derivative, never back. A critical number is a suspect, and classifying suspects is what the next section is for.

Both kinds of critical number in one function

Find every critical number of $f(x)=\lvert x-2\rvert+1$ and of $g(x)=x^{3}-6x^{2}+5$.

Given

  • $f(x)=\lvert x-2\rvert+1$

  • $g(x)=x^{3}-6x^{2}+5$

  • both are defined for every real $x$

Find

the critical numbers of each, and which kind they are

Solution
The corner: a critical number the equation f′ = 0 cannot find
$$f'(x)=-1\ \text{for}\ x<2,\qquad f'(x)=+1\ \text{for}\ x>2$$

the two one sided slopes disagree, so no single number can be the derivative at $2$

$$f'(2)\ \text{does not exist, and}\ 2\ \text{is in the domain}$$

both halves of the definition are satisfied, so $x=2$ is critical

The smooth one: solve the equation
$$g'(x)=3x^{2}-12x=3x(x-4)$$

factor rather than reach for the quadratic formula; the roots are then visible

$$g'(x)=0\ \text{at}\ x=0\ \text{and}\ x=4$$

$g'$ exists everywhere, so these two are the complete list

Answer $$\boxed{f:\ x=2\ \text{(no derivative)};\qquad g:\ x=0,\ x=4\ \text{(zero derivative)}}$$
Check

The corner is not a technicality: $\lvert x-2\rvert\ge 0$ with equality only at $x=2$, so $f(x)\ge 1=f(2)$ for every $x$ and the absolute minimum really does sit exactly where the derivative failed to exist.

Solving $f'(x)=0$ finds one species of critical number. Asking where $f'$ fails to exist finds the other, and on a graph with a corner or a vertical tangent that is where the answer usually is.

Critical numbers when f′(x) = 0 turns out to be a cubic

Find every critical number of $f(x)=9x^{4}+2x^{3}-57x^{2}+36x$.

Given

  • $f(x)=9x^{4}+2x^{3}-57x^{2}+36x$

  • $f$ is a polynomial, so its domain is every real number

Find

the complete list of critical numbers

Solution

The derivative comes out a cubic with no common variable factor and no grouping that works, so the quadratic formula is not available and guessing is not a method. The rational root theorem turns the hunt into a finite candidate list, and a single hit drops the degree to two, where factoring by hand is routine again.

Differentiate, then strip the constant factor
$$f'(x)=36x^{3}+6x^{2}-114x+36$$

term by term; nothing is factored yet, and in this shape the roots are invisible

$$f'(x)=6\bigl(6x^{3}+x^{2}-19x+6\bigr)$$

a nonzero constant can never make the product zero, so the critical numbers are the roots of the bracket alone; smaller end coefficients also shorten the candidate list in the next step

Write the candidate list off the two end coefficients
$$\text{if}\ \tfrac{p}{q}\ \text{is a root, then}\ p\mid 6\ \text{and}\ q\mid 6$$

the rational root theorem: the numerator divides the constant term and the denominator divides the leading coefficient, with the fraction in lowest terms

$$\pm1,\ \pm2,\ \pm3,\ \pm6,\ \pm\tfrac12,\ \pm\tfrac32,\ \pm\tfrac13,\ \pm\tfrac23,\ \pm\tfrac16$$

eighteen numbers, and the list is exhaustive: no rational number outside it can be a root, which is what makes the search finite

Test candidates until one hits, then stop
$$6(1)^{3}+(1)^{2}-19(1)+6=-6\neq 0$$

small integers first; each test costs one line, and $x=1$ is the cheapest of all

$$6(-2)^{3}+(-2)^{2}-19(-2)+6=-48+4+38+6=0$$

so $x=-2$ is a root; one root is all the division needs, so there is no reason to keep testing the other sixteen

Divide the found factor out
$$6x^{3}+x^{2}-19x+6=(x+2)\bigl(6x^{2}-11x+3\bigr)$$

synthetic division on the coefficient row $6,\ 1,\ -19,\ 6$ with $-2$: bring down $6$, then $1-12=-11$, then $-19+22=3$, then $6-6=0$

$$\text{remainder}=0$$

the last entry is the remainder, so a nonzero value there would mean either the wrong root or an arithmetic slip; it is a free check, not decoration

Finish the leftover quadratic and collect the list
$$6x^{2}-11x+3=(3x-1)(2x-3)$$

the discriminant $121-72=49$ is a perfect square, which is the signal that rational factors exist and are worth hunting for

$$f'(x)=6(x+2)(3x-1)(2x-3)$$

the fully factored derivative; every root is now readable, and so is the sign of $f'$ on each piece

$$x=-2,\qquad x=\tfrac13,\qquad x=\tfrac32$$

$f'$ is a polynomial, so it exists at every real number and the second species of critical number contributes nothing here

Answer $$\boxed{x=-2,\qquad x=\tfrac13,\qquad x=\tfrac32}$$
Check

Independent check from the coefficients rather than by dividing again. For $6x^{3}+x^{2}-19x+6$ the three roots must add up to $-\tfrac16$ and multiply to $-1$. Ours add to $-2+\tfrac13+\tfrac32=-\tfrac16$ and multiply to $-2\cdot\tfrac13\cdot\tfrac32=-1$. Both agree, so no root was invented and none was lost.

One differentiation, two candidate tests, one synthetic division, one quadratic factorisation.

A cubic derivative is neither a dead end nor a quadratic formula problem. One rational root, taken from a finite list, lowers the degree by one and hands the rest back to arithmetic you already do. The same three moves have already been run once, in the implicit differentiation section: asking where $y^{3}-3xy=9$ has slope $3$ produces the cubic $2y^{3}-y^{2}+9=0$, and there too it is the candidate list, one division, and the discriminant of the leftover quadratic that finish the question. If the arithmetic here felt new, that example is the second reading of it, on a curve instead of on a derivative.

A critical number that exists but cannot be written down

Show that $f(x)=x\cos x$ has exactly one critical number in $\left(0,\tfrac{\pi}{2}\right)$, and say what that settles about $f$ on $\left[0,\tfrac{\pi}{2}\right]$.

Given

  • $f(x)=x\cos x$

  • the interval $\left[0,\tfrac{\pi}{2}\right]$, with $x$ in radians

Find

how many critical numbers lie strictly inside the interval, and what they decide

Solution

Setting the derivative to zero gives $x\tan x=1$, and no rearrangement of that produces $x$ in closed form. So we stop trying to solve it. The question a paper asks about such an equation is never what is $c$; it is how many roots there are, and that can be answered with a sign change and a monotonicity remark about an elementary function.

Differentiate, then admit the equation is unsolvable
$$f'(x)=\cos x-x\sin x$$

product rule; $f'$ exists at every real number, so critical numbers can only come from the equation $f'=0$

$$f'(x)=\cos x\,(1-x\tan x)$$

pull out $\cos x$, which is strictly positive on the open interval; the sign of $f'$ therefore rides entirely on the bracket

$$f'(x)=0\iff x\tan x=1$$

no algebra separates $x$ from $\tan x$ here, and trying to is where the marks are lost, not won

Existence: catch the root between two numbers you can compute
$$g(x)=x\tan x,\qquad g(0)=0,\qquad g(1)=\tan 1\approx 1.557$$

name the left side so the claim becomes a statement about one continuous function; $g$ is continuous on $[0,1]$ because $\cos x$ does not vanish there

$$g(0)<1<g(1)\ \Rightarrow\ \exists\,c\in(0,1):\ g(c)=1$$

the Intermediate Value Theorem, which delivers a root without ever producing its decimal expansion

Uniqueness: g climbs, so it can pass 1 only once
$$0\le x_{1}<x_{2}<\tfrac{\pi}{2}\ \Rightarrow\ x_{1}\tan x_{1}<x_{2}\tan x_{2}$$

on this interval both $x$ and $\tan x$ are non negative and both increase, and a product of two such factors increases; no calculus is needed for this step

$$g\ \text{is strictly increasing, so}\ g(x)=1\ \text{has at most one solution}$$

a strictly increasing function takes each value at most once; together with the previous step, exactly one $c$, and it lies in $(0,1)$

Cash the critical number in without knowing its value
$$f\ \text{is continuous on}\ \left[0,\tfrac{\pi}{2}\right]\ \Rightarrow\ \text{an absolute maximum is attained}$$

the Extreme Value Theorem; this is what guarantees there is something to find before any candidate is examined

$$f(0)=0,\quad f\left(\tfrac{\pi}{2}\right)=0,\quad f(1)=\cos 1\approx 0.540>0$$

both endpoints give $0$ while an interior point gives something positive, so the maximum cannot be at an endpoint

$$\max_{[0,\pi/2]}f=f(c)=c\cos c$$

an interior maximum of a differentiable function sits at a critical number by Fermat, and there is only one, so it has to be $c$; naming the answer through $c$ is a complete answer

Answer $$\boxed{\text{exactly one }c\in\left(0,\tfrac{\pi}{2}\right),\ \text{and }\max f=c\cos c}$$
Check

Independent check from the sign of $f'$, which the argument above never used. From $f'(x)=\cos x\,(1-x\tan x)$ with $\cos x>0$, the derivative is positive while $g(x)<1$ and negative once $g(x)>1$, so $f$ rises then falls, exactly the shape a single interior maximum forces. Numerically that pins $c$ down as well: $f'(0.8)\approx 0.123>0$ and $f'(0.9)\approx-0.083<0$, so $c$ lies between $0.8$ and $0.9$ and $c\cos c\approx 0.561$.

Two evaluations of an elementary function, one monotonicity remark, no equation solved.

When $f'(x)=0$ has no closed form solution, the answer is an existence argument, not a number. A sign change gives at least one root, a monotonicity remark cuts it down to exactly one, and the final answer is allowed to carry the unknown $c$ inside it.

Checkpoint
§05.5 — counting critical numbers●●●○○

A cubic whose derivative factors as a perfect square. The count is the whole question.

Given

  • $f(x)=x^{3}-3x^{2}+3x$

  • the domain is all real numbers

Find

  1. (a) How many critical numbers does $f$ have?

Hint 1/4

Two separate searches: the zeros of $f'$, and the points where $f'$ fails to exist. A polynomial makes the second search short.

Hint 2/4

Differentiate, then factor the result before solving; the shape of the factorisation is what decides the count.

Hint 3/4

Here $f'(x)=3x^{2}-6x+3=3(x-1)^{2}$, which is zero exactly when $x=1$, and $f'$ exists for every real $x$.

Hint 4/4

So there is exactly one critical number, $x=1$.

Show solution
Differentiate and factor
$$f'(x)=3x^{2}-6x+3=3(x-1)^{2}$$

the perfect square is the point of the exercise: a repeated root is one number, not two

$$f'(x)=0\iff x=1$$

a square vanishes only where its base does

Check the other species
$$f'\ \text{is a polynomial, defined everywhere}$$

so nothing is contributed by the failure to exist case

Answer $$\boxed{\text{exactly one, at }x=1}$$
Check

Sign test either side: $f'(0)=3>0$ and $f'(2)=3>0$. The derivative touches zero at $x=1$ without changing sign, which is consistent with a repeated root, and tells you in advance that this critical number is not an extremum.

A repeated root is one critical number that happens to be a weak one: flat for an instant, no turn.

⚠ Calling a point outside the domain critical

the derivative misbehaves there, and the domain check gets skipped

wrong$$f(x)=\tfrac1x:\ x=0\ \text{is a critical number}$$
right$$0\notin\operatorname{dom}(f)\ \Rightarrow\ \text{no critical numbers}$$
⚠ Solving f′ = 0 and stopping

the equation is the visible half of the definition, the corner case is the silent half

wrong$$\text{critical numbers}=\{x:f'(x)=0\}$$
right$$\text{critical numbers}=\{x\in\operatorname{dom}f:\ f'(x)=0\ \text{or}\ f'(x)\ \text{undefined}\}$$
⚠ Treating an unsolvable f′ = 0 as an f′ with no zeros

the equation is the only visible route to a critical number, so when the algebra stalls the search feels finished

wrong$$x\tan x=1\ \text{cannot be solved}\ \Rightarrow\ \text{no critical number}$$
right$$g(0)<1<g(1)\ \text{with}\ g\ \text{increasing}\ \Rightarrow\ \text{exactly one}\ c$$
⚠ Reading the candidates of $F=f^{2}$ off $F(x)=0$

the square is the visible object and its roots are the easy thing to write down, so the chain rule step that produces the rest of the candidates gets skipped; differentiating in factored form instead keeps every candidate on the page, and a factor that never vanishes, such as a positive power of $x^{2}+3$ or $\sec^{2}x$, contributes no candidate and no failure of differentiability, which is worth a sentence of its own

wrong$$F=f^{2}:\quad \text{candidates}=\{x:\ F(x)=0\}$$
right$$F'=2ff':\quad \text{candidates}=\{x:\ f(x)=0\}\cup\{x:\ f'(x)=0\}$$

5.6The Closed Interval Method

Compare $f$ at every interior critical number and at both endpoints; the largest and smallest of those values are the absolute extrema.

Existence from one theorem, a short list of interior suspects from another: put them together and an infinite search collapses into a table with four rows.

MethodMethod 5.6: Absolute extrema on a closed interval
Conditions

  • $f$ is continuous on $[a,b]$

  • $c_1,\dots,c_k$ are the critical numbers lying strictly inside $(a,b)$

$$\boxed{\ \max_{[a,b]}f=\max\{f(c_1),\dots,f(c_k),f(a),f(b)\},\qquad \min_{[a,b]}f=\min\{f(c_1),\dots,f(c_k),f(a),f(b)\}\ }$$

Evaluate the function at every critical number inside the interval and at both endpoints; the largest number on that list is the absolute maximum and the smallest is the absolute minimum.

Why the list is complete

Continuity on a closed bounded interval guarantees the extremes are attained, so the search cannot come up empty. Wherever the maximum sits, it is either an endpoint or interior; if it is interior it is also a local maximum, and Fermat forces $f'=0$ there unless $f'$ fails to exist. Either way the point is critical. Nothing outside the list can win, and the list is finite, so what remains is arithmetic.

Looks like this, but is not

The method compares derivatives, since that is what the work was spent computing.

It compares values of $f$. The derivative is used once, to produce the candidate list, and then it is finished. Handing in $f'(c)$ as the maximum value throws away every mark on an otherwise correct solution, and it is invisible to the student who does it, because all the hard work was right.

Absolute extrema of x³ − 6x² + 5 on [−1, 5]

Find the absolute maximum and minimum values of $f(x)=x^{3}-6x^{2}+5$ on $[-1,5]$.

Given

  • $f(x)=x^{3}-6x^{2}+5$

  • the closed interval $[-1,5]$

Find

both absolute extreme values and where they occur

Solution
Check the hypothesis, then build the candidate list
$$f\ \text{is a polynomial, hence continuous on}\ [-1,5]$$

one line, and it is what licenses everything after it

$$f'(x)=3x^{2}-12x=3x(x-4)\ \Rightarrow\ x=0,\ x=4$$

both roots lie strictly inside $(-1,5)$, so both are candidates

Evaluate f, not f′, at all four candidates
$$f(-1)=-1-6+5=-2$$

left endpoint

$$f(0)=5$$

critical number

$$f(4)=64-96+5=-27$$

critical number

$$f(5)=125-150+5=-20$$

right endpoint, and note it does not win either extreme

Compare the four numbers
$$\max\{-2,5,-27,-20\}=5,\qquad \min\{-2,5,-27,-20\}=-27$$

arithmetic; the calculus finished two lines ago

Answer $$\boxed{\max=5\ \text{at}\ x=0,\qquad \min=-27\ \text{at}\ x=4}$$
Check

Independent check by sign of $f'$: $f'$ is positive on $(-1,0)$, negative on $(0,4)$ and positive on $(4,5)$, so the graph rises, falls, rises. That shape forces a peak at $0$ and a trough at $4$, and it predicts $f(5)>f(4)$, which the numbers confirm ($-20>-27$).

One differentiation, four evaluations, one comparison.

Write the four values in a row before deciding anything. The comparison is where the answer is, and it is also where haste costs marks.

Absolute extrema of 3x⁴ − 4x³ − 3x² − 36x on [−1, 3]

Find the absolute maximum and minimum values of $f(x)=3x^{4}-4x^{3}-3x^{2}-36x$ on $[-1,3]$.

Given

  • $f(x)=3x^{4}-4x^{3}-3x^{2}-36x$

  • the closed interval $[-1,3]$

Find

both absolute extreme values and where they occur

Solution

Step two of the method says solve $f'(x)=0$, and here that equation is a cubic that will not come apart by grouping. Rather than abandon the method we spend four lines factoring it. The payoff is double: the candidate list turns out to have one entry, and the quadratic left behind has no real root at all, which is a sentence the marking scheme pays for.

Licence, then the derivative in factored form
$$f\ \text{is a polynomial, hence continuous on}\ [-1,3]$$

one line, and it is what licenses the whole search

$$f'(x)=12x^{3}-12x^{2}-6x-36=6\bigl(2x^{3}-2x^{2}-x-6\bigr)$$

pull the constant out before hunting roots: it cannot vanish, and the smaller end coefficients cut the candidate list roughly in half

$$\tfrac{p}{q}\ \text{with}\ p\mid 6,\ q\mid 2:\quad \pm1,\ \pm2,\ \pm3,\ \pm6,\ \pm\tfrac12,\ \pm\tfrac32$$

the rational root theorem applied to the bracket; twelve candidates, and nothing outside the list can be a rational root

One root, then the division
$$2(1)^{3}-2(1)^{2}-1-6=-7\neq 0$$

$x=1$ is the cheapest test and it misses, so move down the list

$$2(2)^{3}-2(2)^{2}-2-6=16-8-2-6=0$$

$x=2$ hits, and one root is enough to divide with

$$2x^{3}-2x^{2}-x-6=(x-2)\bigl(2x^{2}+2x+3\bigr)$$

synthetic division on the row $2,\ -2,\ -1,\ -6$ with $2$: $-2+4=2$, then $-1+4=3$, then $-6+6=0$, and that final zero confirms the root

Say out loud what the leftover quadratic does
$$b^{2}-4ac=2^{2}-4(2)(3)=-20<0$$

a negative discriminant means no real root, so this factor contributes no further critical number; an unchecked factor is an assumption, not a result

$$2x^{2}+2x+3=2\left(x+\tfrac12\right)^{2}+\tfrac52>0$$

completing the square says more than the discriminant does: the factor is not merely nonzero but positive everywhere, so it cannot flip the sign of $f'$ either

$$\text{critical numbers in}\ (-1,3):\ x=2\ \text{only}$$

$f'$ is a polynomial, so nothing arrives from the failure to exist half of the definition

Evaluate f at the candidate and at both endpoints
$$f(-1)=3+4-3+36=40$$

left endpoint; at a negative input the two odd power terms flip sign, so $-4x^{3}$ contributes $+4$ and $-36x$ contributes $+36$, and that is what makes this the large value

$$f(2)=48-32-12-72=-68$$

the only interior candidate

$$f(3)=243-108-27-108=0$$

right endpoint; it wins neither extreme, which is worth noticing rather than assuming

$$\max\{40,-68,0\}=40,\qquad \min\{40,-68,0\}=-68$$

three values of $f$, never of $f'$; the calculus finished two lines ago

Answer $$\boxed{\max=40\ \text{at}\ x=-1,\qquad \min=-68\ \text{at}\ x=2}$$
Check

Independent check on the shape, read off the sign of $f'$ rather than off the table of values. Since $f'(x)=6(x-2)\bigl(2x^{2}+2x+3\bigr)$ and the quadratic factor is positive everywhere, $f'$ carries the sign of $x-2$: negative on $[-1,2)$, positive on $(2,3]$. So $f$ falls and then rises, which forces the minimum at $x=2$ and pushes the maximum out to an endpoint. Which endpoint wins is the one thing the sign line does not decide — a fall and then a rise says nothing about how far each one goes — so the two heights still have to be compared, and $40>0$ settles it.

One differentiation, two candidate tests, one synthetic division, one discriminant, three evaluations.

When the leftover quadratic has a negative discriminant, write that sentence down instead of skipping past it. It does two jobs at once: no extra critical number for the candidate list, and no extra sign change for the derivative.

Checkpoint
§05.6 — a search where the maximum happens twice●●●○○

A short closed interval search. The interesting part is the shape of the answer, not the arithmetic.

Given

  • $f(x)=x+\dfrac{4}{x}$

  • the closed interval $[1,4]$

Find

  1. (a) Report the absolute maximum and minimum values of $f$ on $[1,4]$ and state where each occurs.

Hint 1/4

Three jobs in order: continuity on the interval, the candidate list, the comparison of values.

Hint 2/4

Candidates are the critical numbers inside $(1,4)$ together with the two endpoints; differentiate $x+4x^{-1}$ to find the first group.

Hint 3/4

With $f(x)=x+4/x$ on $[1,4]$: $f'(x)=1-\dfrac{4}{x^{2}}=0$ gives $x=2$ inside the interval, so the candidates are $1$, $2$ and $4$.

Hint 4/4

The values are $f(1)=5$, $f(2)=4$, $f(4)=5$, so the maximum is $5$ and the minimum is $4$.

Show solution
Hypothesis and candidates
$$f\ \text{is continuous on}\ [1,4]$$

the only bad point of $x+4/x$ is $x=0$, which is not in this interval

$$f'(x)=1-\frac{4}{x^{2}}=0\ \Rightarrow\ x^{2}=4\ \Rightarrow\ x=2$$

the root $x=-2$ solves the equation but lies outside the interval, so it is not a candidate here

Evaluate and compare
$$f(1)=1+4=5,\quad f(2)=2+2=4,\quad f(4)=4+1=5$$

three values, and two of them tie

$$\max=5\ \text{at}\ x=1\ \text{and}\ x=4,\qquad \min=4\ \text{at}\ x=2$$

the maximum value is a single number even though it is attained at two different places

Answer $$\boxed{\max=5\ \text{at}\ x=1,4;\qquad \min=4\ \text{at}\ x=2}$$
Check

Independent check with the arithmetic and geometric mean inequality: $x+\frac4x\ge 2\sqrt{4}=4$ for every $x>0$, with equality exactly when $x=4/x$, that is at $x=2$. The minimum we found is forced, and no search could beat it.

The extreme value is unique; the location need not be. Report the value once and list every place it happens.

⚠ Never evaluating the endpoints

the endpoints take no calculus to find, so they do not feel like part of the work

wrong$$\max_{[-1,5]}f=\max\{f(0),f(4)\}=5$$
right$$\max_{[-1,5]}f=\max\{f(-1),f(0),f(4),f(5)\}=5$$
⚠ Comparing values of f′ instead of values of f

the derivative is the object you spent the effort on, so it stays in hand

wrong$$\text{compare } f'(c_1),f'(c_2),\dots$$
right$$\text{compare } f(c_1),f(c_2),f(a),f(b)$$
⚠ Asserting the sign of a leftover quadratic instead of testing it

a factor with no real root looks like something that can be ignored rather than something that has to be accounted for

wrong$$f'=6(x-2)\bigl(2x^{2}+2x+3\bigr)\ \Rightarrow\ \text{only}\ x=2\ \text{matters}$$
right$$b^{2}-4ac=-20<0,\ a>0\ \Rightarrow\ 2x^{2}+2x+3>0\ \text{everywhere}$$

5.7The Mean Value Theorem

Turns a bound on $f'$ into a bound on $f$ itself: some interior $c$ has $f'(c)$ equal to the average slope.

Every method so far started from a formula for $f$. This theorem starts from what you know about $f'$ and ends with a conclusion about $f$, and it works even when no formula for $f$ is available at all.

TheoremTheorem 5.7: The Mean Value Theorem
Conditions

  • $f$ is continuous on the closed interval $[a,b]$

  • $f$ is differentiable on the open interval $(a,b)$; the endpoints are exempt, which is why a vertical tangent at an end does no harm

  • Two further faces of this theorem carry most of its marks. With $f(a)=f(b)$ it is Rolle's theorem, which caps how many zeros a function can have; and on an interval where $f'$ keeps one sign it converts that sign into monotonicity of $f$, and the same move applied to $f'$ and $f''$ makes $f'$ monotone. The papers name both steps MVT, and the block should teach all three faces, not only the passage from a bound on $f'$ to a bound on $f$.

$$\boxed{\ \exists\,c\in(a,b):\quad f'(c)=\frac{f(b)-f(a)}{b-a}\ }$$

Somewhere strictly between the two ends, the instantaneous rate equals the average rate over the whole trip. Drive from Ankara to Istanbul averaging 100 km per hour and at some instant your speedometer read exactly 100.

Built out of the two theorems we already have

Take first the case $f(a)=f(b)$, which is Rolle's theorem. Continuity on the closed interval hands us an absolute maximum and minimum; if either lies strictly inside, Fermat forces $f'=0$ there, and if both sit at the ends then $f$ is constant and any interior point serves. For the general statement, subtract the secant: with $m=\frac{f(b)-f(a)}{b-a}$ put $h(x)=f(x)-[f(a)+m(x-a)]$. Then $h(a)=h(b)=0$, so Rolle produces a $c$ with $h'(c)=0$, and since $h'(c)=f'(c)-m$ that is exactly $f'(c)=m$. Tilting the picture until the secant is horizontal is the whole idea.

Looks like this, but is not

The two ends of $f(x)=\lvert x\rvert$ on $[-1,1]$ have the same height, so somewhere in between the tangent must be horizontal.

There is no such point. The slope is $-1$ on the whole left half and $+1$ on the whole right half, and the only place left is the corner at $0$, where $f'$ does not exist. That is precisely the hypothesis this function fails, and losing it costs the conclusion entirely.

Finding the c the theorem promises for x³ − x on [0, 2]

Verify that $f(x)=x^{3}-x$ satisfies the hypotheses of the Mean Value Theorem on $[0,2]$ and find every $c$ the conclusion allows.

Given

  • $f(x)=x^{3}-x$

  • the interval $[0,2]$

Find

all values of $c$ in $(0,2)$ with $f'(c)$ equal to the average rate

Solution
Check the hypotheses before using the conclusion
$$f\ \text{is a polynomial: continuous on}\ [0,2],\ \text{differentiable on}\ (0,2)$$

polynomials satisfy both hypotheses everywhere, so this step is short, but skipping it on a function with a corner would be fatal

Compute the average rate first
$$\frac{f(2)-f(0)}{2-0}=\frac{6-0}{2}=3$$

$f(2)=8-2=6$ and $f(0)=0$; this number is the target slope

Solve f′(c) = 3
$$f'(x)=3x^{2}-1$$

the derivative

$$3c^{2}-1=3\ \Rightarrow\ c^{2}=\tfrac43\ \Rightarrow\ c=\pm\tfrac{2}{\sqrt3}$$

solve, then filter

$$c=\tfrac{2}{\sqrt3}\approx 1.155$$

only the positive root lies in $(0,2)$; the theorem promises a point inside the open interval, so a root outside it is not an answer

Answer $$\boxed{c=\tfrac{2}{\sqrt3}\approx 1.155}$$
Check

Independent existence argument: $f'(x)=3x^{2}-1$ is continuous and climbs from $f'(0)=-1$ to $f'(2)=11$, so it must pass through $3$ somewhere in between, and being increasing on $(0,2)$ it passes through exactly once. That predicts a single $c$, which matches what the algebra produced.

The theorem is an existence statement. Finding $c$ is a bonus the exam sometimes asks for; the uses that matter never need its value.

From a two sided bound on f′ to a window for f

A function $f$ is differentiable at every real number, $f(1)=-4$, and $2\le f'(x)\le 5$ for every $x$. Nothing else about $f$ is known. Find every value $f(5)$ could have, and every value $f(-1)$ could have.

Given

  • $f$ is differentiable at every real number

  • $f(1)=-4$

  • $2\le f'(x)\le 5$ for every real $x$

Find

the sharpest window for $f(5)$ and the sharpest window for $f(-1)$

Solution

There is no formula for $f$, so nothing can be differentiated, evaluated or graphed. The Mean Value Theorem is the only tool that carries a fact about $f'$ everywhere into a fact about $f$ at one particular point, and it does that without needing to know where its $c$ is.

Turn the unknown value into a difference
$$f(5)-f(1)=f'(c_{1})(5-1)=4f'(c_{1}),\qquad c_{1}\in(1,5)$$

differentiability everywhere supplies both hypotheses on $[1,5]$; the unknown location of $c_{1}$ costs nothing because the bound holds at every point

Multiply by a positive step, then shift by the known value
$$2\le f'(c_{1})\le 5\ \Rightarrow\ 8\le 4f'(c_{1})\le 20$$

multiplying by the positive number $4$ leaves both inequalities pointing the same way

$$-4+8\le f(5)\le -4+20\ \Rightarrow\ 4\le f(5)\le 16$$

adding the known $f(1)=-4$ to every part is legal for any real number, positive or not

Repeat to the left, where the step is negative
$$f(-1)-f(1)=f'(c_{2})(-1-1)=-2f'(c_{2}),\qquad c_{2}\in(-1,1)$$

the same theorem on $[-1,1]$; the step is $-2$ here, and that minus sign is the whole difficulty of this half

$$2\le f'(c_{2})\le 5\ \Rightarrow\ -10\le -2f'(c_{2})\le -4$$

multiplying an inequality by a negative number reverses it, so the upper bound on $f'$ becomes the lower bound on the difference

$$-14\le f(-1)\le -8$$

add $f(1)=-4$ again; note the window sits below the known value, as it must

Show neither window can be shrunk
$$f(x)=-4+2(x-1)\ \Rightarrow\ f(5)=4,\ f(-1)=-8$$

a line of slope $2$ satisfies every hypothesis and sits at one end of both windows

$$f(x)=-4+5(x-1)\ \Rightarrow\ f(5)=16,\ f(-1)=-14$$

slope $5$ reaches the other end, so both endpoints of both windows are actually attained and no narrower answer is correct

Answer $$\boxed{4\le f(5)\le 16,\qquad -14\le f(-1)\le -8}$$
Check

Independent consistency check, from monotonicity rather than from the algebra. Because $f'\ge 2>0$ everywhere, $f$ increases, so $f(-1)<f(1)=-4<f(5)$ has to hold whatever $f$ is. Our windows put $f(-1)$ in $[-14,-8]$, entirely below $-4$, and $f(5)$ in $[4,16]$, entirely above it. A window that straddled $-4$ would have exposed a reversed inequality.

Two applications of the theorem, two multiplications, two straight lines to prove sharpness.

One known value of $f$ plus a bound on $f'$ traps $f$ along the whole line. Read the step $b-a$ before you multiply: to the right of the known point it is positive and the inequalities keep their direction, to the left it is negative and both of them flip.

Checkpoint
§05.7 — which hypothesis is missing●●○○○

The Mean Value Theorem has two hypotheses and they are easy to conflate, so it is worth one thirty second test.

Given

  • $f$ is continuous on $[0,2]$

  • nothing else is known about $f$

Find

  1. (a) True or false: this alone guarantees a point $c$ in $(0,2)$ with $f'(c)=\dfrac{f(2)-f(0)}{2}$.

Hint 1/4

Count the hypotheses of the theorem, then count how many the statement actually gives you.

Hint 2/4

The theorem needs continuity on the closed interval and differentiability on the open one; the conclusion mentions $f'(c)$, so a derivative has to exist somewhere.

Hint 3/4

Here only continuity is given. The absolute value function on $[-1,1]$, shifted to $[0,2]$, is continuous and has no derivative at its corner.

Hint 4/4

So the statement is false: continuity alone is not enough.

Show solution
Produce a counterexample rather than argue in general
$$f(x)=\lvert x-1\rvert\ \text{on}\ [0,2]$$

continuous everywhere, with a corner placed inside the interval

$$\frac{f(2)-f(0)}{2}=\frac{1-1}{2}=0$$

the target slope is zero, so a horizontal tangent is what would be needed

$$f'(x)=-1\ \text{on}\ (0,1),\qquad f'(x)=1\ \text{on}\ (1,2)$$

no point has slope $0$, and at $x=1$ the derivative does not exist at all

Answer $$\boxed{\text{False}}$$
Check

Check that the counterexample is honest: it satisfies the hypothesis given (continuity on the closed interval) and fails only the hypothesis that was withheld, which is what a counterexample has to do.

When a theorem has two hypotheses, the fastest way to test a claim is to break exactly one of them and see whether the conclusion survives.

⚠ Accepting a c outside the open interval

the equation for $c$ usually has two roots and only one of them is admissible

wrong$$c=\pm\tfrac{2}{\sqrt3}\ \text{both accepted on}\ (0,2)$$
right$$c=\tfrac{2}{\sqrt3}\in(0,2)\ \text{only}$$
⚠ Using the theorem where the function has a corner

continuity is checked, differentiability is assumed to come with it

wrong$$f(x)=\lvert x\rvert\ \text{on}\ [-1,1]:\ \exists c,\ f'(c)=0$$
right$$f'(0)\ \text{does not exist, so the theorem does not apply}$$
Estimating a value by hand

the question asks for a decimal of something like $\sqrt{26}$, $(1.98)^{4}$ or $\sin(0.1)$, and there is a friendly point nearby

  1. Name the function and the target

    Write $f$ and the input $x$ you actually want. For $\sqrt{26}$ that is $f(x)=\sqrt{x}$ and $x=26$.

  2. Choose the base point

    Take the nearest input $a$ where $f(a)$ and $f'(a)$ are both exact. Nearest matters: the error grows like the square of $x-a$.

  3. Build the line

    $L(x)=f(a)+f'(a)(x-a)$, with $f'$ evaluated at $a$, not left as a function.

  4. Evaluate and judge

    Compute $L(x)$, then check the direction of the error by squaring, cubing or comparing with a known bound.

Where it goes wrong

  • Choosing a base point where $f(a)$ is not exact, which imports a second error.

  • Using the target itself as the base point, which produces the number you were trying to avoid computing.

  • Reporting the estimate without any statement about its size of error, when the question asked how good it is.

Absolute extrema on a closed interval

the words absolute, largest, smallest, maximum or minimum appear together with a closed interval $[a,b]$

  1. Check continuity on the interval

    One line, and it is the licence for everything after it. A polynomial or a sum of continuous pieces takes a sentence; a quotient or a piecewise definition needs a real check.

  2. Collect the critical numbers inside

    Solve $f'(x)=0$ and also ask where $f'$ fails to exist. Discard anything outside $(a,b)$. When that equation comes out a cubic, the next box does the factoring; when it has no closed form solution at all, the expected answer is an existence argument by sign change, not a number.

  3. Evaluate f at every candidate

    Critical numbers first, then $f(a)$ and $f(b)$. Values of $f$, never values of $f'$.

  4. Compare and report both parts

    Largest value is the absolute maximum, smallest is the absolute minimum. Give the value and the location, and list every location if a value repeats.

Where it goes wrong

  • Forgetting the endpoints, which are candidates even though no calculus points at them.

  • Keeping a critical number that lies outside the interval.

  • Comparing derivative values instead of function values at the last step.

  • Answering with the location when the value was asked for.

Getting a cubic apart when nothing factors by inspection

solving $f'(x)=0$ leaves a cubic (or higher) polynomial, or the sign of such a polynomial has to be read, and neither a common factor nor grouping opens it up

  1. Pull out every common factor first

    A constant or a power of $x$ in front cannot make the product zero on its own, except that a factor of $x$ does hand you the root $x=0$ for free. Removing the rest shrinks the end coefficients, and the candidate list in the next step is built from exactly those two numbers.

  2. Write the candidate list, fractions included

    For integer coefficients, every rational root $p/q$ in lowest terms has $p$ dividing the constant term and $q$ dividing the leading coefficient. Write the list out. It is finite and it is complete: no rational number outside it can be a root.

  3. Test candidates in order of cheapness, and stop at the first hit

    Small integers first, then the halves and thirds. One root is all you need; there is no prize for finding a second one by substitution when division will hand it to you.

  4. Divide out the factor with synthetic division

    Lay the coefficients in a row, inserting a $0$ for any missing power, and run bring down, multiply, add. The last entry is the remainder: a nonzero value there means either the wrong root or a slip in the arithmetic, so it is a free check on the previous step.

  5. Settle the leftover quadratic out loud

    Either factor it, or compute $b^{2}-4ac$. A negative discriminant means no further real root, and because the quadratic then keeps the sign of its leading coefficient everywhere, it also contributes no sign change. Write that sentence down: it is the difference between a candidate list you have proved complete and one you have merely stopped adding to. The same factored form is what a sign line for $f'$ is read from later, when the shape of the graph is the question. The same closing sentence is what settles a count of points on an implicitly defined curve, and the implicit differentiation section has already run these five steps twice: once on $2y^{3}-y^{2}+9=0$, where the negative discriminant of the leftover quadratic is what turns "a point" into "the only point", and once where tangency handed over a repeated root and two synthetic divisions finished the cubic. Read those two if the steps here look like a new technique rather than a name for one you have already used.

Where it goes wrong

  • Testing only $\pm1$ and concluding that the cubic has no rational root; the theorem allows fractions, and their denominators come from the leading coefficient.

  • Reading the candidates off $f(x)=0$ instead of $f'(x)=0$. The roots of a function and the critical numbers of that function have nothing to do with each other, and this one scores zero rather than partial credit.

  • Forgetting to write a $0$ for a missing power in the synthetic division row, which shifts every entry after it.

  • Stopping at the first root and reporting one critical number, when the leftover quadratic still had two real roots waiting.

  • Asserting that the leftover quadratic is always positive because it looks that way. State the discriminant or complete the square; the assertion on its own carries no marks.

  • Trying to solve an equation that has no closed form solution at all. If no candidate works and the expression is not a polynomial, the question is asking how many roots there are, not what they are.

Counting the exact number of real roots

the question asks for the exact number of real roots or how many solutions an equation has, and the polynomial does not factor, so the roots themselves are out of reach

  1. Set up and record the licences

    Move everything to one side and name it $f$. Write one line saying $f$ is continuous and differentiable, which for a polynomial is everywhere. Every theorem below is going to ask for exactly this, and each use costs nothing once it is on the page.

  2. Floor: sign changes of f

    Evaluate $f$ at a handful of convenient integers until the signs change. Each sign change hands you at least one root by the Intermediate Value Theorem, and roots found in disjoint intervals are distinct. This half can only ever produce the words at least.

  3. Control the zeros of the derivative

    Differentiate and ask how many zeros $f'$ has. The standard route is $f''$: if $f''$ keeps one sign, then $f'$ is strictly monotone and so has at most one zero. Prove the sign of $f''$, by completing the square or by a negative discriminant, rather than asserting it.

  4. Ceiling: convert zeros of f' into a cap on zeros of f

    If $f$ had one more root than $f'$ has zeros plus one, Rolle applied between consecutive roots would produce too many zeros of $f'$. So the number of distinct roots of $f$ is at most the number of zeros of $f'$ plus one. State the contradiction explicitly; this is where about half the marks are.

  5. Close the gap and answer

    If the floor and the ceiling agree, say the exact number and where each root lives. If they do not, sharpen the weaker half: hunt one more sign change of $f$, or use the Intermediate Value Theorem on $f'$ to show that its single possible zero really occurs.

Where it goes wrong

  • Stopping after the sign changes and answering with a number, when that half only proves at least.

  • Asserting that $f''>0$ because it looks positive, instead of completing the square or quoting the discriminant.

  • Capping $f$ at the number of zeros of $f'$ and forgetting the plus one.

  • Trying to solve $f(x)=0$; the polynomial is chosen so that this cannot be done, and the count is the whole answer.

  • Naming the theorem loosely: the course accepts MVT here, but the hypothesis you must check is that the two endpoint values are equal, which is Rolle.

Proving an inequality by monotonicity, then squeezing a number out of it

the question says show that $f(x)>c$ for $x>a$, and usually adds a second part that deduces a numerical fact such as one constant being smaller than another

  1. Put the claim on one function

    Move everything to one side so the statement reads $f(x)\gt f(a)$ for $x>a$, or $f(x)\gt 0$. One function, one interval, one direction. Nothing later works if two functions are still being compared.

  2. Differentiate into a form whose sign is readable

    Factor $f'$, or write it as a square divided by something visibly positive. The aim is not a simplified expression but an expression whose sign can be asserted in one sentence.

  3. Fix the sign of f' on the interval

    Say why $f'\gt 0$ there. If $f'\ge 0$ with equality only at isolated points, $f$ is still strictly increasing; write that sentence, because it is marked separately and it is the usual place a correct solution loses a point.

  4. Anchor at the endpoint

    Compute $f(a)$ exactly, then write the link as a full sentence: $f$ is increasing on $[a,\infty)$ and $f(a)$ equals this number, therefore $f(x)$ exceeds it for $x>a$. The grading key pays for this sentence even when the justification above it is thin.

  5. Feed in the one number the second part wants

    Substitute the specific input that makes the inequality say something arithmetic, then rearrange to the numerical claim. Choose that input only after the inequality is proved, otherwise the argument is circular.

Where it goes wrong

  • Proving $f'>0$ and never computing $f(a)$, so the inequality has nothing to start from.

  • Writing $f'>0$ when $f'$ vanishes at a point, without explaining why strict increase survives.

  • Leaving the link from increasing to the inequality implicit; it carries its own mark.

  • Substituting the target number before the inequality holds, which assumes what was to be shown.

  • Reporting the chain as a chain of symbols with no sentence; one mark on these questions is for clarity of argument.

Absolute extrema when the interval is not closed

an absolute maximum or minimum is asked for on $[a,b)$, $(a,b)$, or an unbounded interval, so the Closed Interval Method cannot be quoted as written

  1. Say what the Extreme Value Theorem does not give

    One line: the interval is not closed (or not bounded), so nothing is guaranteed to be attained and does not exist is a legal answer that has to be argued. The theorem is a guarantee, not a test, so its failure forbids nothing either.

  2. Collect the interior candidates

    Find every critical number and keep only those lying inside the interval. An endpoint the interval excludes is not a candidate, however tempting its value looks.

  3. Build the sign chart of f'

    With no finite list to compare, the sign of $f'$ across the interval does the comparing: it tells you which candidate is beaten by its neighbours and which one wins outright.

  4. Take a one sided limit at every missing end

    Compute $\lim_{x\to b^{-}}f(x)$, or the limit at infinity. A value that is approached but never taken is not an extremum; it is only an infimum or a supremum.

  5. Report each of the two extremes separately

    For an attained extreme, give the value and the location. For one that is only approached, say the extremum does not exist and name the reason, which is that the interval leaves out the point where it would have occurred.

Where it goes wrong

  • Quoting the Closed Interval Method and evaluating $f$ at an endpoint the interval does not contain.

  • Reporting the limit at the missing end as the maximum or the minimum value.

  • Concluding that no extremum exists merely because the interval is not closed; typically one of the two survives and the other does not.

  • Assuming an unbounded interval cannot carry an attained extremum.

  • Answering with a single verdict when the question asks about the maximum and the minimum, which behave differently here.

Predicted change: dy for y = x² at x = 2 with dx = 1

Compute the differential $dy$ for $y=x^{2}$ at $x=2$ when $dx=1$.

Given

  • $y=x^{2}$

  • $x=2$

  • $dx=1$

Find

$dy$

Solution
Use the tangent, which is what dy means
$$dy=f'(x)\,dx=2x\,dx$$

the derivative first, in general form

$$dy=2(2)(1)=4$$

substitute the base point and the step; nothing about $f(3)$ is used

Answer $$\boxed{dy=4}$$
Check

Read it off the tangent line: it has slope $4$ at $(2,4)$, so over a run of $1$ it rises $4$. The number is a property of the line, not of the curve.

True change: Δy for y = x² from x = 2 to x = 3

Compute the actual change $\Delta y$ for $y=x^{2}$ when $x$ moves from $2$ to $3$.

Given

  • $y=x^{2}$

  • $x=2$

  • $dx=1$, so the new input is $3$

Find

$\Delta y$

Solution
Use the function twice, which is what Δy means
$$\Delta y=f(x+dx)-f(x)=f(3)-f(2)$$

no derivative appears anywhere in this computation

$$\Delta y=9-4=5$$

the curve genuinely climbs five units over that step

Answer $$\boxed{\Delta y=5}$$
Check

Check against the figure: the curve ends one unit above the tangent at $x=3$, and $5-4=1$.

Same function, same base point, same step, and two different numbers: $4$ is what the tangent promises, $5$ is what the function delivers, and the difference $1$ is the curvature showing up.

How to tell them apart

If the computation uses $f'$, you are computing $dy$. If it evaluates $f$ at two inputs, you are computing $\Delta y$. A question that gives you a tolerance wants $dy$; a question that gives you two positions wants $\Delta y$.

Scaffolding comes off
The common skeleton

  1. Check that $f$ is continuous on the closed interval, since that is what guarantees there is anything to find.

  2. Differentiate, then collect every critical number: solutions of $f'(x)=0$ and points where $f'$ fails to exist. Keep only those strictly inside the interval.

  3. Evaluate $f$, never $f'$, at each surviving critical number and at both endpoints.

  4. Compare the list of values: largest is the absolute maximum, smallest the absolute minimum. Report the value and the location.

1 · fully worked

Fully worked: 2x³ − 3x² − 12x + 1 on [−2, 3]

Find the absolute extrema of $f(x)=2x^{3}-3x^{2}-12x+1$ on $[-2,3]$.

Given

  • $f(x)=2x^{3}-3x^{2}-12x+1$

  • the closed interval $[-2,3]$

Find

both absolute extreme values and their locations

Solution
Step 1 · continuity
$$f\ \text{is a polynomial, so it is continuous on}\ [-2,3]$$

the hypothesis of the guarantee; without it the rest is guesswork

Step 2 · candidate list
$$f'(x)=6x^{2}-6x-12=6(x-2)(x+1)$$

pull out the $6$ first, then factor; the quadratic formula would work but costs more arithmetic

$$x=2\ \text{and}\ x=-1,\ \text{both inside}\ (-2,3)$$

both survive the filter; a root outside the interval would be discarded here

Step 3 · evaluate f at all four candidates
$$f(-2)=-16-12+24+1=-3$$

left endpoint

$$f(-1)=-2-3+12+1=8$$

critical number

$$f(2)=16-12-24+1=-19$$

critical number

$$f(3)=54-27-36+1=-8$$

right endpoint

Step 4 · compare
$$\max\{-3,8,-19,-8\}=8,\qquad \min\{-3,8,-19,-8\}=-19$$

the arithmetic step where careless sign work undoes correct calculus

Answer $$\boxed{\max=8\ \text{at}\ x=-1,\qquad \min=-19\ \text{at}\ x=2}$$
Check

Sign check on $f'=6(x-2)(x+1)$: positive on $(-2,-1)$, negative on $(-1,2)$, positive on $(2,3)$. Up, down, up — so the peak must be at $-1$ and the trough at $2$, and the right endpoint must exceed the trough, which it does ($-8>-19$).

One factorisation, four evaluations, one comparison.

Every problem of this type is these four steps. What changes is only how hard step 2 is.

2 · you write the reasoning

Same skeleton, easier function: find the absolute extrema of $f(x)=x^{2}-4x+1$ on $[0,3]$. The steps are written out; your job is to supply the reason for each one before opening it.

  1. $f$ is continuous on $[0,3]$.

    reasoning

    A polynomial is continuous everywhere, so the Extreme Value Theorem applies and the two extremes are known to exist before any searching starts.

  2. $f'(x)=2x-4$.

    reasoning

    The derivative is needed only to produce candidates, so it is computed once and then used twice: for the equation and for the question of where it fails to exist.

  3. $2x-4=0$ gives $x=2$, which lies in $(0,3)$.

    reasoning

    The root has to be filtered against the interval. It is inside, so it stays; a root outside would be discarded here, not later.

  4. $f(0)=1$, $f(2)=-3$, $f(3)=-2$.

    reasoning

    Values of $f$, not of $f'$. Three candidates means three evaluations: one critical number and the two endpoints, which are candidates without any calculus.

  5. Maximum $1$ at $x=0$; minimum $-3$ at $x=2$.

    reasoning

    The comparison is arithmetic. The maximum lands at an endpoint, which is normal, and the endpoint is not a local maximum, which does not matter at all.

3 · find the buried error

Harder function, and a solution written by someone else. Two of the four steps below contain an error. Find both before reading on: the task is find the absolute extrema of $f(x)=x+2\cos x$ on $[0,\pi]$.

  1. Step 1. $f$ is continuous on $[0,\pi]$, being a sum of a polynomial and a cosine, so the search is licensed.

  2. Step 2. $f'(x)=1-2\sin x$. Setting it to zero gives $\sin x=\tfrac12$, so the only critical number in the interval is $x=\tfrac{\pi}{6}$.

  3. Step 3. Candidates are $0$, $\tfrac{\pi}{6}$ and $\pi$: $f(0)=0+2(1)=2$, $f(\tfrac{\pi}{6})=\tfrac{\pi}{6}+\sqrt3\approx 2.26$, and $f(\pi)=\pi+2\approx 5.14$.

  4. Step 4. Comparing $2$, $2.26$ and $5.14$: the absolute maximum is about $5.14$ at $x=\pi$ and the absolute minimum is $2$ at $x=0$.

the two buried errors (2)
⚠ step 2

$\sin x=\tfrac12$ has two solutions in $[0,\pi]$, namely $\tfrac{\pi}{6}$ and $\tfrac{5\pi}{6}$, and the second one was dropped.

The first quadrant solution is the one the inverse sine returns, and the second quadrant twin has to be remembered rather than computed.

right

Keep both: $x=\tfrac{\pi}{6}$ and $x=\tfrac{5\pi}{6}$, both inside $(0,\pi)$. The second one turns out to carry the minimum, with $f(\tfrac{5\pi}{6})=\tfrac{5\pi}{6}-\sqrt3\approx 0.886$.

⚠ step 3

$\cos\pi=-1$, so $f(\pi)=\pi-2\approx 1.14$, not $\pi+2\approx 5.14$.

The sign of the cosine at $\pi$ is dropped once the arithmetic speeds up, and the resulting number is large enough to look like a maximum, which stops anyone from questioning it.

right

Evaluate carefully: $f(\pi)=\pi+2\cos\pi=\pi-2\approx 1.14$. With the corrected candidates the true answer is a maximum of $\tfrac{\pi}{6}+\sqrt3\approx 2.256$ at $x=\tfrac{\pi}{6}$ and a minimum of $\tfrac{5\pi}{6}-\sqrt3\approx 0.886$ at $x=\tfrac{5\pi}{6}$.

4 · the bare problem
§05.6 — the bare problem●●●○○

No scaffolding this time. The skeleton is the same as in the three problems above.

Given

  • $f(x)=x^{4}-8x^{2}+3$

  • the closed interval $[-1,3]$

Find

  1. (a) Find the absolute maximum and minimum values of $f$ on $[-1,3]$, and state where each one occurs.

Hint 1/4

Four steps, in this order: continuity, candidates, values, comparison. Write the four headings before computing anything.

Hint 2/4

Candidates are the critical numbers inside the open interval together with both endpoints; here $f'(x)=4x^{3}-16x$, which factors.

Hint 3/4

With $f(x)=x^{4}-8x^{2}+3$ on $[-1,3]$: $f'(x)=4x(x-2)(x+2)$, so the roots are $0$, $2$ and $-2$, of which $-2$ lies outside the interval. Candidates: $-1$, $0$, $2$, $3$.

Hint 4/4

The values are $f(-1)=-4$, $f(0)=3$, $f(2)=-13$, $f(3)=12$, so the maximum is $12$ and the minimum is $-13$.

Show solution
Continuity and candidates
$$f\ \text{is a polynomial: continuous on}\ [-1,3]$$

the licence

$$f'(x)=4x^{3}-16x=4x(x-2)(x+2)$$

factor completely so that the filtering step is visible

$$x=0,\ x=2\ \text{are inside};\ x=-2\ \text{is not}$$

the discarded root is the point of this problem: solving is not the same as selecting

Evaluate and compare
$$f(-1)=1-8+3=-4$$

left endpoint

$$f(0)=3$$

critical number

$$f(2)=16-32+3=-13$$

critical number

$$f(3)=81-72+3=12$$

right endpoint

$$\max=12\ \text{at}\ x=3,\qquad \min=-13\ \text{at}\ x=2$$

compare the four values, not the four derivatives

Answer $$\boxed{\max=12\ \text{at}\ x=3,\qquad \min=-13\ \text{at}\ x=2}$$
Check

Shape check from the sign of $f'=4x(x-2)(x+2)$ inside the interval: $f'(-0.5)=7.5>0$, $f'(1)=-12<0$, $f'(2.5)=22.5>0$. So $f$ rises to $x=0$, falls to $x=2$, then rises to $x=3$. That shape puts a local maximum at $0$ and the trough at $2$, and it forces $f(3)>f(2)$, which the four values confirm.

The discarded root $x=-2$ is worth a moment: it solves $f'(x)=0$ honestly, and it is still not a candidate, because candidacy is about the interval, not about the equation.

Full exam-style question

Exam style: one cubic, three questionsexam format

Let $f(x)=x^{3}-9x$. (a) Find the absolute maximum and minimum values of $f$ on $[-1,4]$. (b) Write the linearization of $f$ at $a=2$, use it to estimate $f(2.1)$, and compare with the exact value. (c) Verify that $f$ satisfies the hypotheses of the Mean Value Theorem on $[0,3]$ and find every $c$ the conclusion provides.

Given

  • $f(x)=x^{3}-9x$

  • part (a) uses the interval $[-1,4]$

  • part (b) uses the base point $a=2$

  • part (c) uses the interval $[0,3]$

Find

the two extreme values with locations, the linear estimate with its error, and the value of $c$

Solution
(a) Continuity, candidates, values
$$f'(x)=3x^{2}-9=3(x^{2}-3)\ \Rightarrow\ x=\pm\sqrt3$$

only $x=\sqrt3\approx 1.732$ lies inside $(-1,4)$; $-\sqrt3\approx-1.732$ is outside and is discarded

$$f(-1)=-1+9=8,\qquad f(\sqrt3)=3\sqrt3-9\sqrt3=-6\sqrt3\approx-10.39,\qquad f(4)=64-36=28$$

three candidates, three values of $f$

$$\max=28\ \text{at}\ x=4,\qquad \min=-6\sqrt3\ \text{at}\ x=\sqrt3$$

one extreme comes from the endpoint list and the other from the critical number list, which is the configuration worth recognising

(b) The linearization at 2, and how far it misses
$$f(2)=8-18=-10,\qquad f'(2)=12-9=3$$

the two numbers the line is made of

$$L(x)=-10+3(x-2)\ \Rightarrow\ L(2.1)=-10+0.3=-9.7$$

the step is $0.1$, so the estimate costs one multiplication

$$f(2.1)=9.261-18.9=-9.639$$

the exact value, for comparison

$$f(2.1)-L(2.1)=0.061$$

the estimate is low by $0.061$, about $0.6\%$ of the value

(c) Mean Value Theorem on [0, 3]
$$f\ \text{is a polynomial: continuous on}\ [0,3],\ \text{differentiable on}\ (0,3)$$

both hypotheses hold, and saying so is worth marks

$$\frac{f(3)-f(0)}{3-0}=\frac{0-0}{3}=0$$

$f(3)=27-27=0$ and $f(0)=0$, so the endpoints have equal height and this is Rolle's case

$$3c^{2}-9=0\ \Rightarrow\ c=\pm\sqrt3\ \Rightarrow\ c=\sqrt3\approx 1.732$$

again only the positive root lies inside the interval

Answer $$\boxed{\text{(a) }\max 28\ \text{at }4,\ \min -6\sqrt3\ \text{at }\sqrt3;\quad \text{(b) }f(2.1)\approx-9.7,\ \text{exact }-9.639;\quad \text{(c) }c=\sqrt3}$$
Check

Two independent checks. The parts agree with each other: the $c=\sqrt3$ that part (c) produces is the same point that carries the minimum in part (a), which is what should happen for a function whose only interior critical number in that range is $\sqrt3$. And the error in part (b) is predicted by the curvature: half of $f''(2)=12$ times $(0.1)^{2}$ is $0.06$, against the measured $0.061$.

One derivative reused three times. Notice how little new work each part needed once $f'$ was on the page.

A question that bundles the three ideas onto one function costs far less time once you notice that a single derivative answers all three parts. Compute $f'$ first, then read each part off it.

Practice

A · concept 3 questions
1§05.5 — what a zero derivative does not tell you●●○○○

One sentence, and the whole content of Fermat's theorem is deciding whether it runs in both directions.

Given

  • $f$ is differentiable everywhere

  • $c$ is a number with $f'(c)=0$

Find

  1. (a) True or false: $f$ must have a local maximum or a local minimum at $c$.

Hint 1/4

You are being asked whether the converse of a theorem is also true. Try to remember a graph with a flat spot that is not a peak or a valley.

Hint 2/4

Fermat's theorem says local extremum plus differentiability implies $f'(c)=0$. Nothing in it licenses the reverse reading.

Hint 3/4

Take $f(x)=x^{3}$ and $c=0$: $f'(x)=3x^{2}$, so $f'(0)=0$, while $f(x)<0$ for $x<0$ and $f(x)>0$ for $x>0$.

Hint 4/4

So the statement is false; $x^{3}$ at $0$ is the standard counterexample.

Show solution
Attack it with a counterexample, not with a general argument
$$f(x)=x^{3},\qquad f'(x)=3x^{2},\qquad f'(0)=0$$

the hypothesis of the claim is satisfied

$$f(-0.1)=-0.001<0=f(0)<0.001=f(0.1)$$

the value at $0$ is beaten on the right and beats the left, so it is neither a local maximum nor a local minimum

Answer $$\boxed{\text{False}}$$
Check

Cross check with the sign of $f'$: $3x^{2}>0$ on both sides of $0$, so $f$ is increasing on both sides and cannot turn around there. A turn requires a sign change in $f'$, and a squared factor cannot provide one.

Zero derivative is a permit to be an extremum, not a proof of being one. Classifying critical numbers needs a separate test.

2§05.3 — absolute minimum from a short description●●●○○

A function is described only by a handful of values and its local behaviour. That is enough, and deciding why it is enough is the exercise.

Given

  • $f$ is continuous on $[1,6]$

  • its only local extrema are a local maximum of $8$ at $x=2$ and a local minimum of $1$ at $x=5$

  • $f(1)=3$ and $f(6)=3$

Find

  1. (a) What is the absolute minimum value of $f$ on $[1,6]$, and where is it attained?

Hint 1/4

An absolute minimum on a closed interval is either at an endpoint or strictly inside. Ask what an interior absolute minimum automatically is.

Hint 2/4

An interior absolute minimum is in particular a local minimum, so it has to be one of the local extrema you were told about.

Hint 3/4

The candidates are therefore $f(1)=3$, the local minimum $1$ at $x=5$, and $f(6)=3$; the local maximum $8$ at $x=2$ cannot be the smallest.

Hint 4/4

The smallest of $3$, $1$ and $3$ is $1$, so the absolute minimum value is $1$ at $x=5$.

Show solution
Argue that the list of candidates is complete
$$\text{interior absolute min}\ \Rightarrow\ \text{local min}$$

if a point strictly inside beats every point of the interval, it certainly beats its neighbours, so it appears in the list of local minima we were given

$$\text{candidates}:\ f(1)=3,\ f(5)=1,\ f(6)=3$$

the two endpoints plus the only interior local minimum

Compare
$$\min\{3,1,3\}=1\ \text{at}\ x=5$$

the local maximum at $x=2$ is irrelevant to this question and is not compared

Answer $$\boxed{\text{value }1,\ \text{at }x=5}$$
Check

Sanity check on the shape: from $3$ up to $8$, down to $1$, back up to $3$. That path is continuous, respects every value given, and never goes below $1$, so no smaller value is possible.

Endpoints and interior local minima are the only competitors for an absolute minimum. Every other feature of the graph is decoration for this question.

3§05.4 — which one is guaranteed a maximum●●●○○

Four functions on four sets. Exactly one of them is covered by the guarantee, and the others each fail it in a different way.

Given

  • $\tan x$ on $[0,\pi/4]$

  • $\tan x$ on $[0,\pi/2)$

  • $1/x$ on $(0,2]$

  • $x^{2}$ on $[0,\infty)$

Find

  1. (a) For which one does the Extreme Value Theorem guarantee an absolute maximum?

Hint 1/4

Check the two hypotheses on each line separately: is the function continuous at every point of the set, and is the set a closed bounded interval.

Hint 2/4

The guarantee needs continuity on a closed and bounded interval, and it needs both at once.

Hint 3/4

On $[0,\pi/4]$ the tangent function is continuous, since the trouble at $\pi/2$ is far outside, and the interval is closed with finite ends.

Hint 4/4

So the first line is the one covered by the theorem.

Show solution
Test each pair against both hypotheses
$$[0,\pi/4]:\ \text{continuous}\ \checkmark,\ \text{closed and bounded}\ \checkmark$$

the maximum is in fact $\tan(\pi/4)=1$, attained at the right endpoint

$$[0,\pi/2):\ \text{not closed, and}\ \tan x\to\infty$$

values grow past every ceiling as $x$ approaches $\pi/2$

$$(0,2]:\ \text{not closed at}\ 0,\ \text{and}\ 1/x\to\infty$$

the same escape route as the example in the notes

$$[0,\infty):\ \text{closed but not bounded}$$

$x^{2}$ has no ceiling, so no largest value exists

Answer $$\boxed{\tan x\ \text{on}\ [0,\pi/4]}$$
Check

Verify the winner directly rather than trusting the theorem: $\tan$ is increasing on $[0,\pi/4]$, so its largest value is at the right end and equals $1$. The theorem promised existence and the direct argument produced the value.

Three different ways to fail one theorem: an open end, an infinite end, and a point of discontinuity. Learn the three shapes and you can check any hypothesis in seconds.

B · computation 5 questions
1§05.1 — a reciprocal, estimated by hand●●○○○

A linearization is worth building whenever the exact computation is a division you would rather not do on paper.

Given

  • $f(x)=\dfrac1x$

  • base point $a=4$

  • target $x=4.1$

Find

  1. (a) Write the linearization $L$ of $f$ at $a=4$.

  2. (b) Use it to estimate $1/4.1$, and say whether the estimate is above or below the true value.

Hint 1/4

The line needs two numbers from the base point, and then the step $x-a$ does the rest.

Hint 2/4

$L(x)=f(a)+f'(a)(x-a)$, and for $f(x)=x^{-1}$ the derivative is $f'(x)=-x^{-2}$.

Hint 3/4

With $a=4$: $f(4)=0.25$ and $f'(4)=-\dfrac{1}{16}=-0.0625$; the target $4.1$ gives a step of $x-a=0.1$.

Hint 4/4

So $L(4.1)=0.25-0.00625=0.24375$, and the true value is $0.243902\ldots$

Show solution
Two numbers from the base point
$$f(4)=\tfrac14,\qquad f'(x)=-\frac{1}{x^{2}}\ \Rightarrow\ f'(4)=-\tfrac{1}{16}$$

the negative slope is the whole character of this function and dropping the sign would push the estimate the wrong way

Build and evaluate
$$L(x)=\tfrac14-\tfrac{1}{16}(x-4)$$

point plus slope times step

$$L(4.1)=0.25-0.00625=0.24375$$

the step is $0.1$

Answer $$\boxed{1/4.1\approx 0.24375}$$
Check

Multiply back: $0.24375\times 4.1=0.999375$, just under $1$. So the estimate is slightly below $1/4.1$, and only by about six parts in ten thousand.

The reciprocal curve bends upward, so its tangent lies below it and the estimate comes out low. Knowing the direction of the error is often worth as much as the estimate.

2§05.2 — a tolerance pushed through a volume●●○○○

A workshop measures a ball bearing and needs to quote the uncertainty in its volume, not in its radius.

Given

  • $V=\dfrac43\pi r^{3}$

  • measured radius $r=6$ cm

  • tolerance $dr=0.1$ cm

Find

  1. (a) Estimate the resulting uncertainty $dV$ in cm³.

  2. (b) Express it as a percentage of the volume.

Hint 1/4

The tolerance is a change in the input, so you want the change in the output that the tangent predicts.

Hint 2/4

$dV=\dfrac{dV}{dr}\,dr$, and differentiating $\dfrac43\pi r^{3}$ with respect to $r$ gives $4\pi r^{2}$.

Hint 3/4

With $r=6$ and $dr=0.1$: $dV=4\pi(6)^{2}(0.1)=14.4\pi$, and $V=\tfrac43\pi(216)=288\pi$.

Hint 4/4

So $dV=14.4\pi\approx 45.2$ cm³, which is $14.4/288=5\%$ of the volume.

Show solution
Differentiate first, substitute after
$$dV=4\pi r^{2}\,dr$$

the derivative of $\tfrac43\pi r^{3}$; substituting $r=6$ before differentiating would turn $V$ into a constant and $dV$ into zero

$$dV=4\pi(36)(0.1)=14.4\pi\approx 45.2\ \text{cm}^{3}$$

now the numbers

Relative form
$$\frac{dV}{V}=\frac{4\pi r^{2}\,dr}{\tfrac43\pi r^{3}}=3\,\frac{dr}{r}=3\left(\frac{0.1}{6}\right)=5\%$$

the same factor of three that a cube showed, because both are volumes and the exponent is what matters

Answer $$\boxed{dV\approx 45.2\ \text{cm}^{3}\approx 5\%\ \text{of}\ V}$$
Check

Compare with the exact change: $\tfrac43\pi(6.1^{3}-6^{3})=\tfrac43\pi(10.981)\approx 46.0$ cm³. The differential is under it by about $2\%$ of itself, which is the expected size for a step of one sixtieth of the radius.

Absolute uncertainty carries units and depends on the size of the object; relative uncertainty is a pure number and travels between problems.

3§05.6 — a quartic with a tie at the top●●●○○

A closed interval search whose answer is symmetric. The arithmetic is short; the reporting is the step that decides the answer.

Given

  • $f(x)=x^{4}-2x^{2}+3$

  • the closed interval $[-2,2]$

Find

  1. (a) Find the absolute maximum and minimum values of $f$ on $[-2,2]$.

  2. (b) State every location at which each is attained.

Hint 1/4

Same four steps as always. Expect the symmetry of the function to show up in the answer.

Hint 2/4

Candidates are the critical numbers inside $(-2,2)$ together with the endpoints; differentiate and factor.

Hint 3/4

With $f(x)=x^{4}-2x^{2}+3$: $f'(x)=4x^{3}-4x=4x(x-1)(x+1)$, so the critical numbers are $-1$, $0$, $1$, all inside, and the endpoints are $\pm2$.

Hint 4/4

The values are $f(\pm2)=11$, $f(\pm1)=2$, $f(0)=3$, so the maximum is $11$ and the minimum is $2$.

Show solution
Candidates
$$f'(x)=4x^{3}-4x=4x(x-1)(x+1)$$

factor completely; three roots appear at once

$$x=-1,\ 0,\ 1\ \text{inside};\ \text{endpoints}\ \pm2$$

five candidates in total

Values
$$f(\pm 2)=16-8+3=11$$

even function, so the two endpoints tie

$$f(\pm 1)=1-2+3=2$$

the two troughs tie as well

$$f(0)=3$$

a local maximum, but not the absolute one

Report both parts
$$\max=11\ \text{at}\ x=\pm2,\qquad \min=2\ \text{at}\ x=\pm1$$

one value each, two locations each

Answer $$\boxed{\max=11\ \text{at}\ x=\pm2,\qquad \min=2\ \text{at}\ x=\pm1}$$
Check

Independent route: write $f(x)=(x^{2}-1)^{2}+2$. A square is at least $0$, so $f(x)\ge 2$ with equality exactly when $x^{2}=1$, and on $[-2,2]$ the square $(x^{2}-1)^{2}$ is largest when $x^{2}=4$, giving $9+2=11$. No calculus was needed for either half.

When the function can be written as a square plus a constant, that form answers the question faster than any search. Look for it before differentiating.

4§05.5 — critical numbers of a quotient●●●○○

A quotient whose denominator never vanishes, so the only thing that can happen is the numerator of $f'$ hitting zero.

Given

  • $f(x)=\dfrac{x}{x^{2}+1}$

  • the domain is all real numbers

Find

  1. (a) Find every critical number of $f$.

  2. (b) Explain why the list is complete.

Hint 1/4

Two searches as always: where $f'$ is zero, and where $f'$ fails to exist. Look at the denominator before you start.

Hint 2/4

Quotient rule: $\left(\dfrac{u}{v}\right)'=\dfrac{u'v-uv'}{v^{2}}$, and a fraction is zero exactly when its numerator is.

Hint 3/4

With $u=x$ and $v=x^{2}+1$: $f'(x)=\dfrac{(1)(x^{2}+1)-x(2x)}{(x^{2}+1)^{2}}=\dfrac{1-x^{2}}{(x^{2}+1)^{2}}$.

Hint 4/4

The numerator vanishes at $x=\pm1$, and the denominator never does, so those two are the complete list.

Show solution
Differentiate
$$f'(x)=\frac{(1)(x^{2}+1)-x(2x)}{(x^{2}+1)^{2}}=\frac{1-x^{2}}{(x^{2}+1)^{2}}$$

quotient rule, then simplify the numerator before solving anything

Both searches, and why the second one is empty
$$1-x^{2}=0\ \Rightarrow\ x=\pm 1$$

a fraction vanishes exactly when its numerator does

$$(x^{2}+1)^{2}\ge 1>0\ \text{for every real}\ x$$

the denominator never vanishes, so $f'$ exists everywhere and the second species of critical number contributes nothing

Answer $$\boxed{x=1\ \text{and}\ x=-1}$$
Check

Check by values: $f(1)=0.5$, $f(-1)=-0.5$, and $f(x)\to 0$ as $\lvert x\rvert$ grows. So $f$ really does turn around twice, once at each critical number, which is what two sign changes of $1-x^{2}$ predict.

For a quotient, differentiate first and then read the two searches off the two levels of the fraction: numerator zero gives one kind, denominator zero gives the other.

5§05.7 — finding c for a square root●●●○○

The Mean Value Theorem needs differentiability only on the open interval, and this function is built to show why that matters.

Given

  • $f(x)=\sqrt{x}$

  • the closed interval $[0,4]$

Find

  1. (a) Check the hypotheses on $[0,4]$, mentioning what happens at $x=0$.

  2. (b) Find every $c$ in $(0,4)$ with $f'(c)$ equal to the average rate of change.

Hint 1/4

Check the two hypotheses one at a time and notice which interval each one is about.

Hint 2/4

Continuity is required on $[0,4]$ and differentiability only on $(0,4)$; the conclusion is $f'(c)=\dfrac{f(b)-f(a)}{b-a}$.

Hint 3/4

Here $f(4)=2$ and $f(0)=0$, so the average rate is $\dfrac{2-0}{4-0}=\dfrac12$, and $f'(x)=\dfrac{1}{2\sqrt{x}}$.

Hint 4/4

Solving $\dfrac{1}{2\sqrt c}=\dfrac12$ gives $\sqrt c=1$, so $c=1$.

Show solution
Hypotheses, read precisely
$$f\ \text{is continuous on}\ [0,4]$$

the square root is continuous from the right at $0$, which is all the closed interval asks for

$$f'(x)=\frac{1}{2\sqrt{x}}\ \text{exists for every}\ x\in(0,4)$$

the failure at $x=0$ is harmless because $0$ is an endpoint and the hypothesis only covers the open interval

Average rate, then solve
$$\frac{f(4)-f(0)}{4-0}=\frac{2-0}{4}=\tfrac12$$

the target slope

$$\frac{1}{2\sqrt c}=\tfrac12\ \Rightarrow\ \sqrt c=1\ \Rightarrow\ c=1$$

and $c=1$ lies in $(0,4)$, so it is admissible

Answer $$\boxed{c=1}$$
Check

Independent check: $f'$ is strictly decreasing on $(0,4)$, from very large near $0$ down to $f'(4)=0.25$. A strictly decreasing continuous function crosses the level $0.5$ exactly once, so exactly one $c$ exists, matching what the algebra found.

A vertical tangent at an endpoint never blocks this theorem. Blocking it requires a failure strictly inside the interval.

C · exam level 3 questions
1§05.6 — a trigonometric search under exam conditions●●●●○

The kind of closed interval problem that appears with a trigonometric function instead of a polynomial, where the arithmetic at the end is where the answers separate.

Given

  • $f(x)=\sin x+\cos x$

  • the closed interval $[0,\pi/2]$

Find

  1. (a) What is the absolute maximum value of $f$ on $[0,\pi/2]$?

Hint 1/4

The usual four steps. Expect one critical number inside, and remember that both endpoints are candidates.

Hint 2/4

Differentiate and solve $f'(x)=0$; then compare values of $f$, not of $f'$.

Hint 3/4

Here $f'(x)=\cos x-\sin x=0$ gives $\tan x=1$, so $x=\pi/4$; the candidates are $0$, $\pi/4$ and $\pi/2$.

Hint 4/4

The values are $f(0)=1$, $f(\pi/4)=\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}=\sqrt2$ and $f(\pi/2)=1$, so the maximum is $\sqrt2$.

Show solution
Candidates
$$f'(x)=\cos x-\sin x=0\ \Rightarrow\ \tan x=1\ \Rightarrow\ x=\tfrac{\pi}{4}$$

dividing by $\cos x$ is safe here because $\cos x=0$ would force $\sin x=0$ too, which never happens

$$\text{candidates}:\ 0,\ \tfrac{\pi}{4},\ \tfrac{\pi}{2}$$

one critical number plus both endpoints

Values
$$f(0)=0+1=1,\qquad f\!\left(\tfrac{\pi}{2}\right)=1+0=1$$

the two endpoints tie, which is a hint that the interior point is the winner

$$f\!\left(\tfrac{\pi}{4}\right)=\tfrac{\sqrt2}{2}+\tfrac{\sqrt2}{2}=\sqrt2\approx 1.414$$

and $\sqrt2>1$, so the maximum sits strictly inside

Answer $$\boxed{\max=\sqrt2\ \text{at}\ x=\tfrac{\pi}{4}}$$
Check

Independent route: $(\sin x+\cos x)^{2}=1+2\sin x\cos x=1+\sin 2x\le 2$, with equality when $\sin 2x=1$, that is $x=\pi/4$. So the maximum of $f$ is exactly $\sqrt2$, no search required.

The identity route also shows why the two endpoints tie: $\sin 2x$ is zero at both ends of this interval.

2§05.1 — find the error in someone else's estimate●●●○○

A student estimates $\sqrt[3]{8.06}$ with a linearization and writes four steps. Exactly one step is wrong.

Given

  • Step 1: take $f(x)=x^{1/3}$ with base point $a=8$, so $f(8)=2$.

  • Step 2: $f'(x)=\tfrac13x^{-2/3}$, so $f'(8)=\tfrac13\cdot\tfrac14=\tfrac{1}{12}$.

  • Step 3: $L(x)=2+\tfrac{1}{12}(x-8)$.

  • Step 4: $L(8.06)=2+\tfrac{1}{12}(8.06)=2.672$, so $\sqrt[3]{8.06}\approx 2.672$.

Find

  1. (a) Which step contains the error, and what is the correct final estimate?

Hint 1/4

Read the four steps against the definition of the linearization, one symbol at a time. Also ask whether the final number is even plausible.

Hint 2/4

In $L(x)=f(a)+f'(a)(x-a)$ the last factor is the step away from the base point, not the input itself.

Hint 3/4

With $a=8$ and $x=8.06$ the step is $x-a=0.06$, so the correct evaluation is $L(8.06)=2+\tfrac{1}{12}(0.06)$.

Hint 4/4

That gives $2.005$, so the error is in Step 4.

Show solution
Audit the steps that are right, so the search narrows
$$f(8)=2\ \checkmark,\qquad f'(8)=\tfrac13(8)^{-2/3}=\tfrac13\cdot\tfrac14=\tfrac{1}{12}\ \checkmark$$

$8^{2/3}=4$, so Steps 1 and 2 are sound

$$L(x)=2+\tfrac{1}{12}(x-8)\ \checkmark$$

Step 3 states the line correctly, which localises the fault to the evaluation

Repair the evaluation
$$L(8.06)=2+\tfrac{1}{12}(8.06-8)=2+\tfrac{0.06}{12}$$

the last factor is the step from the base point; using $8.06$ there adds the whole distance from the origin

$$=2+0.005=2.005$$

the estimate

Answer $$\boxed{\text{Step 4 is wrong};\quad \sqrt[3]{8.06}\approx 2.005}$$
Check

Plausibility check, which would have caught it instantly: $8.06$ is barely more than $8$, so its cube root must be barely more than $2$. An answer of $2.672$ would mean $2.672^{3}\approx 19$, nowhere near $8.06$.

Before trusting an estimate, ask whether it is even in the right neighbourhood. A one line plausibility check catches most linearization slips.

3§05.7 — a bound on the derivative becomes a bound on the value●●●●○

This is the shape in which the Mean Value Theorem is actually examined: nothing is known about the formula for $f$, only a cap on how fast it can change.

Given

  • $f$ is differentiable at every real number

  • $f(2)=5$

  • $f'(x)\le 3$ for every $x$

Find

  1. (a) How large can $f(6)$ be?

  2. (b) Give a function that attains your bound, to show it cannot be improved.

Hint 1/4

You know something about $f'$ everywhere and something about $f$ at one point, and you want $f$ at another point. Which theorem connects those three things?

Hint 2/4

The Mean Value Theorem gives $f(6)-f(2)=f'(c)(6-2)$ for some $c$ between $2$ and $6$; you do not need to know which $c$.

Hint 3/4

With $f(2)=5$ and $f'\le 3$ everywhere: $f(6)-5=4f'(c)\le 4(3)=12$.

Hint 4/4

So $f(6)\le 17$, and the line $f(x)=5+3(x-2)$ reaches exactly $17$.

Show solution
Convert the unknown value into a difference
$$f(6)-f(2)=f'(c)(6-2)\ \text{for some}\ c\in(2,6)$$

differentiability everywhere gives both hypotheses on $[2,6]$, so the theorem applies; the unknown $c$ is harmless because the cap holds at every point

Insert the cap and unwrap
$$f(6)-5=4f'(c)\le 4(3)=12$$

multiplying an inequality by the positive number $4$ keeps its direction

$$f(6)\le 17$$

the bound

Show the bound is sharp
$$f(x)=5+3(x-2)\ \Rightarrow\ f(2)=5,\ f'(x)=3\le 3,\ f(6)=17$$

a function that satisfies every hypothesis and hits the bound exactly, so no smaller bound is correct

Answer $$\boxed{f(6)\le 17,\ \text{attained by}\ f(x)=5+3(x-2)}$$
Check

Test the bound against a different qualifying function: $f(x)=5+3\sin(x-2)$ has $f(2)=5$ and $f'(x)=3\cos(x-2)\le 3$, so every hypothesis holds, and $f(6)=5+3\sin 4\approx 2.73$, comfortably under $17$. Qualifying functions stay under the bound and the straight line meets it, which is what sharp means.

No formula for $f$ was ever needed. A bound on the derivative plus one known value is enough to trap the function everywhere else, and that is what makes this theorem more than a curiosity.

D · interleaved 3 questions
1§05.2 — a balloon, twice●●●●○

A spherical balloon is being inflated. Two questions about it, and part of the exercise is noticing that they are not the same question.

Given

  • $V=\dfrac43\pi r^{3}$, with $r$ in centimetres

  • in part (a) the radius grows from $5.00$ cm to $5.02$ cm

  • in part (b) the radius grows at $0.3$ cm per second, and we look at the instant when $r=5$ cm

Find

  1. (a) Estimate the change in volume, in cm³.

  2. (b) Find how fast the volume is growing at that instant, in cm³ per second.

Hint 1/4

Two questions, two different unknowns: one is a change, the other is a rate. Read what each part gives you before choosing a tool.

Hint 2/4

For a change caused by a small step, $dV=\dfrac{dV}{dr}\,dr$; for a rate in time, differentiate the same relation with respect to $t$ to get $\dfrac{dV}{dt}=\dfrac{dV}{dr}\cdot\dfrac{dr}{dt}$.

Hint 3/4

Both use $\dfrac{dV}{dr}=4\pi r^{2}$, which is $4\pi(25)=100\pi$ at $r=5$. Part (a) has $dr=0.02$ cm; part (b) has $\dfrac{dr}{dt}=0.3$ cm/s.

Hint 4/4

So $dV=100\pi(0.02)=2\pi\approx 6.3$ cm³ and $\dfrac{dV}{dt}=100\pi(0.3)=30\pi\approx 94.2$ cm³/s.

Show solution
One derivative serves both parts
$$\frac{dV}{dr}=4\pi r^{2}=4\pi(25)=100\pi$$

the sensitivity of volume to radius at this instant, and it is the only calculus in the problem

(a) A step in the input calls for a differential
$$dV=100\pi\,(0.02)=2\pi\approx 6.3\ \text{cm}^{3}$$

no time appears anywhere in this part, which is the signal that a differential and not a rate is wanted

(b) A speed in the input calls for the chain rule in time
$$\frac{dV}{dt}=\frac{dV}{dr}\cdot\frac{dr}{dt}=100\pi(0.3)=30\pi\approx 94.2\ \text{cm}^{3}/\text{s}$$

the same factor $100\pi$, now multiplied by a speed instead of a step, so the units come out per second

Answer $$\boxed{dV=2\pi\ \text{cm}^{3};\qquad dV/dt=30\pi\ \text{cm}^{3}/\text{s}}$$
Check

Units settle both answers: cm² times cm gives cm³ in part (a), and cm² times cm per second gives cm³ per second in part (b). And the exact change in (a) is $\tfrac43\pi(5.02^{3}-5^{3})\approx 6.31$ cm³, against our $6.28$.

Same relation, same derivative, two different questions. The word that decides which is which is per second.

2§05.6 — extrema of a product with a square root●●●●○

A search on a closed interval where the derivative takes more work than the search itself.

Given

  • $f(x)=x\sqrt{4-x^{2}}$

  • the domain, and the interval, is $[-2,2]$

Find

  1. (a) Find every critical number of $f$ inside $(-2,2)$.

  2. (b) Find the absolute maximum and minimum values of $f$ on $[-2,2]$.

Hint 1/4

Differentiating this needs two rules at once. Decide which is the outer structure before writing anything.

Hint 2/4

It is a product, and the second factor needs the chain rule: $\left(\sqrt{u}\right)'=\dfrac{u'}{2\sqrt u}$.

Hint 3/4

With $f(x)=x\sqrt{4-x^{2}}$: $f'(x)=\sqrt{4-x^{2}}+x\cdot\dfrac{-2x}{2\sqrt{4-x^{2}}}=\dfrac{4-2x^{2}}{\sqrt{4-x^{2}}}$.

Hint 4/4

The numerator vanishes at $x=\pm\sqrt2$, and $f(\pm\sqrt2)=\pm 2$ while $f(\pm 2)=0$.

Show solution
Differentiate: product on the outside, chain on the inside
$$f'(x)=\sqrt{4-x^{2}}+x\cdot\frac{-2x}{2\sqrt{4-x^{2}}}$$

product rule first, and the second factor's derivative comes from the chain rule with inner function $4-x^{2}$

$$f'(x)=\frac{(4-x^{2})-x^{2}}{\sqrt{4-x^{2}}}=\frac{4-2x^{2}}{\sqrt{4-x^{2}}}$$

putting everything over one denominator is what makes the zeros visible

Both species of critical number
$$4-2x^{2}=0\ \Rightarrow\ x=\pm\sqrt2$$

inside the interval, so both are candidates

$$f'\ \text{fails to exist at}\ x=\pm2$$

these are the endpoints, so they are already on the candidate list for their own reason

Evaluate and compare
$$f(\sqrt2)=\sqrt2\cdot\sqrt{2}=2,\qquad f(-\sqrt2)=-2$$

since $4-(\sqrt2)^{2}=2$

$$f(-2)=f(2)=0$$

the square root vanishes at both ends

$$\max=2\ \text{at}\ x=\sqrt2,\qquad \min=-2\ \text{at}\ x=-\sqrt2$$

compare the four values

Answer $$\boxed{\max=2\ \text{at}\ x=\sqrt2,\qquad \min=-2\ \text{at}\ x=-\sqrt2}$$
Check

Symmetry check: $f(-x)=-f(x)$, so the graph is odd and the minimum must be the negative of the maximum, at the mirrored point. Our two answers obey that exactly. A second check: $f(x)^{2}=x^{2}(4-x^{2})\le 4$ by the arithmetic and geometric mean inequality, so $\lvert f\rvert\le 2$ and both bounds are reached.

When the derivative is a single fraction, the numerator gives one species of critical number and the denominator gives the other. Building that fraction is worth the extra algebra.

3§05.4 — a function that is defined everywhere and still escapes●●●●○

A closed interval, a function defined at every point of it, and a question about whether the guarantee applies.

Given

  • $f(x)=\dfrac{1}{x-2}$ for $x\neq 2$, and $f(2)=0$

  • the closed interval $[0,3]$

  • so $f$ has a value at every point of $[0,3]$

Find

  1. (a) Which statement about the absolute extrema of $f$ on $[0,3]$ is correct?

Hint 1/4

Check the hypotheses of the guarantee first, at every point of the interval, including the awkward one.

Hint 2/4

The Extreme Value Theorem needs continuity at every point of a closed bounded interval; at $x=2$ compare the limit with the value.

Hint 3/4

Here $f(2)=0$ while $f(x)\to+\infty$ as $x\to 2^{+}$ and $f(x)\to-\infty$ as $x\to 2^{-}$, so $f$ is defined but not continuous at $2$.

Hint 4/4

Values climb past every ceiling just right of $2$ and fall below every floor just left of it, so neither extreme exists.

Show solution
Test the hypothesis at the only point in doubt
$$\lim_{x\to 2^{+}}\frac{1}{x-2}=+\infty\neq 0=f(2)$$

the point $2$ is in the domain, so continuity there is a fair question, and the answer is no

$$\text{EVT does not apply}$$

the interval is closed and bounded, so it is the other hypothesis that failed

Defeat every candidate explicitly
$$\text{for any}\ M>0:\ x=2+\frac{1}{M+1}\in[0,3]\ \text{gives}\ f(x)=M+1$$

so no proposed maximum survives

$$\text{for any}\ m<0:\ x=2-\frac{1}{\lvert m\rvert+1}\ \text{gives}\ f(x)=-(\lvert m\rvert+1)$$

so no proposed minimum survives either

Answer $$\boxed{\text{neither extreme value exists}}$$
Check

Try the recipe on numbers: propose $M=50$; then $x=2+1/51\approx 2.0196$ lies in $[0,3]$ and $f(x)=51>50$. Propose $m=-50$; then $x=2-1/51$ gives $-51<-50$.

Closed and bounded was never the problem here. One interior point of discontinuity was enough, and it took out both extrema at once.

Shaped like the real papers 5 questions
1§05.7 — counting the real roots of a quartic●●●●●

The most repeated question in this course's Midterm 1 window, and it carries about half of a large question. Budget twelve minutes: the floor and the ceiling are two separate arguments and each needs its own theorem named.

Given

  • $f(x)=x^{4}+4x^{3}+12x^{2}+8x-6$

  • $f$ is a polynomial, so it is continuous and differentiable on all of $\mathbb{R}$

  • the roots are not rational, so factoring is not available

Find

  1. (a) Find the exact number of distinct real roots of $f$.

  2. (b) Name the theorem that licenses the step from "$f'$ has exactly one real zero" to "$f$ has at most two real zeros", and say in one sentence why its hypotheses hold here.

Hint 1/4

You are not asked to find the roots, and you cannot: this quartic has no rational root and no obvious factorisation. So the count has to come from information about $f$ that is cheaper to obtain than the roots themselves.

Hint 2/4

A sign change of a continuous function forces a root, which gives a floor. For a ceiling, use the fact that between two roots of $f$ the derivative must vanish somewhere, so count the zeros of $f'$ first, and control those by looking at $f''$.

Hint 3/4

$f'(x)=4x^{3}+12x^{2}+24x+8$ and $f''(x)=12(x^{2}+2x+2)$. Useful values: $f(-2)=10$, $f(-1)=-5$, $f(0)=-6$, $f(1)=19$, and $f'(-1)=-8$, $f'(0)=8$.

Hint 4/4

$f''>0$ makes $f'$ strictly increasing, so $f'$ has at most one zero, and its sign change on $(-1,0)$ makes it exactly one. That caps $f$ at two roots, while the two sign changes of $f$ force at least two.

Show solution
Floor: two sign changes, two roots
$$f(-2)=16-32+48-16-6=10>0,\qquad f(-1)=1-4+12-8-6=-5<0$$

$f$ is a polynomial, hence continuous, so the Intermediate Value Theorem puts a root inside $(-2,-1)$

$$f(0)=-6<0,\qquad f(1)=1+4+12+8-6=19>0$$

a second sign change puts a second root inside $(0,1)$, and the two roots are distinct because the two intervals are disjoint

$$\#\{\text{real roots}\}\ \ge\ 2$$

the Intermediate Value Theorem only ever produces existence, so this half of the argument can never give the exact count

The derivative vanishes at most once
$$f'(x)=4x^{3}+12x^{2}+24x+8$$

differentiate term by term

$$f''(x)=12x^{2}+24x+24=12\bigl(x^{2}+2x+2\bigr)=12\bigl[(x+1)^{2}+1\bigr]\ \ge\ 12>0$$

completing the square turns "looks positive" into a proof; the discriminant $24^{2}-4(12)(24)=-576<0$ says the same thing

$$f''>0\ \text{on}\ \mathbb{R}\ \Longrightarrow\ f'\ \text{is strictly increasing}$$

if $f'(u)=f'(v)$ with $u<v$, Rolle applied to $f'$ would give a zero of $f''$ between them, and there is none

$$\Longrightarrow\ f'\ \text{has at most one real zero}$$

a strictly increasing function takes every value at most once, in particular the value $0$

And it does vanish once
$$f'(-1)=-4+12-24+8=-8<0,\qquad f'(0)=8>0$$

$f'$ is continuous, so the Intermediate Value Theorem puts a zero of $f'$ inside $(-1,0)$

$$f'\ \text{has exactly one real zero}$$

at most one from the previous block, at least one from here

Ceiling: Rolle caps the count at two
$$\text{suppose}\ a<b<c\ \text{were three roots of}\ f$$

argue by contradiction, since the ceiling cannot be seen directly

$$f(a)=f(b)=0\ \Longrightarrow\ f'(\xi_{1})=0\ \text{for some}\ \xi_{1}\in(a,b);\quad f(b)=f(c)=0\ \Longrightarrow\ f'(\xi_{2})=0\ \text{for some}\ \xi_{2}\in(b,c)$$

Rolle's theorem on $[a,b]$ and on $[b,c]$; a polynomial is continuous on each closed interval and differentiable on each open one, and the endpoint values agree because both are roots

$$\xi_{1}<b<\xi_{2}\ \Longrightarrow\ f'\ \text{has two distinct zeros, contradicting the previous block}$$

so three roots is impossible, and the same argument rules out any larger number

$$2\ \le\ \#\{\text{real roots}\}\ \le\ 2\ \Longrightarrow\ \#\{\text{real roots}\}=2$$

floor and ceiling meet, which is what "exact number" demands

(b) The licence, in one sentence
$$\text{Rolle's theorem}$$

$f$ is a polynomial, so on the closed interval between any two of its roots it is continuous, on the open interval it is differentiable, and its values at the two ends are equal (both are $0$), which are exactly the three hypotheses; the course's own solutions call this step MVT, since Rolle is the case of the Mean Value Theorem with equal endpoint values

Check

Independent route through the shape of the graph. The single zero $r$ of $f'$ lies near $-0.404$, and $f'<0$ to its left and $f'>0$ to its right, so $f$ falls then rises and $f(r)$ is the global minimum. Numerically $f(-0.404)\approx-7.51<0$, while $f\to+\infty$ in both directions, so a curve of that shape meets the axis exactly twice. Same answer, no theorem quoted.

2§05.5 — critical numbers of a square, then extrema on a half open interval●●●●●

The shape of the largest application question in this course: one function, three parts, and the middle part carries the biggest slice of the marks even though it produces no value of $F$ at all. Fifteen minutes.

Given

  • $F(x)=f(x)^{2}$, where $f(x)=\dfrac{x^{2}-9}{x^{2}+3}$

  • part (a) concerns all of $\mathbb{R}$

  • parts (b) and (c) concern the half open interval $[-1,3)$

Find

  1. (a) How many critical numbers does $F$ have? List them, say which come from $f(x)=0$ and which from $f'(x)=0$, and state why no critical number comes from a failure of differentiability.

  2. (b) Find the absolute maximum value of $F$ on $[-1,3)$ and the point where it is attained.

  3. (c) Does $F$ attain an absolute minimum on $[-1,3)$? Answer in one sentence, and say what the Extreme Value Theorem does and does not give you here.

Hint 1/4

Do not expand. $F$ is built from $f$ by a single outer operation, and the whole of part (a) is about what that outer operation does to the places where the graph is flat.

Hint 2/4

The chain rule gives $F'=2ff'$, so a critical number of $F$ is any point of the domain where $f=0$ or $f'=0$, plus any point where $F'$ fails to exist. Listing the roots of $F(x)=0$ and stopping is the standard way to lose the whole part.

Hint 3/4

$f'(x)=\dfrac{24x}{(x^{2}+3)^{2}}$, so $F'(x)=\dfrac{48x(x^{2}-9)}{(x^{2}+3)^{3}}$, whose denominator is never $0$. Also $F(-1)=4$ and $F(0)=9$.

Hint 4/4

The critical numbers are $-3$, $0$ and $3$. On $[-1,3)$ the sign of $F'$ is positive on $(-1,0)$ and negative on $(0,3)$, so $F$ climbs to $9$ at $x=0$ and then falls towards $0$ without ever reaching it.

Show solution
(a) Differentiate the square, not the expansion
$$f'(x)=\frac{2x(x^{2}+3)-(x^{2}-9)(2x)}{(x^{2}+3)^{2}}=\frac{2x\bigl[(x^{2}+3)-(x^{2}-9)\bigr]}{(x^{2}+3)^{2}}=\frac{24x}{(x^{2}+3)^{2}}$$

quotient rule, then factor $2x$ out of the numerator so the cancellation is visible

$$F'(x)=2f(x)f'(x)=2\cdot\frac{x^{2}-9}{x^{2}+3}\cdot\frac{24x}{(x^{2}+3)^{2}}=\frac{48x(x^{2}-9)}{(x^{2}+3)^{3}}$$

chain rule on the outer square; keeping the factored form is what makes the count readable

$$(x^{2}+3)^{3}\ \ge\ 27>0\quad\text{for every real }x$$

the denominator never vanishes, so $F'$ is defined at every real number and no critical number arises from a failure of differentiability; this is the sentence the grading keys pay for

$$48x(x^{2}-9)=0\iff x=0\ \text{or}\ x=\pm 3$$

three critical numbers: $\pm3$ are the zeros of $f$ and $0$ is the zero of $f'$

$$F(x)=0\iff x=\pm 3$$

the roots of $F$ are only two of the three, so a candidate list read off $F(x)=0$ misses $x=0$, which is exactly where the maximum turns out to be

(b) The search on the half open interval
$$\text{candidates in}\ [-1,3):\quad x=-1\ (\text{included endpoint}),\qquad x=0\ (\text{critical})$$

$-3$ lies outside the interval and $3$ is excluded from it, so neither is a candidate

$$F(-1)=\left(\frac{1-9}{1+3}\right)^{2}=(-2)^{2}=4,\qquad F(0)=\left(\frac{0-9}{0+3}\right)^{2}=(-3)^{2}=9$$

values of $F$, never values of $F'$

$$x\in(-1,0):\ x<0,\ x^{2}-9<0\ \Rightarrow\ F'>0;\qquad x\in(0,3):\ x>0,\ x^{2}-9<0\ \Rightarrow\ F'<0$$

the Closed Interval Method cannot be quoted here, so the sign of $F'$ has to do the comparing instead

$$\max_{[-1,3)}F=9\ \ \text{at}\ x=0$$

$F$ rises on $(-1,0)$ and falls on $(0,3)$, so $x=0$ beats every other point; value and location reported separately

(c) The minimum that is approached and never reached
$$F(x)>0\ \text{for every}\ x\in[-1,3)$$

$F=f^{2}$ vanishes only where $f$ does, that is at $x=\pm3$, and neither point belongs to the interval

$$\lim_{x\to 3^{-}}F(x)=\left(\frac{9-9}{9+3}\right)^{2}=0$$

$F$ is continuous at $3$, but $3$ is not in the interval, so the value $0$ is approached and never taken

$$\inf_{[-1,3)}F=0\ \text{is not attained}\ \Longrightarrow\ \text{no absolute minimum}$$

given any $x_{0}$ in the interval there is an $x$ nearer to $3$ with $0<F(x)<F(x_{0})$, so no candidate can be smallest

$$[-1,3)\ \text{is bounded but not closed}\ \Longrightarrow\ \text{EVT gives no guarantee}$$

the theorem promises both extreme values only on a closed bounded interval; losing that hypothesis does not forbid an extreme value, and here the maximum survives while the minimum does not, which is precisely why the theorem is a guarantee and not a test

Check

Spot checks against each claim. Towards the open end, $F(2.9)=\left(\frac{-0.59}{11.41}\right)^{2}\approx 2.7\times10^{-3}$ and $F(2.99)\approx 2.5\times10^{-5}$: positive and shrinking, which is the infimum $0$ being approached but not reached. Away from the peak, $F(-0.5)=\left(\frac{-8.75}{3.25}\right)^{2}\approx 7.25<9$, consistent with $x=0$ being the maximum. Closing the interval to $[-1,3]$ would restore the minimum, value $0$ at $x=3$, and that single change is the whole content of part (c).

3§05.1 — the region where the tangent line is still good enough●●●●○

The official exercise set owns this topic and asks it twice in this shape: not "estimate a number" but "where is the estimate good" and "which of two estimates is better and why". The reason is the marked object in the second half. Ten minutes.

Given

  • $f(x)=\dfrac{1}{x}$

  • $L$ is the linearization of $f$ at the base point $a=2$

Find

  1. (a) Find every $x$ for which the error $\lvert f(x)-L(x)\rvert$ is smaller than $\tfrac14$.

  2. (b) Of the two estimates $L(1.9)$ and $L(2.1)$, which one is closer to the value it estimates? One sentence, carrying the reason.

Hint 1/4

The question is about the gap between two graphs, so the first thing on the page should be that gap written as a single expression in $x$, before the inequality is touched at all.

Hint 2/4

$L(x)=f(a)+f'(a)(x-a)$, with $f'$ evaluated at $a$. Once $f(x)-L(x)$ is over a common denominator, the inequality becomes a quadratic one, and quadratic inequalities are settled by factoring and reading signs, not by cancelling.

Hint 3/4

$f(2)=\tfrac12$ and $f'(x)=-x^{-2}$ give $f'(2)=-\tfrac14$, so $L(x)=\tfrac12-\tfrac14(x-2)=1-\tfrac{x}{4}$ and $f(x)-L(x)=\dfrac{(x-2)^{2}}{4x}$.

Hint 4/4

For $x>0$ the inequality $\dfrac{(x-2)^{2}}{4x}<\tfrac14$ is $(x-1)(x-4)<0$; for $x<0$ the gap is never that small. For part (b), compare $\dfrac{h^{2}}{4(2+h)}$ with $\dfrac{h^{2}}{4(2-h)}$.

Show solution
The line, and the gap in closed form
$$f(2)=\tfrac12,\qquad f'(x)=-\frac{1}{x^{2}}\ \Rightarrow\ f'(2)=-\tfrac14$$

the derivative is evaluated at the base point before the line is written, otherwise $L$ is not a line

$$L(x)=\tfrac12-\tfrac14(x-2)=1-\frac{x}{4}$$

$L(x)=f(a)+f'(a)(x-a)$, with the height $f(a)$ kept

$$f(x)-L(x)=\frac{1}{x}-1+\frac{x}{4}=\frac{4-4x+x^{2}}{4x}=\frac{(x-2)^{2}}{4x}$$

one common denominator; the numerator is a perfect square centred at $a$, which is why the estimate is good near $2$ and why the error grows quadratically away from it

(a) The branch that contains the base point
$$x>0\ \Longrightarrow\ \left\lvert f(x)-L(x)\right\rvert=\frac{(x-2)^{2}}{4x}$$

numerator and denominator are both positive, so the absolute value does nothing

$$\frac{(x-2)^{2}}{4x}<\frac14\iff (x-2)^{2}<x$$

multiplying by the positive number $4x$ preserves the direction of the inequality, and this is the step that needs $x>0$ stated first

$$x^{2}-5x+4<0\iff (x-1)(x-4)<0\iff 1<x<4$$

the product of two factors is negative exactly between their roots

$$x=1:\ \frac{1}{4}= \frac14;\qquad x=4:\ \frac{4}{16}=\frac14$$

at both endpoints the error equals the threshold, so the inequality is strict and the interval is open

(a) The other branch contributes nothing
$$x<0,\ t=-x>0\ \Longrightarrow\ \left\lvert f(x)-L(x)\right\rvert=\frac{(x-2)^{2}}{4\lvert x\rvert}=\frac{(t+2)^{2}}{4t}$$

rewrite with a positive variable so the size can be estimated

$$\frac{(t+2)^{2}}{4t}=\frac{t}{4}+1+\frac{1}{t}\ \ge\ 1+2\sqrt{\frac{t}{4}\cdot\frac{1}{t}}=2$$

expand and apply the arithmetic-geometric mean inequality to the two variable terms; the bound $2$ is reached at $t=2$

$$2>\tfrac14\ \Longrightarrow\ \text{no}\ x<0\ \text{qualifies}$$

the tangent line drawn at $2$ says nothing usable about the other branch of the hyperbola, and skipping this check would leave the answer set unproved

$$\boxed{\,1<x<4\,}$$

the full solution set

(b) Equal steps, unequal errors
$$E(2\pm h)=\frac{h^{2}}{4(2\pm h)},\qquad 0<h<2$$

substitute $x=2\pm h$ into the exact gap; the numerator is the same on both sides, so only the denominator decides

$$4(2-h)<4(2+h)\ \Longrightarrow\ E(2+h)<E(2-h)$$

the same positive numerator over the larger denominator is the smaller number, so the step to the right is always the better estimate

$$h=0.1:\quad E(2.1)=\frac{0.01}{8.4}\approx 0.00119\ <\ E(1.9)=\frac{0.01}{7.6}\approx 0.00132$$

the numerical answer to the question as asked

$$f''(x)=\frac{2}{x^{3}}>0\ \text{on}\ x>0$$

the same conclusion read off the concavity: $f$ is concave up there, so the tangent line lies below the curve on both sides, and the curvature that pulls the curve away from the line is larger at $1.9$ than at $2.1$

Check

Direct check against the true values. $L(2.1)=1-0.525=0.475$ while $1/2.1=0.476190\ldots$, a miss of $0.00119$; $L(1.9)=1-0.475=0.525$ while $1/1.9=0.526315\ldots$, a miss of $0.00132$. Both estimates fall short of the truth, matching $f-L=(x-2)^{2}/(4x)\ge 0$ on $x>0$. And at the far edge of the answer interval, $x=3.9$ gives error $\frac{(1.9)^{2}}{15.6}\approx 0.231<0.25$, while $x=4.1$ gives $\frac{(2.1)^{2}}{16.4}\approx 0.269>0.25$, so the boundary really sits at $4$.

4§05.7 — which theorem lets a derivative skip nothing●●●●●

This course states Darboux's theorem for you inside the question and then spends the marks on whether you can tell when it applies and when it does not. Short to write, easy to answer confidently and wrongly. Eight minutes.

Given

  • Darboux's theorem, as this course states it: if $g$ is differentiable at every point of $[a,b]$, then $g'$ takes every value between $g'(a)$ and $g'(b)$ somewhere on $[a,b]$

  • $h(x)=2x+1$ for $x\ge 0$, and $h(x)=2x-1$ for $x<0$

Find

  1. (a) Is there a function $g$, differentiable at every real number, with $g'=h$? Decide, and justify.

  2. (b) Suppose $f$ is differentiable on $[0,4]$ with $f'(0)=-2$ and $f'(4)=5$. Must $f'(c)=1$ for some $c$ in $(0,4)$? Decide, name the theorem you used, and say in one sentence why applying the Intermediate Value Theorem to $f'$ would not have been enough.

Hint 1/4

Being somebody's derivative is a restriction, and continuity is not the dividing line: some discontinuous functions are derivatives and some are not. Read the theorem you were handed as a test that a candidate function has to pass.

Hint 2/4

A function with a jump skips values, and the theorem says a derivative cannot skip a value lying between two of its own values. So write down the set of values the candidate actually takes. For part (b), notice which hypotheses you were given and which you were not.

Hint 3/4

For $x\ge 0$, $h(x)=2x+1\ge 1$; for $x<0$, $h(x)=2x-1<-1$. On $[-1,1]$ the endpoint values are $h(-1)=-3$ and $h(1)=3$. In part (b), $-2<1<5$ and $f$ is differentiable at every point of $[0,4]$.

Hint 4/4

$h$ takes no value in $(-1,1)$, so $0$ lies between two of its values and is never attained, and $h$ fails the test. In part (b) the hypothesis of the theorem is exactly what was given, so the answer is yes, and $c$ cannot be an endpoint because neither $f'(0)$ nor $f'(4)$ equals $1$.

Show solution
(a) The set of values the candidate takes
$$x\ge 0\ \Rightarrow\ h(x)=2x+1\ge 1;\qquad x<0\ \Rightarrow\ h(x)=2x-1<-1$$

each branch is a line, and each branch stays clear of the gap between $-1$ and $1$

$$h(\mathbb{R})\subseteq(-\infty,-1)\cup[1,\infty)$$

so no value in the open interval $(-1,1)$ is attained anywhere

(a) Run the test on a concrete interval
$$h(-1)=-3,\qquad h(1)=3,\qquad -3<0<3$$

the value $0$ lies strictly between two values that $h$ does take, which is the situation the theorem speaks about

$$h(c)=0\ \text{has no solution, since}\ 0\in(-1,1)$$

excluded by the previous block

$$\text{if}\ g\ \text{were differentiable on}\ [-1,1]\ \text{with}\ g'=h,\ \text{Darboux would force}\ h(c)=0\ \text{for some}\ c\in[-1,1]$$

the hypothesis of the theorem is exactly what $g'=h$ on $[-1,1]$ asserts

$$\Longrightarrow\ \text{no such}\ g\ \text{exists}$$

the contradiction kills the assumption, and it kills it on every interval containing $0$, not just this one

(a) The same conclusion by hand, as a check on the reasoning
$$g(x)=x^{2}+x+C_{1}\ (x\ge 0),\qquad g(x)=x^{2}-x+C_{2}\ (x<0)$$

these are the only antiderivatives available on each branch separately

$$\text{continuity at}\ 0\ \Rightarrow\ C_{1}=C_{2},\qquad\text{then}\ g'(0^{-})=-1\ne 1=g'(0^{+})$$

the two one sided slopes disagree, so $g$ is not differentiable at $0$ no matter how the constants are chosen; the obstruction sits at the single point $0$

(b) The hypothesis is already on the page
$$f\ \text{differentiable on}\ [0,4],\qquad f'(0)=-2,\qquad f'(4)=5$$

this is precisely the hypothesis of Darboux's theorem as stated; nothing further is needed and nothing further was offered

$$-2<1<5\ \Longrightarrow\ \exists\,c\in[0,4]\ \text{with}\ f'(c)=1$$

the theorem applied to the value $1$

$$f'(0)=-2\ne 1\ \text{and}\ f'(4)=5\ne 1\ \Longrightarrow\ c\in(0,4)$$

the endpoints rule themselves out by their own values, so the $c$ is interior, which is what was asked

(b) Why the Intermediate Value Theorem is not available
$$\text{IVT hypothesis}:\ f'\ \text{continuous on}\ [0,4]$$

that was never assumed, and it does not follow from $f$ being differentiable

$$g(x)=x^{2}\sin(1/x)\ (x\ne 0),\ g(0)=0\ \Rightarrow\ g'(x)=2x\sin(1/x)-\cos(1/x)\ (x\ne0),\ g'(0)=0$$

a function differentiable everywhere whose derivative has no limit at $0$, so derivatives really can be discontinuous

$$\text{Darboux needs only that}\ f'\ \text{exists on}\ [0,4]$$

which is why this course states the theorem separately instead of leaning on the Intermediate Value Theorem: it delivers the intermediate value property without asking for continuity

Check

The two parts check each other. In (a) what fails is not smoothness but a jump of size $2$ at a single point, and the hand computation with $C_{1}$ and $C_{2}$ locates the failure at exactly that point without using the theorem at all. In (b), the example $g(x)=x^{2}\sin(1/x)$ shows that any argument leaning on continuity of $f'$ would have rested on an assumption nobody made, while the conclusion still holds. Read together: a derivative may be wildly discontinuous, but it may never jump over a value.

5§05.5 — a cubic derivative, factored and then counted●●●●○

Half of this question is a count and half is a comparison, and they are marked separately. The count is the half that is lost: a candidate list with no factorisation behind it earns nothing, and the sentence that closes the list is the one about the leftover quadratic. Budget ten minutes.

Given

  • $f(x)=3x^{4}-2x^{2}+8x$

  • the closed interval $[-2,1]$

  • $f$ is a polynomial, so it is continuous on $[-2,1]$ and differentiable everywhere

Find

  1. (a) How many critical numbers does $f$ have? Show the factorisation that proves your list is complete.

  2. (b) Find the absolute maximum and the absolute minimum of $f$ on $[-2,1]$, and say where each one occurs.

Hint 1/4

Both parts run off one object: the derivative in factored form. Part (a) counts its real zeros, part (b) compares three heights of $f$. Neither part needs a graph, and neither needs the quartic itself to be factored.

Hint 2/4

Rational root theorem: for integer coefficients, a rational root $p/q$ in lowest terms has $p$ dividing the constant term and $q$ dividing the leading coefficient. And a quadratic with $b^{2}-4ac<0$ has no real zero at all, so it contributes no candidate.

Hint 3/4

Differentiating $f(x)=3x^{4}-2x^{2}+8x$ gives $f'(x)=12x^{3}-4x+8=4(3x^{3}-x+2)$. For the bracket, $p$ divides $2$ and $q$ divides $3$, so the candidates are $\pm1,\ \pm2,\ \pm\tfrac13,\ \pm\tfrac23$. The interval, again, is $[-2,1]$.

Hint 4/4

There is exactly one critical number, $x=-1$, and on $[-2,1]$ the maximum is $24$ at $x=-2$ while the minimum is $-7$ at $x=-1$.

Show solution

The interval is closed and $f$ is a polynomial, so the Closed Interval Method applies and all the work sits in its second step. Nothing opens this cubic by inspection: there is no common variable factor, no grouping, and the quadratic formula does not apply to a cubic. The rational root theorem is the only entry point, and one hit drops the degree to two.

Differentiate, then strip the constant factor
$$f'(x)=12x^{3}-4x+8$$

term by term; the absent $x^{2}$ term is a feature of this $f$, not a simplification we performed, and it has to be carried as a $0$ later

$$f'(x)=4\bigl(3x^{3}-x+2\bigr)$$

a nonzero constant cannot make a product vanish, so the critical numbers are the roots of the bracket alone; the smaller end coefficients also shorten the candidate list about to be written

Write the candidate list, then stop at the first hit
$$\tfrac{p}{q}\ \text{with}\ p\mid 2,\ q\mid 3:\quad \pm1,\ \pm2,\ \pm\tfrac13,\ \pm\tfrac23$$

the rational root theorem applied to the bracket; eight numbers, and nothing outside the list can be a rational root, which is what makes the search finite instead of hopeful

$$3(1)^{3}-1+2=4\neq 0$$

the cheapest test first, and it misses; each test costs one line, so cheap ones go first

$$3(-1)^{3}-(-1)+2=-3+1+2=0$$

$x=-1$ is a root, and one root is all the division needs; testing the remaining six candidates would buy nothing that the division does not hand over

Divide, then settle the leftover quadratic out loud
$$3x^{3}-x+2=(x+1)\bigl(3x^{2}-3x+2\bigr)$$

synthetic division on the coefficient row $3,\ 0,\ -1,\ 2$ with $-1$: bring down $3$, then $0-3=-3$, then $-1+3=2$, then $2-2=0$. The $0$ for the missing $x^{2}$ term is what keeps the columns aligned

$$b^{2}-4ac=(-3)^{2}-4(3)(2)=-15<0$$

no real zero from this factor, so the list of critical numbers closes at one entry; a factor left unchecked is an assumption rather than a result, and this is the sentence that turns the count into an answer

$$3x^{2}-3x+2=3\left(x-\tfrac12\right)^{2}+\tfrac54>0$$

completing the square says more than the discriminant does: the factor is not merely nonzero but positive everywhere, so it cannot flip the sign of $f'$ either

$$\text{critical numbers of}\ f:\ x=-1\ \text{only}$$

$f'$ is a polynomial, so it exists at every real number and the fails to exist half of the definition contributes nothing here

Evaluate f at the candidate and at both endpoints
$$f(-2)=48-8-16=24$$

left endpoint; endpoints are candidates even though no calculus points at them

$$f(-1)=3-2-8=-7$$

the only interior candidate, and it lies inside $(-2,1)$, so it is kept

$$f(1)=3-2+8=9$$

right endpoint; it wins neither extreme, which is a result rather than an omission

$$\max\{24,-7,9\}=24,\qquad \min\{24,-7,9\}=-7$$

values of $f$, never of $f'$; the calculus ended when the candidate list closed

Answer $$\boxed{\text{one critical number}\ x=-1;\quad \max=24\ \text{at}\ x=-2,\quad \min=-7\ \text{at}\ x=-1}$$
Check

Independent check from expansion rather than from the division. Multiplying out gives $(x+1)(3x^{2}-3x+2)=3x^{3}-3x^{2}+2x+3x^{2}-3x+2=3x^{3}-x+2$, the bracket we began with, so no root was invented and none was lost. As a separate guard on part (b), two values not used in the search: $f(0)=0$ and $f\left(-\tfrac32\right)=-\tfrac{21}{16}$ both sit above $-7$ and below $24$, which is what a single interior minimum and an endpoint maximum have to look like.

Part (a) is not a warm up for part (b); on a marking scheme it is usually the larger half. A count of critical numbers is worth marks only when the factorisation sits underneath it, and the line that closes the count is the one about the leftover quadratic, not the one that finds the root.

Mistake ledger (16 entries)
⚠ Dropping the height and keeping only the slope

the slope feels like the calculus part, so $f(a)$ gets treated as scenery

wrong$$L(x)=f'(a)(x-a)$$
right$$L(x)=f(a)+f'(a)(x-a)$$
⚠ Leaving the derivative as a function of x

the formula is copied before $f'$ has been evaluated anywhere

wrong$$L(x)=f(a)+f'(x)(x-a)$$
right$$L(x)=f(a)+f'(a)(x-a)$$
⚠ Reporting the quantity when the question asks for its change

the formula for $V$ is more familiar than the formula for $dV$

wrong$$dV=s^{3}=1000\ \text{cm}^{3}$$
right$$dV=3s^{2}\,ds=15\ \text{cm}^{3}$$
⚠ Substituting the numbers before differentiating

the same haste that flattens a related rates problem into a constant

wrong$$V=10^{3}=1000\ \Rightarrow\ dV=0$$
right$$dV=3s^{2}\,ds\ \Rightarrow\ dV=15$$
⚠ Answering with the location when the value was asked for

the search produces $c$ last, so $c$ is the number still in your hand

wrong$$\max_{[-2,5]}f=5$$
right$$\max_{[-2,5]}f=9\ \text{at}\ x=5$$
⚠ Assuming the absolute maximum must be one of the local maxima

the interior of the graph is where the interesting shape is, so the ends stop being looked at

wrong$$\max=\max\{\text{local maxima}\}$$
right$$\max=\max\{\text{local maxima},\,f(a),\,f(b)\}$$
⚠ Reading the theorem backwards

an implication read as an equivalence, which is easy to do at speed

wrong$$f\ \text{not continuous}\ \Rightarrow\ \text{no maximum}$$
right$$f\ \text{not continuous}\ \Rightarrow\ \text{the theorem says nothing}$$
⚠ Applying the theorem on an interval that is not closed

the bracket at the end of the interval is small and gets copied wrongly

wrong$$\max_{(0,1]}\frac1x\ \text{exists}$$
right$$\frac1x\ \text{has no maximum on}\ (0,1]$$
⚠ Calling a point outside the domain critical

the derivative misbehaves there, and the domain check gets skipped

wrong$$f(x)=\tfrac1x:\ x=0\ \text{is a critical number}$$
right$$0\notin\operatorname{dom}(f)\ \Rightarrow\ \text{no critical numbers}$$
⚠ Solving f′ = 0 and stopping

the equation is the visible half of the definition, the corner case is the silent half

wrong$$\text{critical numbers}=\{x:f'(x)=0\}$$
right$$\text{critical numbers}=\{x\in\operatorname{dom}f:\ f'(x)=0\ \text{or}\ f'(x)\ \text{undefined}\}$$
⚠ Never evaluating the endpoints

the endpoints take no calculus to find, so they do not feel like part of the work

wrong$$\max_{[-1,5]}f=\max\{f(0),f(4)\}=5$$
right$$\max_{[-1,5]}f=\max\{f(-1),f(0),f(4),f(5)\}=5$$
⚠ Comparing values of f′ instead of values of f

the derivative is the object you spent the effort on, so it stays in hand

wrong$$\text{compare } f'(c_1),f'(c_2),\dots$$
right$$\text{compare } f(c_1),f(c_2),f(a),f(b)$$
⚠ Accepting a c outside the open interval

the equation for $c$ usually has two roots and only one of them is admissible

wrong$$c=\pm\tfrac{2}{\sqrt3}\ \text{both accepted on}\ (0,2)$$
right$$c=\tfrac{2}{\sqrt3}\in(0,2)\ \text{only}$$
⚠ Using the theorem where the function has a corner

continuity is checked, differentiability is assumed to come with it

wrong$$f(x)=\lvert x\rvert\ \text{on}\ [-1,1]:\ \exists c,\ f'(c)=0$$
right$$f'(0)\ \text{does not exist, so the theorem does not apply}$$
⚠ Treating an unsolvable f′ = 0 as an f′ with no zeros
wrong$$x\tan x=1\ \text{cannot be solved}\ \Rightarrow\ \text{no critical number}$$
right$$g(0)<1<g(1)\ \text{with}\ g\ \text{increasing}\ \Rightarrow\ \text{exactly one}\ c$$
⚠ Asserting the sign of a leftover quadratic instead of testing it
wrong$$f'=6(x-2)\bigl(2x^{2}+2x+3\bigr)\ \Rightarrow\ \text{only}\ x=2\ \text{matters}$$
right$$b^{2}-4ac=-20<0,\ a>0\ \Rightarrow\ 2x^{2}+2x+3>0\ \text{everywhere}$$
Formula card
Linearization at a
$$L(x)=f(a)+f'(a)(x-a)$$

$f$ differentiable at $a$; the estimate degrades as $\lvert x-a\rvert$ grows

Differential
$$dy=f'(x)\,dx$$

$f$ differentiable at $x$; $dx$ is a chosen finite step

Absolute and local extrema
$$f(c)\ge f(x)\ \text{on the domain, or only near }c$$

$c$ is in the domain; local compares only on an open interval around $c$

Extreme Value Theorem
$$f\ \text{continuous on}\ [a,b]\ \Rightarrow\ \max\ \text{and}\ \min\ \text{are attained}$$

continuity at every point, and a closed bounded interval

Fermat's theorem
$$\text{local extremum at}\ c\ \text{and}\ f'(c)\ \text{exists}\ \Rightarrow\ f'(c)=0$$

$c$ interior to the domain; the converse is false

Rational root theorem
$$\tfrac{p}{q}\ \text{a root of}\ a_{n}x^{n}+\dots+a_{0}\ \Rightarrow\ p\mid a_{0},\ q\mid a_{n}$$

integer coefficients, and $p/q$ written in lowest terms

Closed Interval Method
$$\max_{[a,b]}f=\max\{f(c_1),\dots,f(c_k),f(a),f(b)\}$$

$f$ continuous on $[a,b]$; the $c_i$ are the critical numbers inside $(a,b)$

Sign of a quadratic with no real root
$$b^{2}-4ac<0\ \Rightarrow\ ax^{2}+bx+c\ \text{keeps the sign of}\ a\ \text{everywhere}$$

$a\neq 0$ and the discriminant is strictly negative

Mean Value Theorem
$$f'(c)=\frac{f(b)-f(a)}{b-a}\ \text{for some}\ c\in(a,b)$$

continuous on $[a,b]$, differentiable on $(a,b)$

Rolle's theorem
$$f(a)=f(b)\ \Longrightarrow\ f'(c)=0\ \text{for some }c\in(a,b)$$

Valid when $f$ is continuous on $[a,b]$, differentiable on $(a,b)$, and the two endpoint values agree. Use it whenever two points of equal height have to produce a flat tangent between them: capping how many roots a function can have, or forcing a zero of $f'$ out of nothing. It is the Mean Value Theorem with $f(a)=f(b)$, and the course solutions call this step MVT, so either name is accepted as long as the three hypotheses are checked.

Zero counting bound (Rolle applied between roots)
$$\#\{\text{distinct zeros of }f\ \text{in }I\}\ \le\ \#\{\text{zeros of }f'\ \text{in }I\}+1$$

Valid when $f$ is differentiable on the interval $I$. This is the ceiling half of every exact root count: between any two distinct roots of $f$ lies a zero of $f'$, so $k+1$ roots would force $k$ zeros of the derivative. Use it after you have controlled the zeros of $f'$, and never on its own, since it only ever gives an upper bound.

Sign of the derivative controls monotonicity (MVT corollary)
$$f'\gt 0\ \text{on }(a,b)\ \Longrightarrow\ f\ \text{strictly increasing on }[a,b];\qquad f'\equiv 0\ \text{on }(a,b)\ \Longrightarrow\ f\ \text{constant on }[a,b]$$

Valid when $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$; the reversed inequality gives strictly decreasing. If $f'\ge 0$ and vanishes only at isolated points, $f$ is still strictly increasing, and saying so is worth a mark on its own. Use it to turn a sign you can read off a factored $f'$ into a statement about $f$, which is how the papers get from the derivative to an inequality, to a count of zeros, or to $f'$ itself being monotone via $f''$.

Intermediate Value Theorem
$$f\ \text{continuous on }[a,b],\ N\ \text{between }f(a)\ \text{and }f(b)\ \Longrightarrow\ f(c)=N\ \text{for some }c\in(a,b)$$

Valid when $f$ is continuous on the closed interval; the usual case is $N=0$ with $f(a)$ and $f(b)$ of opposite signs. Use it for the floor half of a root count, and for pinning down that a monotone $f'$ really does vanish once. It produces existence only, never a count, so an argument that stops here can only support the words at least.

Direction of the linear approximation error
$$f''\ge 0\ \text{between }a\ \text{and }x\ \Longrightarrow\ f(x)\ge L(x);\qquad f''\le 0\ \Longrightarrow\ f(x)\le L(x)$$

Valid when $f$ is twice differentiable on the interval joining $a$ to $x$. Concave up puts the tangent line below the curve, so $L$ underestimates; concave down puts it above, so $L$ overestimates. Use it whenever a question asks which way the estimate errs, or which of two equal steps to the left and right of $a$ is the better estimate: at equal distance from $a$, the side on which $\lvert f''\rvert$ is smaller carries the smaller error. The marks sit in this reason, not in the decimal.

Darboux's theorem (derivatives have the intermediate value property)
$$f\ \text{differentiable on }[a,b],\ N\ \text{between }f'(a)\ \text{and }f'(b)\ \Longrightarrow\ f'(c)=N\ \text{for some }c\in[a,b]$$

Valid when $f'$ exists at every point of $[a,b]$; $f'$ is not assumed continuous, and that is the whole point, since a derivative can be discontinuous and so the Intermediate Value Theorem is not available for $f'$. Use it to decide whether a given piecewise function can be somebody's derivative (a function that skips values cannot be one), and on the true or false item claiming that opposite signs of $f'$ at two points force a zero of $f'$ between them, which is true but for this reason and not by continuity.

Check yourself

Close the page and write, from memory: the formula for the tangent line used as an estimator, the difference between $dy$ and $\Delta y$ in one sentence, the two hypotheses of the Extreme Value Theorem, the definition of a critical number in both of its halves, the four steps of the closed interval search, and the statement of the Mean Value Theorem. Then compare with the formula card, and mark only the ones you could not produce.

  • Estimate $\sqrt{102}$ by hand, and say whether your estimate is high or low?

    c-linearization

  • Say what $dy$ and $\Delta y$ mean for one specific function and one specific step, without using the word approximately?

    c-differentials

  • Draw a function on $[0,4]$ whose absolute maximum is at an endpoint while its only local maximum is somewhere inside?

    c-extreme-values

  • Name a continuous function with no absolute maximum, and say which hypothesis it breaks?

    c-evt

  • Give one function whose critical number comes from $f'=0$ and one whose critical number comes from $f'$ failing to exist?

    c-critical-numbers

  • Run the four steps on $f(x)=x^{3}-3x$ over $[0,2]$ without looking anything up?

    c-closed-interval

  • Use a cap on $f'$ and one known value of $f$ to bound $f$ at another point?

    c-mvt

  • Find every critical number of a quartic whose derivative is a cubic that will not factor by inspection, and say why your list is complete?

    c-critical-numbers

  • Turn $2\le f'\le 5$ and one known value of $f$ into upper and lower bounds for $f$ at a point on each side of it?

    c-mvt

Glossary (14 terms)
linearizationdoğrusallaştırma

The tangent line $L(x)=f(a)+f'(a)(x-a)$ used as a formula for approximate values of $f$ near the base point $a$.

differentialdiferansiyel

The quantity $dy=f'(x)\,dx$: the change in output predicted by the tangent line over a chosen input step $dx$.

absolute maximummutlak maksimum

A value $f(c)$ that is greater than or equal to every value $f$ takes on its domain.

local maximumyerel maksimum

A value $f(c)$ that is greater than or equal to the values of $f$ on some open interval around $c$, with no claim about the rest of the domain.

extremumuç değer

A maximum or a minimum, of either the absolute or the local kind.

critical numberkritik nokta

A point $c$ of the domain of $f$ at which $f'(c)=0$ or $f'(c)$ fails to exist.

Fermat's theorem

If $f$ has a local extremum at $c$ and $f'(c)$ exists, then $f'(c)=0$. The converse is false.

Extreme Value Theoremuç değer teoremi

A function continuous on a closed bounded interval attains an absolute maximum and an absolute minimum on that interval.

Closed Interval Methodkapalı aralık yöntemi

The search that compares the values of $f$ at every critical number inside $(a,b)$ and at both endpoints, and reports the largest and smallest.

Rolle's theorem

If $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c$ strictly between $a$ and $b$.

Mean Value Theoremortalama değer teoremi

If $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$, some $c$ in $(a,b)$ has $f'(c)$ equal to the average rate of change over $[a,b]$.

relative errorbağıl hata

An uncertainty expressed as a fraction of the quantity itself, such as $dV/V$, which carries no units and transfers between problems.

base pointdayanak nokta

The input $a$ at which a linearization is built; every number in $L$ depends on it.

critical pointkritik nokta

The same object as a critical number: a point $c$ in the domain of $f$ where $f'(c)=0$ or $f'(c)$ fails to exist. Exam papers use the two names interchangeably, and a count of critical points counts $x$ values, not pairs.

What comes next
§06 · What derivatives say about the shape of a graph, limits at infinity, and curve sketching

This section could tell you that a critical number is a suspect but never which kind. Next comes the machinery that classifies it: what the sign of $f'$ says about rising and falling, what $f''$ says about bending, and how the two together let you draw a graph you have never seen.

Sources
  • James Stewart, Calculus, Metric Version, ninth edition — sections 2.9, 3.1 and 3.2 The three sections named on this week's syllabus line; statements and hypotheses follow this book's numbering and conventions.
  • Standard single variable calculus practice Problem types here follow the usual repertoire for this material; numbers and wording are our own.

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