6 concepts17 worked examples28 exercises3 exam-level6 figures
What are you here for?
07Curve sketching, optimization, and Newton's method
Thirteen metres of fencing and a straight wall to lean the pen against. Three metres out from the wall encloses $21\ \mathrm{m^{2}}$ of grass, three and a half metres encloses $21$ again, four metres encloses $20$. The best pen is hiding somewhere inside that half metre, and no amount of further guessing will ever say exactly where.
By the end of this section you can turn a sentence like that into one function of one variable, produce the exact best value, and prove it is the best rather than merely a candidate; you can also take a function you have never seen and draw its whole graph from two sign lines and its two ends; and when the equation you have to solve has no formula for its roots, you can chase a root to six decimal places with four lines of arithmetic.
In 60 seconds
Three jobs that all run on the same two derivatives: draw the graph from a fixed checklist, turn a word problem into one variable and close the argument that the critical point really is the best, and when the equation refuses to be solved, replace the curve by its tangent and iterate.
an equation with no closed form for its roots, and a decimal answer is acceptable
Three most common mistakes
Reporting the critical number as the answer. The question asked for the largest volume, the cheapest cost or the shortest time; $x=10/3$ is where, not how much, and the two carry different units.
Sketching before the domain and the asymptotes are known. Sample points cannot see a break: six values of $x^{2}/(x-1)$ join into an innocent rising curve that walks straight through $x=1$, where the real graph splits and runs to both infinities.
Stopping Newton's method at the first agreement of two decimals. Agreement of $x_{n}$ and $x_{n+1}$ is evidence that the iteration has settled, not proof that it settled on a root; evaluating $f$ at the last iterate costs one line and settles it.
Midterm 1, Midterm 2 and the Final carry 28 percent each, quizzes 10 percent and homework 6 percent. Optimization answers are graded as arguments: the sentence that rules out the endpoints, or that says why the single critical point must be the absolute one, is a scored line and not scratch work.
How much time do you have?
10 minutes
You leave able to reduce an optimization problem to one variable and finish it, which is the single most repeated task of the week.
In 60 seconds card, Turning a sentence into one function of one variable, Proving it is the best and not merely a candidate, Formula card
45 minutes
Add the sketching checklist, the slanted line a rational graph settles onto, and one full run of Newton's method: enough for a complete question of any of the three types.
everything in the 10 minute path, The picture on the screen is not the graph, The slanted line the graph settles onto, Replacing the curve by its tangent, Full exam style question, Practice C
full read
The failure modes of Newton's method, the scaffolded ladder, and the interleaved set where the type of the question is hidden. This is the part that transfers to a question you have not seen before.
all blocks in order, When the tangent leads you away, Scaffolding comes off, Practice A to D, Mistake ledger
By the end of this section
Check a plotted or sampled picture against the formula — domain gaps, the two sign lines, the limits at the ends — and assemble the result into one finished sketch.
Divide out a rational function to expose its slant asymptote, and decide from the remainder whether the graph ever meets that line.
Translate a worded maximum or minimum problem into one of one variable on a stated interval, before differentiating anything.
Justify that a critical point is the absolute maximum or minimum on the interval at hand, using the endpoints or the sign of the derivative rather than assumption.
Run Newton's iteration from a stated starting guess and report a root to a stated number of decimal places, with a check that it is a root.
Diagnose a failed run of Newton's method from its iterates alone and choose a starting guess that repairs it.
Syllabus coverage
3.6
Graphing with calculus and technology
What a drawn or plotted picture does not certify: it cannot see a number missing from the domain, it puts a turning point or a bend near rather than at the right value, and it shows nothing outside its own window — the slanted line of the next block is exactly such a feature. The order of the sketch itself belongs to the previous section and is recalled here, not rebuilt.
covered
3.7
Optimization problems
Translation to one variable, then the argument that closes it: the Closed Interval Method where the interval is closed, the sign of the derivative where it is not.
covered
3.9
Antiderivatives
The third number on the week's line is the antiderivative material: the differentiation table read backwards, and the constant that comes with it. It opens the next section, and nothing in this one uses it.
deferred
Newton's method
Newton's method: the tangent line iteration, its speed, and how it fails
Carried by the title of the week rather than by a number of its own, and taught here in full: the iteration, the geometry it comes from, the speed it converges at, and the four ways a run fails. Examinable like the numbered material around it.
covered
Bisection and the secant method
Other root finders that need no derivative
Named once, in the block on failure modes, only because they are what you fall back on when the derivative at your guess is zero. Not examinable this term: no question this week asks you to run either of them.
off_syllabus
Recall first
Critical number
A number $c$ in the domain of $f$ with $f'(c)=0$ or with $f'(c)$ undefined.
Every optimization in this section ends by solving $f'=0$, and every wrong answer that survives to the end started by admitting a number that was not in the domain.
First Derivative Test
At a critical number $c$: $f'$ going $+\to-$ gives a local maximum, $-\to+$ gives a local minimum, and no sign change gives neither.
It is the only classification tool that still works when the interval is open, which is the usual situation in a worded problem.
Closed Interval Method
A continuous $f$ on a closed bounded interval $[a,b]$ attains an absolute maximum and minimum; compare the values of $f$ at the critical numbers inside with $f(a)$ and $f(b)$.
Where the physical interval really is closed, this is the shortest complete argument available, and it is the one that catches answers sitting at an endpoint.
Concavity and the sign of the second derivative
$f''>0$ on an interval means the graph bends upward there, $f''\lt 0$ means it bends downward, and an inflection point is where that sign changes.
Step six of the sketching checklist, and the one line version of the classification test at a critical number where $f''$ is cheap.
Limits at infinity
$\lim_{x\to\pm\infty}f(x)=L$ makes $y=L$ a horizontal asymptote; for a rational function, divide numerator and denominator by the highest power of $x$ in the denominator.
Step four of the checklist. It is also what proves that a remainder term dies and a slant asymptote is real.
Equation of the tangent line
The tangent to $y=f(x)$ at $x=a$ is $$y=f(a)+f'(a)(x-a).$$
Newton's method is nothing but this line, solved for its own $x$ intercept. The formula is derived from it in two lines rather than memorised.
Intermediate Value Theorem
If $f$ is continuous on $[a,b]$ and $f(a)$ and $f(b)$ have opposite signs, then $f(c)=0$ for some $c$ between $a$ and $b$.
It is how you find a starting guess for Newton's method that is honestly near a root, instead of guessing and hoping.
Try it yourself first (3 questions)
1§07.0 — one critical number is not one answer●●○○○
Nothing here is graded and nothing here is new; each item is a tool the section leans on. A classmate is minimising a cost function on the interval of all positive numbers, finds exactly one critical number at $c=4$, and writes "so the cheapest design has $x=4$".
Given
$C$ is continuous on $(0,\infty)$
$C'(4)=0$, and $C'(x)\ne 0$ for every other $x>0$
Find
Does $C(4)$ have to be the smallest value of $C$ on that interval?
Hint 1/4
Do not compute. Ask what a critical number is a candidate for, and how many kinds of candidate there are.
Hint 2/4
A critical number can be a local maximum, a local minimum, or neither; only the sign of $C'$ on the two sides decides which.
Hint 3/4
Given again: $C$ is continuous on $(0,\infty)$, $C'(4)=0$, no other critical number. Nothing in that data has told you the sign of $C'$ anywhere.
Hint 4/4
It need not be the minimum: the same data fits a function with a maximum at $4$, or with a flat moment and no turn at all.
Show solutionBuild one function of each kind
$$C(x)=(x-4)^{2}$$
critical at 4, and here the value really is the minimum
$$C(x)=-(x-4)^{2}$$
same critical number, and now 4 is the maximum
$$C(x)=(x-4)^{3}$$
same critical number again, and now it is neither
Read off what the data actually decides
$$C'(4)=0 \Rightarrow \text{horizontal tangent at } 4$$
this is the literal content of the hypothesis
$$\text{sign of } C' \text{ on each side} \Rightarrow \text{the verdict}$$
and no sign of $C'$ was given anywhere
Answer $$\text{No: uniqueness of the critical number decides nothing.}$$
Check
Three functions with identical hypotheses give three different answers, so the hypotheses cannot force one of them.
This is the whole content of one of this section's blocks: finding the candidate is the easy half, and the sentence that convicts it is the graded half.
2§07.0 — the tangent line, written down●○○○○
Newton's method is this formula and nothing else, so it is worth thirty seconds now. A function is known only at one point, together with its slope there.
Given
$f(2)=5$
$f'(2)=-3$
Find
Which equation is the tangent line to $y=f(x)$ at $x=2$?
Hint 1/4
Ask which number is the height and which is the slope; the point is $(2,5)$ and the slope is $-3$.
Hint 2/4
The point slope form of a line through $(a,f(a))$ with slope $f'(a)$ is $y=f(a)+f'(a)(x-a)$.
Hint 3/4
Given again: $f(2)=5$ and $f'(2)=-3$, so $a=2$, $f(a)=5$, $f'(a)=-3$ go into that form in that order.
Hint 4/4
The tangent line is $y=5-3(x-2)$.
Show solutionPut the numbers in the right slots
$$y=f(a)+f'(a)(x-a)$$
point slope form: the constant is a height, the coefficient is a slope
$$y=5+(-3)(x-2)=5-3(x-2)$$
with $a=2$
Answer $$y=5-3(x-2)$$
Check
Check the point: at $x=2$ the right side is $5$, which is $f(2)$, and the coefficient of $x$ is $-3$, which is $f'(2)$.
Newton's method solves exactly this equation for $y=0$, so the two lines above are the entire derivation.
3§07.0 — how far out a rational graph can settle●●●○○
One claim about the two ends of a rational graph. Decide, and be ready with either a reason or a counterexample, since that is what a quiz pays for.
Given
Claim: "A rational function can have a horizontal asymptote at one end and a slant asymptote at the other."
Find
True or false, with justification.
Hint 1/4
Ask what fixes the behaviour at each end: the two ends of a rational function are governed by the same quotient.
Hint 2/4
For a rational function the end behaviour is read from the degrees: numerator degree below or equal to the denominator gives a horizontal asymptote, exactly one above gives a slant one. The degrees do not change between the two ends.
Hint 3/4
Given again: the claim asks for both at once. Since the same degree comparison applies at $+\infty$ and at $-\infty$, the same verdict comes out at both ends.
Hint 4/4
The claim is false for rational functions: whichever case the degrees put you in, they put you in it at both ends.
polynomial division is an identity in $x$, valid for every $x$ in the domain
$$\frac{R(x)}{D(x)}\to 0 \text{ as } x\to+\infty \text{ and as } x\to-\infty$$
the degree gap forces the remainder term to die at both ends
Read the two cases off the quotient
$$\deg q=0 \Rightarrow y=q \text{ horizontal at both ends}$$
a constant quotient is the same constant in both directions
$$\deg q=1 \Rightarrow y=q(x) \text{ slant at both ends}$$
one line, followed out to both infinities
Answer $$\text{False: one division decides both ends at once.}$$
Check
Test it on $f(x)=x^{2}/(x-1)$: dividing gives $x+1+1/(x-1)$, and $y=x+1$ is the asymptote at $+\infty$ and at $-\infty$ alike.
This is why the checklist asks for asymptotes once, not twice.
Notation
symbol
reads as
means
watch out
$x_{n}$
x sub n
the $n$-th guess produced by Newton's iteration, counting the as $x_{0}$
It is a whole sequence of numbers, not one unknown. $x_{3}$ is the fourth number written down, not the third.
$x^{*}$
x star
the exact root being chased, the number the iterates are trying to reach
You almost never have it. Statements about $|x_{n}-x^{*}|$ are about how the method behaves, not quantities you can compute during a run.
$Q(x)$
Q of x
the objective function after the constraint has been used to remove every variable but one
Before the substitution $Q$ has two variables and cannot be differentiated as it stands. The single letter hides which variable survived, so say so in words.
$0\lt x\lt 8$
x is strictly between 0 and 8
the interval on which the objective actually describes a physical object
This interval comes from the picture, not from the algebra. The formula usually stays perfectly happy at values that build a box with negative sides.
$y=mx+b$
y equals m x plus b
a slant asymptote, the line the graph settles onto as $x\to\pm\infty$
The word "asymptote" says the vertical gap tends to zero, not that the gap is never zero. The curve may cross this line at a finite $x$.
$\approx$
is approximately equal to
the left side has been rounded; the two sides are not the same number
Once $\approx$ appears in a chain it may not turn back into $=$. Exact answers such as $2ab$ or $\sqrt[3]{500/\pi}$ stay exact until the last line.
Conventions used here
Where against how much
An optimization answer names both: the value of the variable that achieves the optimum, and the value of the objective there. "The volume is largest when $x=10/3\ \mathrm{cm}$, and that largest volume is $19600/27\approx 725.93\ \mathrm{cm^{3}}$." One without the other is half an answer, and the two have different units.
The most common lost mark in this material is a correct critical number handed in as if it were the maximum.
Closing an optimization argument
Every optimization answer here ends with one sentence naming the reason the critical point wins: comparison with the endpoints on a closed interval, or the sign of $f'$ on each side, or the behaviour of the objective at the open ends of the interval.
A critical number is a candidate. Without that sentence nothing has been proved, and on a graded script nothing has been earned.
Naming an asymptote
An asymptote is named as an equation of a line: $x=1$, $y=0$, $y=x+1$. Never as a number and never as the expression it came from. A graph is allowed to cross a horizontal or slant asymptote; it can never cross a vertical one, because there is no point of the graph there at all.
"The asymptote is $1$" does not say which of the two lines through $1$ is meant, and the two are opposite claims about the picture.
Reporting an iterate
Newton iterates are carried at full calculator precision and quoted to ten decimal places; the final root is rounded once, at the end, to the number of places the question asks for. Rounding an intermediate iterate and feeding it back in destroys the digits the next step was about to produce.
The whole point of the method is that the digit count doubles per step, and rounding early throws away exactly the digits being doubled.
7.1The picture on the screen is not the graph
Use before every sketch: domain, both sign lines and the end limits supply what a sampled plot silently leaves out.
The previous section built the order that produces a sketch; this one starts from a picture and asks which of its features are true.
Solvable with what we have
Build the sign line of $f'$ and classify each critical number.
Build the sign line of $f''$ and locate the inflection points.
Evaluate $\lim_{x\to\pm\infty}f(x)$ for a rational expression.
Plot points by arithmetic.
Not solvable yet
Say how many separate pieces the graph of $\dfrac{x^{2}}{x-1}$ comes in.
Say which side of the line $y=x+1$ each piece runs along far out.
Take $f(x)=\dfrac{x^{2}}{x-1}$, evaluate at $-2,-1,0,2,3,4$ to get $-\tfrac{4}{3},\,-\tfrac{1}{2},\,0,\,4,\,4.5,\,\tfrac{16}{3}$, and join the dots with a smooth curve.
Why it fails
The six values climb steadily, so the curve climbs steadily and walks straight across $x=1$. The real graph is not there at all: $f(0.9)=-8.1$ and $f(1.1)=12.1$, so between two samples the function falls to $-\infty$ and returns from $+\infty$. No finite set of samples can rule that out, which is why the checklist starts with the domain.
MethodWhat calculus adds to a plotted picture
Conditions
a picture of $y=f(x)$ is already in front of you, from a device or from sampled points
the formula for $f$ is available, so the domain, $f'$, $f''$ and the limits can be computed
The same five ingredients are also assembled in the opposite direction, with no picture in front of you and sometimes with no formula either. When the asymptotes, $f'$, $f''$ and a handful of values are supplied instead of $f$, list every number at which $f'$ or $f''$ can change sign, use those numbers as the columns of a table whose rows are $x$, $f'(x)$, $f''(x)$, $f(x)$, and draw the graph from the completed table; a further part may then ask for the constants of a given family of functions that fits the same data. One extra condition governs that direction: a change of concavity across a number missing from the domain is not an inflection point, since an inflection point has to be a point of the graph.
A plot is a finite list of samples joined up. It cannot see a number missing from the domain, it places a turning point near rather than at the right value, and it says nothing about what lies outside its window. The domain, the two sign lines and the limits supply those pieces.
Six correct sample values of $x^{2}/(x-1)$, joined the way anyone would join them, against what the function actually does. Every sample is right; the curve through them is wrong near $x=1$.
Looks like this, but is not
It looks as though the sign line of $f'$ fixes the picture: for $f(x)=\dfrac{x^{2}}{x-1}$ the function is decreasing on $(0,1)$ and decreasing on $(1,2)$, the same verdict on both pieces.
The two pieces look nothing alike. On $(0,1)$ the graph starts at the origin and dives to $-\infty$; on $(1,2)$ it comes down out of $+\infty$ and lands at $(2,4)$. Identical sign, opposite pictures, because of the gap between them.
What a plot of x²/(x²+3) leaves out
A plotted picture of this function shows a smooth rise toward a ceiling. Everything below is what the picture cannot certify.
Given
$f(x)=\dfrac{x^{2}}{x^{2}+3}$
Find
a complete labelled sketch
SolutionDomain, intercepts, symmetry
$$x^{2}+3>0 \text{ for every } x \Rightarrow \text{domain} = \mathbb{R}$$
the denominator never vanishes, so there is no gap and no vertical asymptote to look for
$$f(0)=0$$
the only intercept: $f(x)=0$ forces $x^{2}=0$
$$f(-x)=\frac{(-x)^{2}}{(-x)^{2}+3}=f(x)$$
even, so everything computed for $x\ge 0$ is mirrored and half the work disappears
Sanity check the asymptote from the formula itself: $f(x)=1-\dfrac{3}{x^{2}+3}$, which is visibly below $1$ for every $x$ and rises to $1$ as $|x|$ grows. The rewritten form also re-derives the minimum at $x=0$ in one line, since $3/(x^{2}+3)$ is largest there.
Two quotient rules and one limit. Nothing here needs a device, and nothing a device draws gives the exact numbers.
A picture never gives the ceiling or the two bends exactly; those are the numbers worth writing down.
The full sketch of (x³ − 3x + 2)/x³
Sketch $y=\dfrac{x^{3}-3x+2}{x^{3}}$. Give the domain, every intercept, every asymptote, the intervals of increase and decrease, the local extreme values, the concavity with any inflection point, and say whether the curve ever meets its horizontal asymptote.
Solution
The two earlier sketches had a single power on top, so $f(x)=0$ collapsed to $x=0$; here it is a cubic equation whose repeated root is also the local minimum.
Domain, and the intercepts the numerator hands over
$$D=\{x\in\mathbb{R}:x\ne 0\}$$
only the denominator can fail, and it fails once, so the picture arrives in two branches
$$f(x)=0\iff x^{3}-3x+2=0\quad (x\ne 0)$$
a quotient is zero exactly where its numerator is, and multiplying $f=0$ by $x^{3}$ is legal because $x=0$ is already excluded
$$x^{3}-3x+2=(x-1)(x^{2}+x-2)=(x-1)^{2}(x+2)$$
$x=1$ kills the cubic by inspection, so $(x-1)$ divides out and the quotient factors again
the sign changes at $\tfrac43$, which is in the domain, so it is a genuine inflection point
$$\text{IP}=\Bigl(\tfrac43,\tfrac{5}{32}\Bigr)$$
$x=0$ is not one, because there is no point of the graph there
Answer $$\boxed{\begin{aligned} &\text{VA }x=0,\quad \text{HA }y=1\ \text{(met at }x=\tfrac23),\quad \text{zeros }x=-2,\ x=1\\ &\downarrow(-\infty,0),(0,1);\quad \uparrow(1,\infty);\quad \text{local min } f(1)=0\\ &\text{concave down }(-\infty,0)\text{ and }(\tfrac43,\infty);\ \text{up }(0,\tfrac43);\ \text{IP}\bigl(\tfrac43,\tfrac{5}{32}\bigr)\end{aligned}}$$
Check
Two values taken from the original quotient, not from the table: $f(-1)=-4$ and $f(2)=\tfrac12$. The first sits on the falling left branch below the axis, the second between the local minimum $0$ and the level $1$ — where the sign lines put them.
Once the numerator is anything but a single power, $f(x)=0$ is a polynomial equation, and its multiplicities say whether the graph crosses the axis or bounces off it.
Four polynomials to order, two of which do not exist
For each description, give a polynomial of the smallest possible degree with that property, or write DOES NOT EXIST with one sentence saying what forbids it. (a) a local minimum at $x=-1$ and a local maximum at $x=3$; (b) an inflection point at $x=0$ and no critical number; (c) degree $3$ with two inflection points; (d) exactly two critical numbers, both local minima.
Solution
Every other example runs forwards, from a formula to its features. A part marked all or nothing wants the reverse, and it wants a certificate.
Turn each phrase into a condition on P, P′ or P″
$$\text{local min at }a:\ P'(a)=0\ \text{and}\ P'\ \text{goes}\ -\ \text{to}\ +$$
the equation is not the property; the sign change is the half you check against
$$\text{inflection at }a:\ P''(a)=0\ \text{and}\ P''\ \text{changes sign}$$
a nonzero linear function changes sign once, so two inflection points are out of reach
$$\Longrightarrow\ \textbf{DOES NOT EXIST}$$
degree $4$ is the first that can: $P(x)=x^{4}-6x^{2}$ has $P''=12(x-1)(x+1)$
(d) the obstruction is a theorem
$$a\lt b\ \text{both local minima};\ P\ \text{continuous on}\ [a,b]$$
assume it exists and work between the two minima, where the Extreme Value Theorem applies
$$\max_{[a,b]}P\ \text{is attained, and not at } a \text{ or } b$$
near $a$ the values are $\ge P(a)$, so a maximum at $a$ would force $P$ constant there, which no nonconstant polynomial is
$$\Longrightarrow\ \exists c\in(a,b):\ P'(c)=0$$
an interior maximum is a critical number, and it is a third one
$$\Longrightarrow\ \textbf{DOES NOT EXIST}$$
between two local minima there is always a local maximum, so two critical numbers cannot both be minima
Answer $$\boxed{\begin{aligned} \text{(a)}\ &P(x)=-x^{3}+3x^{2}+9x\ (+C) & \text{(b)}\ &P(x)=x^{3}+3x\ (+C)\\ \text{(c)}\ &\textbf{DOES NOT EXIST} & \text{(d)}\ &\textbf{DOES NOT EXIST}\end{aligned}}$$
Check
Differentiate the constructions back: $-3(x+1)(x-3)$ vanishes at $-1$ and $3$ and nowhere else, and $3x^{2}+3$ is never zero while $6x$ changes sign at the origin. For (c) and (d) the certificate is the argument, and raising the degree repairs (c) at degree $4$ but never repairs (d).
Write the reason first and the formula second. Two obstructions cover almost every impossible part: a degree count on $f'$ or $f''$, and the fact that a maximum on a closed interval has to be somewhere.
Checkpoint
§07.1 — what the picture cannot show●●○○○
Thirty seconds, no paper. Someone plotted six correct values of $x^{2}/(x-1)$ and joined them into one smooth rising curve, missing the fact that the graph splits in two.
Given
The six samples used were at $x=-2,-1,0,2,3,4$
The function is $f(x)=\dfrac{x^{2}}{x-1}$
Find
Which step of the checklist would have caught the mistake first?
Hint 1/4
Ask which single piece of information tells you a point of the $x$ axis has no graph above it at all.
Hint 2/4
Step one is the domain: the set of $x$ for which the formula produces a number. A number missing from the domain cuts the picture into separate pieces.
Hint 3/4
Given again: the samples $-2,-1,0,2,3,4$ all lie in the domain, and the formula $x^{2}/(x-1)$ fails at exactly one number, which is not among them.
Hint 4/4
The domain step: $x=1$ is missing, so the graph is in two pieces before any derivative is computed.
Show solutionTest the steps in order
$$\text{domain}: x-1=0 \text{ at } x=1 \Rightarrow 1 \notin \text{domain}$$
found at step one, before intercepts, before any derivative
$$\text{intercepts}: f(0)=0$$
true, and silent about the break
$$\operatorname{sign} f' \text{ and } \operatorname{sign} f''$$
these come later and would inherit the wrong picture anyway
Answer $$\text{Step 1, the domain.}$$
Check
Independent check: the two one sided values $f(0.9)=-8.1$ and $f(1.1)=12.1$ straddle the gap, which is exactly what a missing domain point produces.
The order is not decoration: every later step is a statement about a picture whose pieces step one has already counted, and a plot counts nothing.
⚠ Joining samples across a hole in the domain
The values on each side are perfectly good numbers and the eye wants a single curve through them.
wrong$$f(0)=0,\ f(2)=4 \Rightarrow \text{one rising curve from } (0,0) \text{ to } (2,4)$$
right$$1\notin \text{domain} \Rightarrow \text{two separate branches, } f(0.9)=-8.1,\ f(1.1)=12.1$$
⚠ Skipping the symmetry line
It is the one step that produces no equation, so it feels like it produces nothing.
wrong$$\text{six sign computations for } f(x)=\frac{x^{2}}{x^{2}+3} \text{ on all of } \mathbb{R}$$
right$$f(-x)=f(x) \Rightarrow \text{compute on } [0,\infty) \text{ and mirror}$$
7.2The slanted line the graph settles onto
Divide a top heavy rational function; the quotient line is the asymptote and the remainder says which side the graph sits on.
Step four of the checklist asked what happens at the far ends, and for $x^{2}/(x-1)$ the honest answer so far has been "it grows" — which is not a line to draw.
RuleRule 7.1: Slant asymptote by division
Conditions
$f=N/D$ is a quotient of polynomials with $\deg N=\deg D+1$; the standard instance is $P/P'$, since differentiating a polynomial drops its degree by exactly one, which is where the slope $1/n$ comes from and is what the derivative rules card states
equivalently: the division leaves a quotient of degree one
No quotient of polynomials is required. For any $f$ defined near that infinity: if $m=\lim_{x\to\infty}f(x)/x$ and $b=\lim_{x\to\infty}\bigl(f(x)-mx\bigr)$ both exist and are finite, then $y=mx+b$ is an asymptote as $x\to\infty$, and the same pair of limits must be computed again, separately, as $x\to-\infty$, where $\sqrt{x^{2}}=-x$ can deliver a different line or a horizontal one. Division of $N$ by $D$ with $\deg N=\deg D+1$ is the special case in which those two limits happen to be readable off a remainder.
$$\boxed{\ f(x)=mx+b+\frac{R(x)}{D(x)}\ \text{ with }\ \frac{R(x)}{D(x)}\xrightarrow[x\to\pm\infty]{}0 \ \Longrightarrow\ \lim_{x\to\pm\infty}\bigl[f(x)-(mx+b)\bigr]=0\ }$$
Divide out; whatever is left over dies at both ends, so far out the graph and the line $y=mx+b$ become indistinguishable — and the leftover term also tells you which side of the line the graph is on.
The finished graph of $\textcolor{#1f6feb}{x^{2}/(x-1)}$ with the two lines it lives between: the vertical asymptote $x=1$ that splits it, and the slant asymptote $\textcolor{#6f42c1}{y=x+1}$ that both branches settle onto.
Looks like this, but is not
$g(x)=\dfrac{x^{3}+1}{x-1}$ looks like the same situation: a polynomial over a polynomial, top degree bigger than bottom, so divide and read off a line.
The division gives $g(x)=x^{2}+x+1+\dfrac{2}{x-1}$. The leftover still dies, but what it leaves behind is a parabola, not a line, so there is no slant asymptote — the graph settles onto $y=x^{2}+x+1$. The condition in the rule is not "top degree bigger", it is "bigger by exactly one".
The full sketch of x²/(x−1)
The function from the opening of this section, taken through all eight steps and finished.
Given
$f(x)=\dfrac{x^{2}}{x-1}$
Find
a complete labelled sketch, with every asymptote named as an equation
SolutionDomain and intercepts
$$x-1=0 \text{ at } x=1 \Rightarrow \text{domain } (-\infty,1)\cup(1,\infty)$$
one number missing, so the picture is in two pieces
$$f(x)=0 \iff x^{2}=0 \iff x=0$$
the origin is both the $x$ intercept and the $y$ intercept
$$f(-x)=\frac{x^{2}}{-x-1}\ne \pm f(x)$$
neither even nor odd, so nothing can be mirrored and both sides must be computed
Rewrite once, and get every end for free
$$x^{2}=(x-1)(x+1)+1$$
long division of $x^{2}$ by $x-1$; check by expanding: $x^{2}-1+1=x^{2}$
$$f(x)=x+1+\frac{1}{x-1}$$
we divide rather than differentiate first, because this one identity supplies the asymptote, the sign of the gap, and an easier $f'$
near $x=1$ the term $1/(x-1)$ dominates and its sign is the sign of $x-1$
$$f(x)-(x+1)=\frac{1}{x-1}\to 0$$
so $y=x+1$ is a slant asymptote, and the gap is positive for $x>1$ and negative for $x\lt 1$: the right branch sits above the line, the left branch below it
Which way it goes
$$f'(x)=1-\frac{1}{(x-1)^{2}}$$
differentiating the rewritten form; the quotient rule on the original gives the same thing with more work
$$f'(x)=0 \iff (x-1)^{2}=1 \iff x=0 \text{ or } x=2$$
both are in the domain, so both are critical numbers
Independent check of the local minimum by a different test: $f''(2)=2/(2-1)^{3}=2>0$, which says minimum, agreeing with the sign change of $f'$ found above. And a scale check on the asymptote: $f(101)=10201/100=102.01$ against $x+1=102$, a gap of $0.01=1/(101-1)$, exactly the remainder term.
Two derivatives, four test values, and one long division.
One division did four jobs: it produced the slant asymptote, the side of the line each branch runs on, a two term $f'$, and a two term $f''$. When a rational function is top heavy, divide first and differentiate afterwards.
Checkpoint
§07.2 — reading the line off the division●●○○○
Thirty seconds. A top heavy rational function, so a slanted line is waiting inside it.
Given
$f(x)=\dfrac{2x^{2}-3x+1}{x-2}$
The division gives $2x^{2}-3x+1=(x-2)(2x+1)+3$
Find
Which line is the slant asymptote of $f$?
Hint 1/4
You are not being asked to divide; the division is written out in the data. Ask which part of it survives far out.
Hint 2/4
After dividing, $f(x)=q(x)+R/(x-2)$, and the asymptote is the quotient $q$, because the remainder term tends to zero.
Hint 3/4
Given again: $2x^{2}-3x+1=(x-2)(2x+1)+3$, so $f(x)=2x+1+\dfrac{3}{x-2}$, and the last term dies as $x\to\pm\infty$.
Hint 4/4
The asymptote is the quotient: $y=2x+1$.
Show solutionTurn the division into the useful form
which is the definition of the line being an asymptote
Answer $$y=2x+1$$
Check
Numerical check far out: at $x=1002$ the function is $2005003/1000=2005.003$ while the line gives $2(1002)+1=2005$ — a gap of $0.003=3/1000$, which is the remainder term exactly.
The quotient is the asymptote and the remainder is the gap; the remainder is worth keeping because its sign says which side of the line the graph is on.
⚠ Calling the whole quotient the asymptote
The division has just been written out and the leftover term is sitting right there on the page.
wrong$$y=x+1+\frac{1}{x-1}$$
right$$y=x+1$$
⚠ Reporting a horizontal asymptote for a top heavy quotient
The habit of comparing degrees stops at "numerator bigger, so no horizontal asymptote" and never asks what replaces it.
wrong$$\frac{x^{2}}{x-1}: \ \text{no asymptote as } x\to\pm\infty$$
7.3Turning a sentence into one function of one variable
Every word problem starts here: name the objective, use the constraint to reach one variable, and state the interval before differentiating.
The pen against the wall from the first paragraph is not a graph problem yet: there is no function in the sentence, and building one is the whole difficulty.
MethodThe optimization recipe
Conditions
the quantity to be made largest or smallest can be written as a formula
the words contain a relation between the variables, which is the constraint
The constraint may be absent. When the object is described parametrically, or when the objective is built from a tangent line or from an angle inside a figure, there is no equation $g(x,y)=0$ to solve and substitute: the geometry or the parametrisation already delivers one variable, and the recipe starts at the differentiation step. The trigger word may be absent too. A physical sentence such as large enough to be seen without a telescope has to be turned into minimise the distance before anything is written, and the interval $I$ may have to be built out of the wording, from phrases such as after today or for the rest of the year, rather than read off a picture.
$$\boxed{\ \underbrace{Q(x,y)}_{\text{objective}}\ \text{ with }\ \underbrace{g(x,y)=0}_{\text{constraint}}\ \Longrightarrow\ Q(x)\ \text{ on }\ I\ \Longrightarrow\ Q'(x)=0\ }$$
Write down what you are trying to make big or small, write down the relation the problem imposes, use the second to knock a variable out of the first, decide honestly which values of the survivor build a real object, and only then differentiate.
The pen from the opening, as a function: $\textcolor{#1f6feb}{A(x)=x(13-2x)}$ against the distance $x$ out from the wall. The three sampled pens sit on the curve; the peak sits between two of them.
Looks like this, but is not
It looks as though the objective $A=xy$ can be differentiated as it stands: $\dfrac{dA}{dx}=y$, set it to zero, and conclude $y=0$.
That computation treats $y$ as a constant, which it is not: the fence forces $y=13-2x$, so changing $x$ changes $y$ too. Its answer, $y=0$, is a pen with no width and no area — the smallest possible, not the largest. Differentiating before substituting does not just lose accuracy, it answers a different question.
The largest pen against a wall with 13 m of fence
The problem from the first paragraph, finished exactly rather than sampled.
Given
$13$ metres of fencing
a straight wall forms the fourth side, so only three sides are fenced
the pen is a rectangle
Find
the dimensions of the largest pen, and its area
SolutionName the variables on a picture
$$x=\text{the two sides running out from the wall},\ y=\text{the side parallel to the wall}$$
naming them on a drawing first is what stops the constraint from being written backwards
$$A=xy$$
the objective: this is the quantity the question is about
Write the constraint and use it
$$2x+y=13$$
the fence covers two sides of length $x$ and one of length $y$; the wall is free
$$y=13-2x$$
we solve for $y$ rather than for $x$ because $y$ appears once and to the first power, so the substitution stays linear
$$A(x)=x(13-2x)=13x-2x^{2}$$
one variable at last, and now differentiation means something
State the interval before differentiating
$$x>0 \ \text{ and }\ y=13-2x>0 \Rightarrow 0\lt x\lt 6.5$$
outside this range one of the sides is negative and there is no pen to speak of
Differentiate and classify
$$A'(x)=13-4x$$
one derivative, no rules needed
$$A'(x)=0 \iff x=3.25$$
inside $(0,6.5)$, so it is admissible
$$A'>0 \text{ for } x\lt 3.25,\quad A'\lt 0 \text{ for } x>3.25$$
the only critical number, with $A$ rising before it and falling after, so this is the absolute maximum on the whole interval
Independent check by a different route: $A(x)=13x-2x^{2}=-2\left(x-\tfrac{13}{4}\right)^{2}+\tfrac{169}{8}$, and a completed square shows the maximum $169/8=21.125$ at $x=13/4$ with no calculus at all. The sampled pens of the opening, $21$ and $21$ and $20$, are all below it, as they must be.
One substitution, one derivative, one sign check — the arithmetic was never the hard part.
Notice that $y$ came out exactly twice $x$. That is not an accident of these numbers: for a rectangle against a wall with any fixed length of fence, half the fence goes on the side parallel to the wall.
A field store with a half-cylinder roof: building both equations off the picture
A field store is put up on a flat concrete slab. The figure shows it from the end and from the side. The lower part is a rectangular box of height $h$; lying on top of it along the whole length is a half-cylinder whose flat rectangular face covers the top of the box exactly, so the radius of the half-cylinder is half the width of the box. Seen from above, the store is twice as long as it is wide. Both ends of the half-cylinder are closed by flat semicircular caps. Galvanised sheet is used for the four vertical walls, for the curved roof and for the two semicircular caps; the floor is the concrete slab and uses no sheet. Exactly $150\ \mathrm{m^{2}}$ of sheet is available and all of it is used. The figure carries no formulas. Find the radius and the wall height that make the enclosed volume as large as possible, and find that volume.
Solution
A composite solid is the one setting where neither the objective nor the constraint can be copied from a formula sheet, so every face has to be argued in or ruled out, and that is exactly where the first marks of the grading key sit.
Name what the picture forces, then count unknowns against equations
$$r=\tfrac12(\text{box width})=\text{radius of the roof},\qquad h=\text{wall height},\qquad \ell=\text{length}$$
the half-cylinder rests on the box, so its radius is not free: it is fixed by the width, and spending one letter on both is the first economy the picture offers
$$\ell=2\cdot(2r)=4r$$
twice as long as it is wide, and the width is $2r$; note that this equation carries a number and is still not the constraint, it is a shape relation the figure gives away for free
two long walls and two short ones; the short walls are as wide as the box, $2r$, not as wide as the radius, and that factor of two is the commonest slip in the setup
half of a cylinder's lateral surface; the flat rectangle under it is where the two pieces meet, so it sits inside the solid and is not sheet, and adding the box top here is what turns a correct derivative into a wrong answer
the two cubic terms partly cancel: a fatter roof adds volume but eats the sheet the walls were going to use, and that competition is the entire problem
$$r>0\ \text{ and }\ h>0\iff 5\pi r^{2}<150\iff 0<r<\sqrt{\tfrac{30}{\pi}}\approx 3.09$$
at the right-hand end the roof alone would swallow all $150\ \mathrm{m^{2}}$ and there is no wall left; past it the formula still returns numbers, but no store exists to have that volume
Differentiate, justify, and translate back into the words of the question
$$V'(r)=100-4\pi r^{2}$$
one derivative, and only now, because until this line the expression had two letters in it
the interval is open, so there are no endpoint values to compare: rising and then falling is what makes this the absolute maximum, and that sentence is itself worth marks
using $r^{2}=25/\pi$ inside the cubic term is quicker and safer than cubing $5/\sqrt{\pi}$; in the words of the question the store is $2r\approx 5.64$ m wide and $4r\approx 11.28$ m long
Check against the two formulas that were built from the figure, not against the reduced one. Walls: $12rh=12\cdot\frac{5}{\sqrt{\pi}}\cdot\frac{5\sqrt{\pi}}{12}=25$. Roof and caps: $5\pi r^{2}=5\pi\cdot\frac{25}{\pi}=125$. Their sum is $150\ \mathrm{m^{2}}$ exactly, so the answer really does use the sheet it was given. Volume from the unreduced form: $8r^{2}h+2\pi r^{3}=\frac{250}{3\sqrt{\pi}}+\frac{250}{\sqrt{\pi}}=\frac{1000}{3\sqrt{\pi}}$, which agrees. Two rivals confirm the maximum: $r=2.5$ gives $V\approx 184.55$, and the degenerate store with no walls, $r=\sqrt{30/\pi}\approx 3.09$ with $h=0$, gives $V\approx 185.41$, both below $188.06$.
A gable-roofed shed: the same picture read backwards, with the slant length missing
A workshop shed is built on a concrete slab, with vertical walls and a gable roof. From the end the shed is $2x$ wide with walls $h$ high, and the roof climbs from each eave to a ridge over the middle of the shed at a pitch of $3$ vertical to $4$ horizontal. From above, the shed is twice as long as it is wide. Steel sheet covers the four walls, the two triangular gables above the end walls and the two flat roof panels; the slab is concrete. The shed has to enclose exactly $252\ \mathrm{m^{3}}$, roof space included. No slant length is marked on the figure and no area formula is given. Find the dimensions that use the least sheet, and how much sheet that is.
Solution
Swapping the roles so that the volume is obeyed and the material is minimised, and hiding one needed length behind a roof pitch, tests whether the setup was understood rather than copied from the previous figure.
Decide from the verbs which quantity plays which role
the roles come from the verbs and not from the shapes: here the area is optimised and the volume is obeyed, the exact reverse of the store problem, while the geometry to be read off the figure is the same kind of work
half the width is the useful letter because the horizontal run from an eave to the ridge is exactly half the width, and the pitch will be applied to that run
Produce the two lengths the figure never labels
$$\text{rise}=\frac{3}{4}\,x$$
a pitch of $3$ in $4$ applied to the run $x$; nothing about the roof can be written down until this height exists, and it appears nowhere in the given data
a roof panel is as wide as the sloping side, not as wide as the horizontal run, so using $x$ here would undercount every panel by a fifth while every later step still looked correct
the roof is a prism, so its volume is the gable triangle times the length, and the roof space counts because the question says the shed encloses that volume
$$\text{walls}:\ 2(4x\,h)+2(2x\,h)=12xh$$
two long walls of length $4x$ and two end walls of width $2x$, all of height $h$; the gable triangles are not part of them
part of the $x^{2}$ cost cancels because a wider shed gets more volume free from its roof and so needs shorter walls; what survives is the honest price of width
falling and then rising is the whole justification that this is the absolute minimum, since the interval has no endpoint values to compare; the sign of $S''$ says the same thing a second way
Back into the unreduced pair, the ones built from the figure: $V=8x^{2}h+3x^{3}=8\cdot 9\cdot\frac{19}{8}+3\cdot 27=171+81=252\ \mathrm{m^{3}}$, the constraint hit exactly, and $S=12xh+\frac{23}{2}x^{2}=85.5+103.5=189\ \mathrm{m^{2}}$, agreeing with $S(3)$. Neighbours confirm the minimum: $S(2.5)=194.95$ and $S(3.5)=193.75$, both above $189$, and the flat-walled extreme $x=\sqrt[3]{84}$ with $h=0$ needs $220.57\ \mathrm{m^{2}}$. A dimension check on the reduced objective: $378/x$ and $7x^{2}$ are both areas once $x$ is a length, so no volume term was left in by accident.
The largest flower bed inside a semicircular lawn of radius 6 m
A lawn is a half disc of radius $6$ m whose straight edge is the segment $AB$ through the centre $O$, so $\lvert AB\rvert=12$ m. Let $C$ be a point of the arc and $\theta=\angle BOC$. (a) For the triangular bed $ABC$, write the area as a function of $\theta$ and find the largest one. (b) For the trapezium $ABCD$ with $C$ and $D$ on the arc, symmetric about $O$, find the largest area and the smallest area, and the $\theta$ where each occurs.
Solution
The earlier semicircle problem worked in $x$, which is right when the figure gives a length. When it gives an angle, $x$ buys a square root in every later line.
Put the moving corner on the circle with one letter
a point of the circle at central angle $\theta$ is $(r\cos\theta,r\sin\theta)$; this parametrisation replaces the constraint, so nothing has to be eliminated later
$$\text{distance from } C \text{ to the line } AB = 6\sin\theta$$
$AB$ lies on the $x$ axis, so the distance is the second coordinate; every area below rests on this
Rebuild the winner from its sides: at $\theta=\tfrac{\pi}{3}$ the parallel sides are $12$ and $6$ and the height is $3\sqrt3$, giving $\tfrac12(18)(3\sqrt3)=27\sqrt3$. The magnitude is right too: the whole half disc is $18\pi\approx 56.5\ \mathrm{m^{2}}$, which no bed inside it may beat.
A corner on a circle costs one letter, and its distance to the diameter is $r\sin\theta$. The smallest value here was $0$, where the bed is not a bed: name a degenerate endpoint instead of dropping it.
Checkpoint
§07.3 — which equation is the constraint●●○○○
Thirty seconds, no algebra. A closed rectangular box with a square base of side $x$ and height $h$ has to hold a fixed volume, and we want the least material.
Given
The base is square with side $x$, and the height is $h$
The volume must be exactly $32\ \mathrm{cm^{3}}$
Surface area of a closed box with square base: $S=2x^{2}+4xh$
Find
Which equation is the constraint, that is, the one used to remove a variable?
Hint 1/4
Ask which sentence in the problem is a requirement imposed on the box, and which one is the thing being measured.
Hint 2/4
The objective is what you are asked to make small; the constraint is the equation the problem forces on the variables, and it is the one you solve for a variable and substitute.
Hint 3/4
Given again: the material is $S=2x^{2}+4xh$ and the requirement is a volume of $32$. A box with a square base of side $x$ and height $h$ has volume $x^{2}h$.
Hint 4/4
The constraint is $x^{2}h=32$; solving it gives $h=32/x^{2}$, which goes into $S$.
Show solutionSort the two equations
$$S=2x^{2}+4xh \quad \text{(objective: the thing to be made small)}$$
the question asks for least material, so this is what is optimised
$$x^{2}h=32 \quad \text{(constraint: the requirement)}$$
every admissible box satisfies it, so it is what eliminates a variable
Dimension check on the substituted objective: $x^{2}$ and $128/x$ are both areas once $x$ is a length, so nothing incompatible has been added.
A quick test when the words are confusing: the constraint is the sentence with a specific number attached to it, and the objective is the sentence with the word largest, smallest, cheapest or fastest.
⚠ Differentiating an objective that still has two variables
The objective is written down first and looks ready; the constraint feels like extra information rather than a required step.
The formula is happy at every real number, so nothing on the page objects until the answer turns out to be a box with a negative side.
wrong$$V(x)=x(30-2x)(16-2x),\ V'=0 \Rightarrow x=\tfrac{10}{3} \text{ or } x=12$$
right$$0\lt x\lt 8 \Rightarrow \text{only } x=\tfrac{10}{3} \text{ is admissible}$$
7.4Proving it is the best and not merely a candidate
Turns a critical point into a proven absolute extremum by a sign argument, and it still works on open intervals.
The pen finished with a line most write-ups leave out: $A'$ is positive before $3.25$ and negative after, so that critical number is the winner and not just a suspect.
TheoremTheorem 7.2: First Derivative Test for Absolute Extreme Values
Conditions
$c$ is a critical number of $f$
$f$ is continuous on an interval $I$ containing $c$
$I$ may be open, half open, or infinite — that is the point of the test
$$\boxed{\begin{aligned} f'>0 \text{ for all } x\lt c \text{ in } I \ \text{ and }\ f'\lt 0 \text{ for all } x>c \text{ in } I &\ \Longrightarrow\ f(c)=\max_{I} f\\ f'\lt 0 \text{ for all } x\lt c \text{ in } I \ \text{ and }\ f'>0 \text{ for all } x>c \text{ in } I &\ \Longrightarrow\ f(c)=\min_{I} f \end{aligned}}$$
If the function climbs over the whole stretch on one side of $c$ and falls over the whole stretch on the other, then nothing anywhere on the interval can beat the value at $c$ — no endpoints required, which is why this is the version that survives on an open interval.
Why the single critical radius has to win: $\textcolor{#1f6feb}{S(r)=2\pi r^{2}+2000/r}$ is unbounded at both ends of $(0,\infty)$, so the one dip in the middle is the absolute minimum even though the interval has no endpoints to check.
Looks like this, but is not
It looks as though finding the critical number is the whole job: for $C(x)=x+\dfrac{4}{x}$ we get $C'(x)=1-\dfrac{4}{x^{2}}=0$ at $x=2$, so the answer is $x=2$ with $C(2)=4$.
Only if the interval contains $2$. If the problem says the design must satisfy $3\le x\le 10$, then $2$ is not admissible at all, $C'>0$ throughout $[3,10]$, and the cheapest allowed design is the left endpoint: $C(3)=3+\tfrac43\approx 4.33$, against $C(10)=10.4$. The critical number is a fact about the formula; the answer is a fact about the formula and the interval together.
The can of 1000 cm³ with the least metal
An interval with no endpoints at all, so the argument has to come from the shape of the objective.
Given
a closed cylindrical can of radius $r$ and height $h$
the volume is fixed at $V=1000\ \mathrm{cm^{3}}$
surface area of a closed cylinder: $S=2\pi r^{2}+2\pi r h$
Find
the radius and height that use the least metal, and how much that is
Independent check by testing two other cans of the same volume: $r=4$ gives $h=1000/(16\pi)\approx 19.89$ and $S\approx 100.5+500=600.5$; $r=7$ gives $S\approx 307.9+285.7=593.6$. Both use more metal than $553.58$, on either side of the optimum, as the sign argument promised.
One substitution, one derivative, one sign argument on an open interval, and one translation back into the language of the question.
The exact answer $h=2r$ survived the arithmetic and is worth more than the decimals: it is the same for every fixed volume, so a can of any size is cheapest when its height equals its diameter.
The cheapest closed tank of 4 m³ when the three faces cost different amounts
A closed tank with a square base of side $x$ metres and height $h$ metres must hold $4\ \mathrm{m^{3}}$. The base plate costs $300$ TL/m², the four welded walls $400$ TL/m², the thin lid $100$ TL/m². The slab it stands on forces $x\le 2.5$ and the drain fitting forces $x\ge 1$. Find the cheapest tank and its cost, and the dearest the two limits still allow.
Solution
A cost is a sum of unit price times area, one term per face. A price column beside an area column is what keeps them paired.
Price the winner face by face: base and lid are $8\ \mathrm{m^{2}}$ costing $1600$ TL, the walls $4(2)(1)=8\ \mathrm{m^{2}}$ at $400$, so $3200$ TL, total $4800$ TL. Independently, $C''(x)=800+\dfrac{12800}{x^{3}}>0$, so $C$ is concave up: its one critical number must be the minimum and the maximum must sit at an endpoint.
One product per face, and never merge faces with different prices. Without the slab the interval would be $(0,\infty)$, with no endpoints to compare and the sign of $C'$ as the only closing argument.
A stone cube with a stone sphere on top, under a 6 m ceiling
A monument is a solid stone cube of edge $x$ metres with a solid stone sphere on top. The ceiling fixes the total height at $6$ m, so the edge plus the diameter is $6$ m, and stone is paid for by volume. (a) Which $x$ uses the least stone, and how much? (b) Which uses the most? Say what (b) assumes about the shape.
Solution
Stacked solids under a fixed total height need no constraint equation: the height sentence removes a variable in one line. The test is at the two ends.
The constraint is the height sentence
$$x+2r=6\ \Longrightarrow\ r=\frac{6-x}{2}$$
the sphere sits on the cube, so the heights add; solve for $r$ and the survivor is the letter the question asks about
read as a closed interval, all $6$ m go to the cube: a monument with no sphere on it
$$\text{if both solids must be present: } 0\lt x\lt 6\ \Longrightarrow\ \text{no maximum}$$
on the open interval $V$ comes as close to $216$ as you like without reaching it, so the answer changes from a number to there is none
Answer $$\boxed{\ \text{(a)}\ x=\frac{6\sqrt{\pi}}{\sqrt6+\sqrt{\pi}}\approx 2.52\ \mathrm{m},\ V_{\min}\approx 38.1\ \mathrm{m^{3}};\quad \text{(b)}\ x=6,\ V=216\ \mathrm{m^{3}}\ \text{if a sphere of radius }0\text{ counts; otherwise no maximum}\ }$$
Check
Both ends check without calculus: a cube of edge $6$ has volume $216$, a sphere of diameter $6$ has $36\pi\approx 113.1$, smaller because a ball fills only $\tfrac{\pi}{6}$ of the box round it. Both match $V(6)$ and $V(0)$, and the interior value $38.1$ is below either.
A critical number is not a maximum until something says so; here it was the worst design on the list. When an endpoint kills one piece of the object, say whether that still counts: the interval is closed or open by that decision.
Checkpoint
§07.4 — closing the argument on an open interval●●○○○
Thirty seconds. A cost function on all positive numbers, with one critical number and known behaviour at the two open ends.
Given
$C$ is continuous on $(0,\infty)$, with exactly one critical number, at $x=5$
$C(x)\to\infty$ as $x\to 0^{+}$ and as $x\to\infty$
Find
What can be concluded about $C(5)$?
Hint 1/4
Ask where else a small value could possibly be, given that the function is huge at both ends and turns only once.
Hint 2/4
On an interval with no endpoints, an absolute extremum can only be at a critical number; Theorem 7.2 upgrades a candidate to an answer once the sign of $C'$ is settled on both sides.
Hint 3/4
Given again: one critical number at $x=5$, and $C\to\infty$ at both ends. A continuous function that starts huge, turns once, and ends huge must be falling before the turn and rising after it.
Hint 4/4
$C(5)$ is the absolute minimum of $C$ on $(0,\infty)$.
Show solutionRule out the alternatives
$$C'\ne 0 \text{ on } (0,5) \Rightarrow C' \text{ keeps one sign there}$$
a continuous derivative cannot change sign without a zero, and there is none
$$C\to\infty \text{ as } x\to 0^{+} \Rightarrow C' \lt 0 \text{ on } (0,5)$$
coming down from arbitrarily large values
$$C\to\infty \text{ as } x\to\infty \Rightarrow C'>0 \text{ on } (5,\infty)$$
going back up to arbitrarily large values
Apply the theorem
$$C'\lt 0 \text{ left of } 5,\ C'>0 \text{ right of } 5 \Rightarrow C(5)=\min_{(0,\infty)}C$$
the minimum half of Theorem 7.2
Answer $$\text{Absolute minimum on } (0,\infty).$$
Check
Independent check by contradiction: if some $C(a)\lt C(5)$ existed, then between $a$ and the far end where $C$ is huge there would have to be a second turn, hence a second critical number, which the data forbids.
"Unbounded at both ends, one critical number" is a complete argument and takes one line. It is the standard closer for the open interval problems in this section.
⚠ Handing in the critical number as the answer
It is the last thing computed and it is a number, so it looks like a result.
wrong$$\text{"The largest volume is } x=\tfrac{10}{3}\ \mathrm{cm}\text{"}$$
right$$\text{"The volume is largest at } x=\tfrac{10}{3}\ \mathrm{cm},\ \text{ and equals } \tfrac{19600}{27}\approx 725.93\ \mathrm{cm^{3}}\text{"}$$
⚠ Using the Closed Interval Method on an open interval
It is the method drilled last week, and it ends with a tidy comparison table.
wrong$$S \text{ on } (0,\infty):\ \text{compare } S(0) \text{ and } S(\infty) \text{ with } S(r_{c})$$
right$$S'\lt 0 \text{ on } (0,r_{c}),\ S'>0 \text{ on } (r_{c},\infty) \Rightarrow S(r_{c})=\min$$
7.5Replacing the curve by its tangent
Iterate $x_{n+1}=x_n-f(x_n)/f'(x_n)$ to get decimals for a root when the equation has no closed form solution.
Every optimization above ended with an equation that the exact-root routine of the previous section opened; change the cost slightly and $f'(x)=0$ becomes $x^{3}-x-1=0$, where that routine offers only $\pm1$ as candidates and neither one is a root.
RuleRule 7.3: Newton's iteration
Conditions
$f$ is differentiable near the root
$f'(x_{n})\ne 0$ at every iterate produced
$x_{0}$ is a starting guess, supplied by a sketch or by a sign change of $f$
From where you are, walk down the tangent line until it hits the axis, and stand there instead; the correction is the height divided by the slope, which is exactly the horizontal distance the tangent needs to reach zero.
One step on $\textcolor{#1f6feb}{f(x)=x^{3}-x-1}$: the $\textcolor{#6f42c1}{\text{tangent}}$ at the point above $x_{0}=1.5$ crosses the axis at $x_{1}=1.3478$, which is closer to the root than the guess we started from.
Looks like this, but is not
It looks as though the correction should simply be the height: if $f(x_{n})=0.875$, move back by $0.875$ and take $x_{n+1}=x_{n}-f(x_{n})=0.625$.
A height and a horizontal distance are different quantities, and the slope is what converts one into the other. Here the tangent is steep, $f'(1.5)=5.75$, so the drop of $0.875$ costs only $0.875/5.75\approx 0.152$ of horizontal travel. Moving the full height overshoots the root at $1.3247$ by more than half a unit and lands at $0.625$, where $f=-1.38$: further from zero than we started.
$n$
$x_{n}$
$\vert f(x_{n})\vert$
distance to the root
0
1.5000000000
$8.8\times 10^{-1}$
$1.8\times 10^{-1}$
1
1.3478260870
$1.0\times 10^{-1}$
$2.3\times 10^{-2}$
2
1.3252003990
$2.1\times 10^{-3}$
$4.8\times 10^{-4}$
3
1.3247181740
$9.2\times 10^{-7}$
$2.2\times 10^{-7}$
4
1.3247179572
$1.9\times 10^{-13}$
$4.4\times 10^{-14}$
Read the exponents in the last column: $-1,-2,-4,-7,-14$. Each one is roughly double the one above it, which is what quadratic convergence means in practice — the error is squared at every step, so the accuracy of the previous line is not improved but doubled. It is also why nobody ever needs a tenth iterate by hand.
Solving x³ − x − 1 = 0 to six decimals
A cubic with one real root and no usable factorisation, so the tangent does the work.
Given
$f(x)=x^{3}-x-1$
$f(1)=-1\lt 0$ and $f(2)=5>0$, so a root lies in $(1,2)$
starting guess $x_{0}=1.5$
Find
the root to six decimal places, with a check that it is a root
SolutionWrite the iteration for this f
$$f'(x)=3x^{2}-1$$
one derivative, computed once and reused at every step
the correction is now smaller than $10^{-6}$, so the run is finished
Stop, and check
$$|x_{4}-x_{3}|=2.2\times 10^{-7}\lt 10^{-6}$$
the stopping test: successive iterates have stopped moving at the requested precision
$$f(1.3247179572)\approx 1.9\times 10^{-13}$$
the check that matters: the value of $f$ at the answer, not the agreement of two guesses
Answer $$\boxed{x^{*}\approx 1.324718}$$
Check
Independent check by bracketing rather than by iterating: $f(1.3247)=-7.7\times 10^{-5}\lt 0$ and $f(1.3248)=+3.5\times 10^{-4}>0$, so the Intermediate Value Theorem puts a root strictly between them, which is where Newton said it was.
One derivative and four divisions. The expensive part of Newton's method is finding a starting guess that is honestly near a root.
Four steps was not luck. Once an iterate is close, each step roughly squares the distance to the root: $2\times 10^{-2}$ became $5\times 10^{-4}$, then $2\times 10^{-7}$, then $4\times 10^{-14}$.
Checkpoint
§07.5 — one step, by hand●●○○○
Thirty seconds with a calculator. Newton's method applied to a function whose root you already know, so the arithmetic is the only thing being tested.
Given
$f(x)=x^{2}-5$
$x_{0}=2$
Find
Compute $x_{1}$.
Hint 1/4
Write down what the formula needs: the height at the guess and the slope at the guess, in that order.
Hint 2/4
$x_{1}=x_{0}-f(x_{0})/f'(x_{0})$, with $f'(x)=2x$ here.
Hint 3/4
Given again: $f(x)=x^{2}-5$ and $x_{0}=2$, so $f(2)=4-5=-1$ and $f'(2)=4$.
Hint 4/4
$x_{1}=2-(-1)/4=2.25$.
Show solutionEvaluate, then step
$$f(2)=4-5=-1,\qquad f'(2)=2(2)=4$$
the height is negative, so the correction will move to the right
$$x_{1}=2-\frac{-1}{4}=2+0.25=2.25$$
subtracting a negative correction; a sign slip here is the most common error in the whole method
Answer $$x_{1}=2.25$$
Check
Independent check: $2.25^{2}=5.0625$, which is nearer to $5$ than $2^{2}=4$ was, so the step improved the guess rather than spoiling it.
When $f(x_{n})$ is negative the iterate moves right, and when it is positive it moves left, on a rising branch. Watching that sign is a free error check on every step.
⚠ Dropping the minus sign in the correction
The formula subtracts, and when $f(x_{n})$ is itself negative two minus signs meet in one line.
wrong$$x_{1}=2-\frac{1}{4}=1.75$$
right$$x_{1}=2-\frac{-1}{4}=2.25$$
⚠ Rounding an iterate before feeding it back in
Seven decimals look like showing off, and the answer is only wanted to four.
Names the three conditions behind quadratic convergence, so you can read a stalling or cycling run and pick a better start.
The run above converged because the starting guess sat on a steep, well behaved piece of the curve; change $x_{0}$ and the same formula can loop for ever without complaining.
NoteNote 7.4: what the method needs, and what it promises
Conditions
$f$ has a continuous second derivative near the root $x^{*}$
$f'(x^{*})\ne 0$, that is, the root is simple
$x_{0}$ is close enough to $x^{*}$
$$\boxed{\ |x_{n+1}-x^{*}|\le C\,|x_{n}-x^{*}|^{2}\quad\text{for some constant } C \text{ near } x^{*}\ }$$
Under those three conditions the new error is roughly the square of the old one, so a guess accurate to three decimals becomes accurate to six. Drop any of the three conditions and the guarantee is gone: the method may still work, but nothing says it must.
A cycle, drawn: for $\textcolor{#1f6feb}{f(x)=x^{3}-2x+2}$ the $\textcolor{#6f42c1}{\text{tangent}}$ at $x_{0}=0$ lands on $1$ and the tangent at $1$ lands back on $0$. The iterates never stop and never move.
Looks like this, but is not
It looks as though starting closer always helps: whatever went wrong, halve the distance to the root and try again.
For $f(x)=x^{1/3}$, whose only root is $0$, the iteration simplifies to $x_{n+1}=x_{n}-\dfrac{x_{n}^{1/3}}{\tfrac13 x_{n}^{-2/3}}=x_{n}-3x_{n}=-2x_{n}$. The distance to the root doubles at every step, from any starting guess whatsoever, however close. What fails is not the distance but the second condition: $f'$ is infinite at the root, so the tangent is vertical there and points nowhere useful.
What the iterates do
What went wrong
The repair
The next step is undefined
$f'(x_{n})=0$: the tangent is horizontal and never meets the axis
Move the guess a little, or bracket a root and restart inside the bracket
Two values alternate for ever
The guess sits on a cycle of the iteration, often between two turning points
Bracket a root with a sign change of $f$ and start inside that bracket
The iterates grow without bound
The guess is on a nearly flat stretch, so each tangent throws you further out
Start where $\vert f'\vert$ is comfortably large, which a rough sketch shows at once
It converges, to the wrong number
There are several roots and the guess was in the wrong one's territory
Bracket the root you actually want, and start on that side of the nearest turning point
Only the first row is an error message; the other three run happily and produce numbers. That is the reason the last line of every Newton answer evaluates $f$ at the final iterate: a run that has settled has not thereby settled on a root.
Diagnosing a cycle, and repairing it
The same equation, once from a guess that loops and once from a guess chosen with a sign change.
Given
$f(x)=x^{3}-2x+2$, so $f'(x)=3x^{2}-2$
a first attempt from $x_{0}=0$ produced $0,\,1,\,0,\,1,\,0,\dots$
Find
why the first attempt failed, and a root to six decimal places
the same squaring of the error as before, once the guess is in a sensible place
$$f(x_{4})\approx -4.2\times 10^{-10}$$
the check: the iterate really is a root, not merely a number the iteration stopped at
Answer $$\boxed{x^{*}\approx -1.769292}$$
Check
Independent check that this is the only real root: $f'(x)=3x^{2}-2$ gives a local maximum at $x=-\sqrt{2/3}\approx-0.8165$ with $f\approx 3.089$ and a local minimum at $x=+0.8165$ with $f\approx 0.911$, both positive, so the graph crosses the axis exactly once — to the left, which is where the repaired run went. That also explains the cycle: the starting guess sat on the hump between the two turning points, where the tangents point away from the only crossing.
Every repair in this block is the same repair: bracket a root with a sign change, check the slope is not near zero inside the bracket, and start there. Guessing $x_{0}$ because it is a round number is what produces the four failure modes below.
Checkpoint
§07.6 — reading the failure off the iterates●●○○○
Thirty seconds. Someone hands you the first four iterates of a run and nothing else — no function, no starting reasoning.
Given
The recorded iterates are $x_{0}=3.0$, $x_{1}=-6.1$, $x_{2}=12.4$, $x_{3}=-25.2$
Every step was computed correctly
Find
Which failure mode is this?
Hint 1/4
Look only at the sizes of the numbers, and ask whether they are settling anywhere at all.
Hint 2/4
The four modes are: undefined step, alternating cycle, unbounded growth, and convergence to an unintended root. Each has its own signature in the list of iterates.
Hint 3/4
Given again: $3.0$, $-6.1$, $12.4$, $-25.2$. The signs alternate, but the sizes are roughly doubling each time, so no pair of values repeats.
Hint 4/4
The iterates are growing without bound: this is divergence, not a cycle.
Show solutionTest the cycle hypothesis
$$\text{a cycle needs } x_{2}=x_{0}$$
that is what a two step cycle means, and $12.4\ne 3.0$
$$|x_{n+1}|/|x_{n}|\approx 2.0,\ 2.0,\ 2.0$$
a constant growth factor, not a repetition
Name it and repair it
$$|x_{n}|\to\infty$$
divergence
$$\text{repair: bracket a root, restart where } |f'| \text{ is large}$$
the standard fix, and the only one that uses information about $f$ rather than about luck
Answer $$\text{Divergence.}$$
Check
Independent check: if this were a cycle the values would have to repeat exactly, and four distinct magnitudes in a row rule that out on their own.
Alternating signs on their own mean nothing. Compare consecutive magnitudes before naming the mode.
⚠ Trusting agreement of two iterates as proof of a root
The stopping test is about agreement, so agreement feels like the finish line.
wrong$$|x_{4}-x_{3}|\lt 10^{-6} \Rightarrow x_{4} \text{ is a root}$$
right$$|x_{4}-x_{3}|\lt 10^{-6} \ \text{ and }\ |f(x_{4})|\lt 10^{-6} \Rightarrow x_{4} \text{ is a root}$$
⚠ Picking the starting guess because it is a round number
Zero and one are easy to evaluate at, and the first step usually succeeds, so the habit survives.
wrong$$x_{0}=0 \text{ for } f(x)=x^{3}-2x+2 \Rightarrow 0,1,0,1,\dots$$
Any question that says sketch, analyse or describe the graph of a function given by a formula. The order was built in the previous section; it is repeated here because every block of this one leans on it.
Domain
Solve for the numbers where the formula fails: a zero denominator, a negative under an even root, a non positive argument of a logarithm. Every such number cuts the picture into separate pieces.
Intercepts
Compute $f(0)$ if $0$ is in the domain, then solve $f(x)=0$. These are the only points you get for free.
Symmetry
Test $f(-x)$. Even means mirror across the $y$ axis, odd means rotate through the origin; either one halves everything that follows.
Asymptotes
One sided limits at every number missing from the domain give the vertical ones; $\lim_{x\to\pm\infty}f(x)$ gives the horizontal one; a division gives the slant one when the numerator is one degree above the denominator.
Sign of $f'$
Build the sign line on the pieces from step one, using the critical numbers as extra breakpoints, and classify each critical number with the First Derivative Test.
Sign of $f''$
A second sign line on the same pieces: positive bends the curve up, negative bends it down, and a change of sign at a point of the domain is an inflection point.
A few exact points
Plot the intercepts, the extrema, the inflection points and one point on each far branch. These are the numbers a picture cannot place exactly, so they are the ones worth writing down.
Draw branch by branch
Draw the asymptotes as dashed lines first, then join the plotted points respecting steps five and six. Each piece of the domain gets its own stroke, and no stroke crosses a vertical asymptote.
Where it goes wrong
Starting at step five because the derivative is the interesting part. The sign line is then built on the wrong number of pieces and every later step inherits the error.
Drawing the curve first and adding the asymptotes afterwards, which always produces a graph that crosses a vertical asymptote.
Treating a number where $f''$ changes sign as an inflection point without checking that it is in the domain.
From a paragraph to a number
Any problem containing the words largest, smallest, cheapest, fastest, closest or least.
Draw and name
Make a picture and put a letter on every length in it, before writing a single equation. Most wrong constraints are written by people who have not drawn the object.
Objective
Write the quantity to be optimised as a formula. This is the sentence with largest or smallest in it.
Constraint
Write the requirement as an equation. This is the sentence with a specific number in it: a fixed volume, a fixed length of fence, a fixed budget.
Eliminate
Solve the constraint for whichever variable appears once and to the first power, and substitute. The objective is now a function of one variable.
Interval
State the set of values that build a real object, from the picture and not from the formula. Note whether it is closed, half open, or infinite, because that decides which closing argument is available.
Critical numbers
Differentiate, solve $Q'=0$, and discard anything outside the interval.
Close it
Closed interval: compare the values at the critical numbers with the values at both endpoints. Open interval: use the sign of $Q'$ on both sides, or the behaviour of $Q$ at the open ends.
Answer the question asked
Translate back: give the dimensions, and give the value of the objective there, each with its unit.
Where it goes wrong
Differentiating an objective that still contains two variables, which silently answers a different question.
Skipping step five, so a critical number that builds a box with a negative side survives to the end.
Stopping after step six, which hands in a candidate rather than an answer.
Running Newton's method by hand
An equation that cannot be solved exactly, when a decimal answer to a stated accuracy is acceptable.
Write it as $f(x)=0$
Move everything to one side. For $\cos x=x$ that means $f(x)=\cos x-x$; the method needs a root, not an equality.
Bracket a root
Find $a$ and $b$ with $f(a)$ and $f(b)$ of opposite signs, so the Intermediate Value Theorem guarantees a root between them. A ten second sketch usually does it.
Choose $x_{0}$
Take a point inside the bracket where $\vert f'\vert$ is comfortably large, and away from any turning point of $f$.
Differentiate once
Compute $f'$ symbolically now; it is reused unchanged at every step.
Iterate
Apply $x_{n+1}=x_{n}-f(x_{n})/f'(x_{n})$, keeping every digit your calculator holds.
Stop
Stop when $\vert x_{n+1}-x_{n}\vert$ is below the tolerance asked for, which is usually after three or four steps.
Check
Evaluate $f$ at the final iterate. A small value of $f$ is the evidence that you found a root; agreement between two iterates is only evidence that the iteration stopped moving.
Where it goes wrong
Rounding an iterate to the accuracy wanted in the answer and feeding it back in, which throws away the digits the next step was about to produce.
Choosing $x_{0}$ because it is a round number rather than because a sign change puts a root nearby.
Reporting the last iterate without evaluating $f$ there, which cannot distinguish a converged run from a cycle.
The slant asymptote of a function that is not a quotient of polynomials
The question says find the slant, oblique or horizontal asymptotes and prove your claim, and the function carries a root, a fractional power or a negative power, so there is no division to perform. This is the shape the asymptote question actually takes on this course's quizzes, and it is also step four of the sketching checklist whenever the function is not rational.
Domain first, then count the ends you owe
Solve for the domain before any limit. A domain such as $(-\infty,-4]\cup[0,\infty)$ has two infinite ends and therefore two separate calculations; a domain $[1,\infty)$ has one, and no answer is owed at $-\infty$.
The slope
At the end you are working on, compute $m=\lim \dfrac{f(x)}{x}$. Divide inside the root by the appropriate power of $x$: $\sqrt{x^{2}+4x}/x=\sqrt{1+4/x}$ for $x>0$, and $\sqrt[3]{x^{3}-x^{2}}/x=\sqrt[3]{1-1/x}$. If this limit is infinite there is no line at that end. If it is $0$ the candidate line is horizontal, but a zero slope is not an asymptote on its own: you still owe $b=\lim f$ and it has to come out finite. For $f(x)=2\sqrt{x}-3$ the slope limit is $0$ while $\lim f=\infty$, so that graph has no asymptote at either end.
Pull the sign out of the root, in writing
For $x\lt 0$, $\sqrt{x^{2}}=|x|=-x$, so a factor of $-1$ enters everything at the $-\infty$ end. An odd root carries no such sign. Write the line down; graders look for it, and it is the single most common place this question is lost.
The intercept
Compute $b=\lim\bigl(f(x)-mx\bigr)$. It arrives as $\infty-\infty$, so repair it first: $A-B=\dfrac{A^{2}-B^{2}}{A+B}$ for a square root, $A-B=\dfrac{A^{3}-B^{3}}{A^{2}+AB+B^{2}}$ for a cube root, then divide top and bottom by the highest power of $x$ present. For an $n$th root use the identity that clears that root, $A-B=\dfrac{A^{n}-B^{n}}{A^{n-1}+A^{n-2}B+\cdots+B^{n-1}}$; the two above are its first cases, and inventing a conjugate for a root that is neither a square nor a cube is where this line silently produces a wrong intercept.
State the line together with its two limits
Write $y=mx+b$ and, beside it, the two limits that produced $m$ and $b$. An asymptote written down without them earns nothing on a paper that says prove your claims.
Do the other end from scratch
Repeat steps two to five at the remaining infinite end. The results need not match: one end can be slanted and the other horizontal, and a symmetric pair $y=mx+b$ and $y=-mx-b$ is often written compactly as $y=|mx+b|$.
Say which side the graph is on
Evaluate the sign of $f(x)-(mx+b)$ for large $|x|$ at that end. Positive puts the branch above its line, negative below. A sketch that puts the branch on the wrong side of the line loses the sketch marks even when both limits were right.
Where it goes wrong
Treating the two limits as two rules. They are one rule with two finite outputs: a finite $m$ with an infinite $b$, and a zero $m$ with an infinite $\lim f$, both mean no asymptote at that end, and both are reported as asymptotes every year.
Reaching for polynomial division out of habit. There is nothing to divide in $x+\sqrt{x^{2}+4x}$, and the degree rule for quotients does not apply to it.
Computing one end and writing as $x\to\pm\infty$ over the result. The sign of $\sqrt{x^{2}}$ flips between the ends, and about half the marks live on the $-\infty$ side.
Stopping once $m$ is found. A finite slope proves nothing on its own, and $\lim(f-mx)=\pm\infty$ is a genuine possibility that kills the asymptote.
Using the conjugate $A+B$ on a cube root. It does not clear the root; only the factorisation of $A^{3}-B^{3}$ does.
Declaring a vertical asymptote at every number missing from the domain. An endpoint of the domain with a finite value there, such as $x=-4$ for $x+\sqrt{x^{2}+4x}$, is just where the graph starts.
Optimisation when the critical equation has infinitely many solutions
The position, height, temperature or brightness in the problem is given by trigonometric formulas in a time variable and the question asks for the best moment. Solving $Q'=0$ then returns a whole family of times rather than one, and the marks are spread along the chain from reading the objective to naming the right month.
Read the objective out of the sentence
No trigger word need appear. Large enough to be seen without a telescope means minimise the distance to it; a signal is strongest where the distance is least. Ask what physical quantity the situation makes best, and write it down as a sentence before any formula. On this course's midterm this sentence alone is worth three marks.
Build the window out of the wording
Phrases such as after today, for the rest of the year, or the flight ends at 11:00 each cut the variable, and none of them arrives as an inequality. Convert them: a paper sat in week 16 with a year of 52 weeks gives $16\lt t\le 52$. Write the inequality down now, because it is what selects the answer at the end.
One formula for the objective
For a distance to a fixed point use $D^{2}=x^{2}+y^{2}$ rather than $D$, and say in one line why you may: $D\ge 0$ and the square root is increasing, so both are smallest at the same place.
Collapse the trigonometry
One Pythagorean identity, $\sin^{2}\theta=1-\cos^{2}\theta$ or $\sec^{2}\theta=1+\tan^{2}\theta$, usually leaves a single trigonometric function. Substitute $u$ for it. The objective becomes a polynomial in $u$, and if $u=\cos\theta$ or $\sin\theta$ then $u$ runs over the closed interval $[-1,1]$ no matter what the time window was.
Extremes in $u$, endpoints included
On $[-1,1]$ use the Closed Interval Method: compare the value at each interior solution of $\frac{dQ}{du}=0$ with the values at $u=-1$ and $u=1$. Both endpoints are attainable values of a sine or cosine, so they are genuine candidates, and the answer frequently sits at one of them.
Back to the family
Solve the resulting trigonometric equation for the original variable and keep the parameter: sine and cosine repeat every $2\pi$, tangent every $\pi$. Carry the family through the substitution, so $\theta=\tfrac{\pi}{4}+k\pi$ with $\theta=\tfrac{\pi t}{32}$ becomes $t=8+32k$ for every integer $k$.
Cut the family down to one member
Feed $k=\ldots,-1,0,1,\ldots$ into the family and keep only what satisfies the window from step two. Say out loud which members you are discarding and why, since that sentence is what shows the window was used.
Answer in the unit the question used
Weeks become a month, minutes after 09:00 become a clock time, and a value of $u$ becomes a distance. This last conversion is marked separately from finding the number.
Where it goes wrong
Writing only the principal solution of $\cos\theta=c$ or $\tan\theta=c$. The family with $k$ in it is worth full marks and one member of it only part.
Taking the window from the algebra instead of from the sentence, which lets a moment before the question's starting date survive to the end as the answer.
Forgetting that $u=\pm 1$ are attainable once the objective is a polynomial in $u=\cos\theta$. The extreme being asked for is often at an endpoint of $[-1,1]$ and at no zero of the derivative at all.
Handing in the value of the variable, $t=40$, when the question asked for a month.
Differentiating the distance itself rather than its square, which drags a root through every later line for no gain.
Building a function to order, and saying when none exists
The question runs backwards: instead of a formula to analyse you are handed a list of features and asked for a function of smallest degree that has them, or for the words DOES NOT EXIST. Such parts are marked all or nothing.
Rewrite every phrase as a condition on f, f′ or f″
Passes through $(a,b)$ gives $f(a)=b$; a local extremum at $a$ gives $f'(a)=0$ plus a sign change; an inflection at $a$ gives $f''(a)=0$ plus a sign change; $y=L$ is a horizontal asymptote gives $\lim_{x\to\infty}f=L$. Keep the equation and the sign change apart: you solve with the first and check with the second.
Count what the family can hold before choosing one
A polynomial of degree $n$ has $\deg f'=n-1$ and $\deg f''=n-2$. Two extremes need two zeros of $f'$, so $n\ge 3$; two inflection points need two sign changes of $f''$, so $n\ge 4$. Start at the smallest $n$ that survives the count, since smallest degree is what is marked.
Build from the derivative down, not from the coefficients up
Write $f'$ factored with the required zeros in it, fix the leading sign from the order of the extremes, integrate, then pin the constant with a value condition. Solving four equations in four coefficients gives the same answer for four times the work.
Check the half that was not an equation
$f'(a)=0$ holds at saddles too. Confirm with the sign of $f'$ around $a$, or with $f''(a)$. A construction that meets every equation and fails one sign change scores what a blank scores.
Before hunting further, look for a structural obstruction
Three cover most impossible parts. A degree count: $f''$ of a cubic is linear, so a cubic cannot have two inflection points. A theorem: between two local minima of a continuous function there is an interior maximum. A limit fact: a nonconstant polynomial has no horizontal asymptote.
Write the verdict in the form the paper wants
Either the formula with the line that certifies it, or DOES NOT EXIST with the sentence that forbids it. An impossible part is answered by its reason.
Where it goes wrong
Offering a function that meets every equation and no sign change, such as $f(x)=x^{3}$ for a local extremum at $0$ because $f'(0)=0$.
Choosing a degree higher than necessary because it is easier to fit; smallest possible degree is part of the question.
Answering an impossible part with a blank, or with a near miss and no comment. The reason is the whole mark.
Reading no critical number as $f'$ undefined rather than $f'$ never zero, which sends the search to corners instead of to a derivative of constant sign.
Most area for 36 m of fence
Fixed perimeter, and the area is what we are allowed to move.
Test a nearby rectangle with the same area: $6\times 13.5$ needs $39$ m of fence and $3\times 27$ needs $60$ m. Both are worse.
Area was the constraint, perimeter the objective — the exact reverse of the problem on the left.
The two problems produce the same square, and they are still different problems: the equation that gets differentiated on the left is the one that gets solved and substituted on the right, and the interval changes from bounded to infinite, which changes the closing argument from a sign check on $(0,18)$ to Theorem 7.2 on $(0,\infty)$.
How to tell them apart
Find the number in the problem. The quantity with the number attached to it is the constraint; the quantity carrying the word largest or smallest is the objective. Swap them and you will differentiate the wrong function and get an answer that is somebody else's correct answer.
Scaffolding comes off
The common skeleton
Draw the object and put a letter on every length.
Write the objective, the quantity that has to come out largest or smallest.
Write the constraint, the requirement with a number attached, and solve it for one variable.
Substitute, so the objective depends on one variable, and state the interval on which that variable builds a real object.
Differentiate, solve for the critical numbers, and throw away the ones outside the interval.
Say in one sentence why the survivor is the absolute one, then report both the dimensions and the value of the objective.
1 · fully worked
The biggest rectangle under the parabola y = 12 − x²
Every step written out, reasons included. Read it once, then cover it and try the next rung.
Given
the rectangle sits on the $x$ axis
its two upper corners are on the curve $y=12-x^{2}$
Find
the dimensions of the rectangle of largest area, and that area
SolutionDraw and name
$$\text{corners at } (\pm x, 0) \text{ and } (\pm x, 12-x^{2})$$
the curve is even, so a rectangle placed symmetrically is the only shape worth considering, and one letter describes it
$$\text{width}=2x,\qquad \text{height}=12-x^{2}$$
read off the picture, not guessed
Objective and constraint
$$A=2x(12-x^{2})=24x-2x^{3}$$
here the constraint, "the corners are on the curve", has already been used in writing the height, so the substitution step is free
$$0\lt x\lt \sqrt{12}$$
beyond $\sqrt{12}$ the height is negative and there is no rectangle
Differentiate and close
$$A'(x)=24-6x^{2}=6(4-x^{2})$$
factoring makes the sign readable at a glance
$$A'(x)=0 \iff x=2$$
only the positive root is in the interval
$$A'>0 \text{ on } (0,2),\quad A'\lt 0 \text{ on } (2,\sqrt{12})$$
rising then falling across the whole interval, so this is the absolute maximum by Theorem 7.2
Independent check with two neighbours: $x=1.8$ gives $A=3.6\times 8.76=31.5$ and $x=2.2$ gives $A=4.4\times 7.16=31.5$. Both fall short of $32$, and symmetrically, which is what a smooth maximum looks like.
The height came out exactly two thirds of the parabola's peak, $8$ out of $12$. That ratio is the same for every parabola $y=k-x^{2}$, which is worth a check the next time this shape appears.
2 · you write the reasoning
Easier on purpose: the numbers are small and the algebra is one line. The steps are already written; your job is the reason column. Say out loud why each line is allowed before opening the reasoning under it. Problem: a rectangle has perimeter $40$ cm; find the largest possible area.
$2x+2y=40 \Rightarrow y=20-x$
reasoning
The perimeter equation is the constraint: it is the requirement with a number in it, and $y$ appears once and to the first power, so it is the cheapest variable to eliminate.
$A(x)=x(20-x)=20x-x^{2}$, for $0\lt x\lt 20$
reasoning
Substituting turns a two variable objective into a one variable function, which is the only form that can be differentiated. The interval comes from the picture: both sides must be positive.
$A'(x)=20-2x=0 \iff x=10$
reasoning
A polynomial is differentiable everywhere, so the zeros of $A'$ are the only critical numbers, and $10$ lies inside the interval, so it is admissible.
$A'>0$ on $(0,10)$ and $A'\lt 0$ on $(10,20)$
reasoning
One test value in each piece is enough because $A'$ is continuous and has no other zero: $A'(5)=10>0$ and $A'(15)=-10\lt 0$.
$x=10$ cm, $y=10$ cm, and the largest area is $100\ \mathrm{cm^{2}}$
reasoning
Rising then falling across the whole interval makes this the absolute maximum, not merely a local one, and the answer names both the dimensions and the area they enclose.
3 · find the buried error
Harder, and this time the work is done for you — badly. Exactly two of the four steps below are wrong. Find both. Problem: an open box is made from a $30\ \mathrm{cm}\times 16\ \mathrm{cm}$ sheet by cutting a square of side $x$ from each corner and folding the flaps up. Find the largest possible volume.
Step 1. $V(x)=x(30-2x)(16-2x)=4x^{3}-92x^{2}+480x$ — the base measures $(30-2x)$ by $(16-2x)$ and the height is $x$.
Step 2. $V'(x)=12x^{2}-184x+480=4(3x^{2}-46x+120)$, which vanishes at $x=\tfrac{10}{3}$ and at $x=12$; both are critical numbers, so both must be tested.
Step 3. $V''(x)=24x-184$, so $V''(10/3)=80-184=-104\lt 0$ (a maximum) and $V''(12)=288-184=104>0$ (a minimum).
Step 4. The maximum is therefore $x=10/3\approx 3.33\ \mathrm{cm}$.
the two buried errors (2)
⚠ step 2
$x=12$ is admitted as a critical number. It is not one: with $x=12$ the shorter side of the base is $16-2(12)=-8$, so there is no box. The interval was never written down, and step five of the skeleton exists precisely to prevent this.
The quadratic really does have two roots, and nothing in the algebra objects. The check that rejects $12$ lives in the picture, not on the page, so it is the step people stop doing under time pressure.
right
The flaps force $30-2x>0$ and $16-2x>0$ and $x>0$, so $0\lt x\lt 8$. The only admissible critical number is $x=10/3$, and the whole of step 3 about $x=12$ can be deleted.
⚠ step 4
The cut size is reported as the volume. $x=10/3\ \mathrm{cm}$ is where the maximum happens; the maximum itself is $V(10/3)=\tfrac{10}{3}\cdot\tfrac{70}{3}\cdot\tfrac{28}{3}=\tfrac{19600}{27}\approx 725.93\ \mathrm{cm^{3}}$.
The critical number is the last thing computed and it is a number, so it looks like the result. The units give it away: a volume cannot be measured in centimetres.
right
Write the final line as a sentence with both halves in it: the volume is largest when $x=10/3\ \mathrm{cm}$, and that largest volume is $19600/27\approx 725.93\ \mathrm{cm^{3}}$.
4 · the bare problem
§07.3 — the same skeleton with no scaffolding●●●○○
No steps and no hints inside the statement. The skeleton at the top of this ladder is the whole method; run it. A tin can with no lid is to be made from a sheet of metal.
Given
a circular cylinder, open at the top, of radius $r$ and height $h$
the volume must be exactly $500\ \mathrm{cm^{3}}$
for an open cylinder the material is $S=\pi r^{2}+2\pi r h$
Find
Find the radius and the height that use the least metal, and state how much metal that is.
Hint 1/4
The plan is fixed: objective, constraint, eliminate, interval, differentiate, close. Decide first which of the two given equations is which.
Hint 2/4
The volume equation is the constraint, so solve it for $h$ and substitute into $S$; the interval is $r>0$, which is open, so the closing argument will be the sign of $S'$ on both sides.
Hint 3/4
Given again: $\pi r^{2}h=500$ and $S=\pi r^{2}+2\pi r h$. Substituting $h=500/(\pi r^{2})$ gives $S(r)=\pi r^{2}+1000/r$ for $r>0$.
Hint 4/4
$S'(r)=2\pi r-1000/r^{2}=0$ gives $r^{3}=500/\pi$, so $r\approx 5.419$ cm, $h=r$, and $S=3\pi r^{2}\approx 276.79\ \mathrm{cm^{2}}$.
Show solutionEliminate and state the interval
$$h=\frac{500}{\pi r^{2}}$$
the constraint, solved for the variable that appears once
Independent check with two other cans of the same volume: $r=4$ needs $\pi(16)+250=300.3\ \mathrm{cm^{2}}$ and $r=7$ needs $\pi(49)+142.9=296.8\ \mathrm{cm^{2}}$. Both are above $276.79$, one on each side.
Closed can: $h=2r$. Open can: $h=r$. Removing the lid halves the height, and the same substitution produced both — which is the point of learning the skeleton rather than the answer.
Full exam-style question
Cutting a 100 cm wire into a square and a triangleexam format
Written the way this appears on a paper: two parts, one of which has its answer at an endpoint. Both parts are marked on the argument, not only on the number.
Given
a wire of length $100\ \mathrm{cm}$ is cut into two pieces
one piece is bent into a square, the other into an equilateral triangle
let $x$ be the length of wire used for the square, so $0\le x\le 100$
the area of an equilateral triangle of side $s$ is $\dfrac{\sqrt3}{4}s^{2}$
Find
the cut that makes the total enclosed area smallest, and the cut that makes it largest
SolutionBuild the objective
$$\text{square: side }\frac{x}{4},\ \text{area } \frac{x^{2}}{16}$$
a square of perimeter $x$ has side $x/4$
$$\text{triangle: side }\frac{100-x}{3},\ \text{area } \frac{\sqrt3}{4}\cdot\frac{(100-x)^{2}}{9}=\frac{\sqrt3(100-x)^{2}}{36}$$
$$\Rightarrow \min \text{ at } x\approx 43.50,\qquad \max \text{ at } x=100$$
the Closed Interval Method, which is available because the interval is closed
Answer in words
$$\text{smallest total area: cut at } 43.50\ \mathrm{cm},\ \text{giving } 271.85\ \mathrm{cm^{2}}$$
square of side $10.87$ cm, triangle of side $18.83$ cm
$$\text{largest total area: do not cut at all, bend the whole wire into the square}$$
the maximum sits at an endpoint, which is exactly what the interior critical number could never have told you
Answer $$\boxed{\text{min } A\approx 271.85\ \mathrm{cm^{2}} \text{ at } x\approx 43.50\ \mathrm{cm};\qquad \text{max } A=625\ \mathrm{cm^{2}} \text{ at } x=100}$$
Check
Independent check on the shape of $A$: $A''(x)=\tfrac18+\tfrac{\sqrt3}{18}>0$ for every $x$, so $A$ is concave up everywhere. A concave up function on a closed interval has its minimum at the interior critical point and its maximum at an endpoint, which is what the three comparisons produced.
One objective, one derivative, one critical number, three evaluations.
The part that costs marks is the maximum. Nothing inside the interval is a candidate for it, and a script that differentiates, finds $43.50$ and stops has answered only half the question.
Practice
A · concept 3 questions
1§07.2 — how close an asymptote has to stay●●○○○
A statement of the kind that opens a quiz. Decide true or false and be ready with a reason or a counterexample, because that is what the marks are for.
Given
Claim: "A graph never touches or crosses its own slant asymptote."
Find
True or false, with justification.
Hint 1/4
Ask what the word asymptote actually claims: something about the gap far out, or something about the gap everywhere?
Hint 2/4
The line $y=mx+b$ is a slant asymptote when $f(x)-(mx+b)\to 0$ as $x\to\pm\infty$. That is a statement about the limit of the gap, and a limit says nothing about the value at any particular $x$.
Hint 3/4
Given again: the claim is about every point. Take $f(x)=x+\dfrac{1}{x^{2}+1}$; the gap $f(x)-x=\dfrac{1}{x^{2}+1}$ tends to $0$, so $y=x$ is a slant asymptote, and the gap is never zero here — but nothing stops a different function's gap from vanishing. Try $f(x)=x+\dfrac{x}{x^{2}+1}$.
Hint 4/4
False: for $f(x)=x+\dfrac{x}{x^{2}+1}$ the gap tends to $0$ and equals $0$ at $x=0$, so the graph crosses the line $y=x$ there.
Show solutionCheck that the line really is an asymptote
$$f(x)-x=\frac{x}{x^{2}+1}$$
the gap between the graph and the candidate line
$$\lim_{x\to\pm\infty}\frac{x}{x^{2}+1}=0$$
divide top and bottom by $x^{2}$: the top tends to $0$, the bottom to $1$
$$\Rightarrow y=x \text{ is a slant asymptote}$$
which is the definition, applied at both ends
Find a crossing
$$f(0)-0=\frac{0}{0+1}=0$$
the gap is zero at the origin
$$\Rightarrow \text{the graph meets } y=x \text{ at } (0,0)$$
so "never touches" is false
Answer $$\text{False.}$$
Check
Independent check on a second point: $f(1)=1+\tfrac12=1.5$ against the line value $1$, a gap of $0.5$; $f(10)\approx 10.099$ against $10$, a gap of $0.099$. The gap really is shrinking towards zero while having been zero at the origin.
Asymptote is a statement about a limit. Crossing is a statement about a value. Neither one constrains the other.
2§07.2 — which quotient hides a slanted line●●○○○
Four rational functions, and only one of them settles onto a slanted line at its two ends. No division is needed to decide: the degrees do it.
Given
The four candidates are $\dfrac{x^{2}+1}{x-1}$, $\dfrac{x^{3}+1}{x-1}$, $\dfrac{x^{2}+1}{x^{2}-1}$ and $\dfrac{x+1}{x^{2}-1}$
In each one the numerator and denominator have no common factor
Find
Which function has a slant asymptote?
Hint 1/4
You are choosing, not computing. Ask what has to be true about the two degrees before a slanted line can appear at all.
Hint 2/4
A slant asymptote appears exactly when $\deg(\text{numerator})=\deg(\text{denominator})+1$: one higher, no more and no less.
Hint 3/4
Given again: the four degree pairs are $(2,1)$, $(3,1)$, $(2,2)$ and $(1,2)$. Only one of those has a gap of exactly one.
Hint 4/4
The pair $(2,1)$ is the one, so $\dfrac{x^{2}+1}{x-1}$ has a slant asymptote — and dividing gives $y=x+1$.
Numerical check at $x=101$: the function gives $10202/100=102.02$ while the line gives $102$, a gap of $0.02=2/100$, exactly the remainder term.
The degree test decides whether to divide at all. Doing the division on a $(3,1)$ pair is not wrong, it just does not produce a line.
3§07.6 — what agreement between two iterates proves●●●○○
A claim about stopping rules, of the kind a quiz uses to separate the procedure from the reason behind it.
Given
Claim: "If two successive Newton iterates satisfy $|x_{n+1}-x_{n}|\lt 10^{-6}$, then $x_{n+1}$ is within $10^{-6}$ of a root."
Find
True or false, with justification.
Hint 1/4
Ask what the two quantities in the claim are: one is the length of a step, the other is a distance from a root. What connects them?
Hint 2/4
The step is $|x_{n+1}-x_{n}|=|f(x_{n})/f'(x_{n})|$. Near a simple root each step covers almost the whole remaining distance, so step and error are comparable. Near a repeated root it covers only a fixed small fraction of it.
Hint 3/4
Given again: the claim asserts step below $10^{-6}$ forces error below $10^{-6}$. Test it on $f(x)=(x-1)^{100}$, whose only root is $x=1$ and whose iteration simplifies to $x_{n+1}=x_{n}-\dfrac{x_{n}-1}{100}$.
Hint 4/4
False: from $x_{n}=1.0001$ the step is $0.0001/100=10^{-6}$, below the tolerance, while the distance to the root is $10^{-4}$, a hundred times larger.
and the iterate is still four decimal places away from the root
Name the condition that failed
$$f'(1)=0$$
the root is repeated, so Note 7.4 does not apply and the error is not squared at each step
Answer $$\text{False.}$$
Check
Independent check of the repair: at $x=1.0001$ the value $f(x)=(10^{-4})^{100}$ is fantastically small, which shows that even evaluating $f$ needs care at a repeated root; the honest test there is the step ratio, which stays at $0.99$ instead of collapsing.
A stopping test measures whether the iteration has slowed down. Only the value of $f$, or a guarantee that the root is simple, measures whether it has arrived.
B · computation 5 questions
1§07.1 — full analysis of x + 1/x●●○○○
A short formula with everything in it: a break in the domain, a symmetry, a slanted line, two turning points and no inflection point. Run the checklist in order.
Given
$f(x)=x+\dfrac{1}{x}$
Find
(a) State the domain and every intercept.
(b) Test for symmetry.
(c) Find every asymptote, each written as the equation of a line.
(d) Find the intervals of increase and decrease and classify each critical point, with its value.
(e) Find the intervals of concavity, and say why there is no inflection point.
Hint 1/4
Do not differentiate yet. The first three parts are read off the formula, and they decide how many pieces the picture has.
Hint 2/4
Domain first, then $f(-x)$ for symmetry, then one sided limits at the missing point for the vertical asymptote and the behaviour of the leftover term for the slanted one. Only then $f'$ and $f''$.
Hint 3/4
Given again: $f(x)=x+1/x$, which is already in divided form, so the leftover term is $1/x$ and it dies at both ends. Differentiating gives $f'(x)=1-1/x^{2}$ and $f''(x)=2/x^{3}$.
Hint 4/4
Domain $x\ne 0$; odd; asymptotes $x=0$ and $y=x$; local maximum $(-1,-2)$ and local minimum $(1,2)$; concave down for $x\lt 0$, up for $x>0$, and $0$ is not in the domain so there is no inflection point.
Sanity check on the strange looking pair of values: the local maximum $-2$ sits below the local minimum $2$. That is not a contradiction, because the two live on different pieces of the domain, separated by the vertical asymptote — and it is the standard trap this function is set to catch.
A local maximum lower than a local minimum is normal whenever a vertical asymptote separates them. Never compare extrema across a gap in the domain.
2§07.2 — slant asymptote, and whether it is met●●○○○
A top heavy quotient. One division answers both parts, which is the reason to divide before doing anything else.
Given
$f(x)=\dfrac{3x^{2}+2x-4}{x+1}$
Find
(a) Find the slant asymptote.
(b) Decide whether the graph ever crosses it, and if so where.
(c) Say which side of the line each branch of the graph runs on.
Hint 1/4
Part (b) is not a separate computation. Ask which single quantity measures the distance between the graph and the line.
Hint 2/4
Divide: $f(x)=q(x)+R/(x+1)$. The line is $y=q(x)$, the gap is the leftover, and the graph crosses the line exactly where the leftover is zero.
Hint 3/4
Given again: $3x^{2}+2x-4=(x+1)(3x-1)-3$, so $f(x)=3x-1-\dfrac{3}{x+1}$ and the gap is $-3/(x+1)$.
Hint 4/4
Asymptote $y=3x-1$; the gap $-3/(x+1)$ is never zero, so there is no crossing; it is negative for $x>-1$ and positive for $x\lt-1$.
Show solutionDivide
$$3x^{2}+2x-4=(x+1)(3x-1)-3$$
check by expanding: $3x^{2}+2x-1-3=3x^{2}+2x-4$
$$f(x)=3x-1-\frac{3}{x+1}$$
we divide rather than guess the line from the leading coefficients, because the constant term of the quotient is part of the answer
Read off the three answers
$$\frac{3}{x+1}\to 0 \Rightarrow y=3x-1 \text{ is a slant asymptote}$$
the leftover dies at both ends
$$f(x)-(3x-1)=-\frac{3}{x+1}\ne 0 \text{ for every } x$$
a nonzero numerator makes the gap nonzero everywhere, so there is no crossing
the sign of the gap is the opposite of the sign of $x+1$
Answer $$\boxed{y=3x-1;\ \text{no crossing};\ \text{below for } x>-1,\ \text{above for } x\lt-1}$$
Check
Numerical check on both sides: $f(9)=\dfrac{243+18-4}{10}=25.7$ against $3(9)-1=26$, so below by $0.3=3/10$; and $f(-11)=\dfrac{363-22-4}{-10}=-33.7$ against $-34$, so above by $0.3$. Both match $|{-3}/(x+1)|$.
When the remainder is a nonzero constant there is never a crossing. When the remainder has a root in the domain, that root is exactly where the graph meets its asymptote.
3§07.3 — the point of a parabola nearest a given point●●●○○
A minimisation where the objective is a distance. There is a standard trick that removes the square root before any differentiation happens, and the first part is there to make you justify it.
Given
the curve $y=x^{2}$
the fixed point $(0,3)$
Find
(a) Explain why minimising the square of the distance gives the same answer as minimising the distance.
(b) Find every point of the curve nearest to $(0,3)$.
(c) Give the shortest distance, exactly and to four decimal places.
Hint 1/4
A point of the curve is $(x,x^{2})$, so everything in this problem is a function of the single variable $x$ already; no constraint has to be eliminated.
Hint 2/4
Distance from $(x,x^{2})$ to $(0,3)$ is $\sqrt{x^{2}+(x^{2}-3)^{2}}$, and squaring is legitimate because $t\mapsto t^{2}$ is increasing on $[0,\infty)$, so it preserves which input is smallest.
Hint 3/4
Given again: the point is $(0,3)$ and the curve is $y=x^{2}$. The square of the distance is $D(x)=x^{2}+(x^{2}-3)^{2}=x^{4}-5x^{2}+9$.
Hint 4/4
$D'(x)=4x^{3}-10x=2x(2x^{2}-5)$, so $x=0$ or $x=\pm\sqrt{5/2}$; comparing $D(0)=9$ with $D(\pm\sqrt{5/2})=2.75$ gives the nearest points $\left(\pm\sqrt{5/2},\,5/2\right)$ at distance $\sqrt{11}/2$.
Show solutionJustify squaring, then square
$$d(x)=\sqrt{x^{2}+(x^{2}-3)^{2}}\ \ge 0$$
distances are never negative, which is what makes the next line legal
$$t\mapsto t^{2} \text{ is increasing on } [0,\infty) \Rightarrow \arg\min d=\arg\min d^{2}$$
an increasing function preserves order, so it preserves the position of the minimum
Independent check by direct distance: from $(1.5811,2.5)$ to $(0,3)$ the horizontal gap is $1.5811$ and the vertical gap is $0.5$, so the distance is $\sqrt{2.5+0.25}=\sqrt{2.75}=1.6583$. The point $(0,0)$ of the curve is $3$ away, which is further, as the comparison said.
Two answers, not one, and the symmetry of the picture is what predicts that before any algebra. If a distance problem is symmetric about a line and the answer is not on that line, expect a pair.
4§07.5 — a cube root by hand●●○○○
Newton's method on a function whose root you can check on a calculator, so that the run itself is what is being practised.
Given
$f(x)=x^{3}-20$
$x_{0}=3$
Find
(a) Write the iteration for this $f$ and simplify it.
(b) Compute $x_{1}$ to $x_{4}$, keeping ten decimal places.
(c) State the root to six decimal places and check it.
Hint 1/4
Write down $f'$ first; it is the only symbolic work in the whole problem and it is reused at every step.
Hint 2/4
$x_{n+1}=x_{n}-\dfrac{f(x_{n})}{f'(x_{n})}$ with $f'(x)=3x^{2}$; simplifying the fraction before substituting numbers saves an arithmetic step each time.
Hint 3/4
Given again: $f(x)=x^{3}-20$ and $x_{0}=3$. The simplified iteration is $x_{n+1}=\dfrac{2x_{n}+20/x_{n}^{2}}{3}$.
Hint 4/4
The iterates are $3$, $2.7407407407$, $2.7146696246$, $2.7144176400$, $2.7144176166$, so $\sqrt[3]{20}\approx 2.714418$.
the check that matters: the answer really is a cube root of $20$
Answer $$\boxed{\sqrt[3]{20}\approx 2.714418}$$
Check
Independent check by bracketing: $2.7^{3}=19.683\lt 20$ and $2.72^{3}=20.1237\gt 20$, so the root is between $2.7$ and $2.72$, which contains the answer.
The corrections went $0.26,\ 0.026,\ 0.00025,\ 2\times 10^{-8}$: each roughly the square of the one before. Watching the corrections is the cheapest way to see whether a run is behaving.
5§07.4 — the largest rectangle in a semicircle●●●○○
A geometric optimisation on an interval whose endpoints give degenerate rectangles, so the closing argument has something to say.
Given
a semicircle of radius $5$, sitting on its diameter along the $x$ axis, so its curve is $y=\sqrt{25-x^{2}}$
a rectangle is inscribed with its base on the diameter and its two upper corners on the curve
Find
(a) Write the area as a function of one variable and state the interval.
(b) Find the dimensions of the largest such rectangle.
(c) Give the largest area, and say why it is the absolute maximum.
Hint 1/4
Put the rectangle symmetrically about the $y$ axis and let $x$ be the half width; then both upper corners are described by the same letter.
Hint 2/4
The corners are at $(\pm x,\sqrt{25-x^{2}})$, so the width is $2x$ and the height is $\sqrt{25-x^{2}}$; maximising $A^{2}$ instead of $A$ removes the square root, which is legitimate because $A\ge 0$.
Hint 3/4
Given again: radius $5$, so $A(x)=2x\sqrt{25-x^{2}}$ for $0\lt x\lt 5$, and $A(x)^{2}=4x^{2}(25-x^{2})=100x^{2}-4x^{4}$.
Hint 4/4
Differentiating the square: $200x-16x^{3}=8x(25-2x^{2})=0$ gives $x=5/\sqrt2$, so the rectangle is $5\sqrt2$ wide and $5/\sqrt2$ tall, with area $25$.
Independent check against the semicircle it sits in: the semicircle has area $\tfrac12\pi(25)\approx 39.27$, and the rectangle uses $25$ of it, about $64$ percent. A number above $100$ percent or below a few percent would have signalled an algebra slip at once.
The answer $25=r^{2}$ holds for every radius: the largest inscribed rectangle in a semicircle of radius $r$ has area $r^{2}$, and is always twice as wide as it is tall.
C · exam level 3 questions
1§07.1 — a full sketch with two vertical asymptotes●●●●○
A midterm sized sketching question. It has a break on each side, a slanted line, an odd symmetry that halves the work, and one critical number that turns out to be no extremum at all.
Given
$f(x)=\dfrac{x^{3}}{x^{2}-4}$
Find
(a) Domain, intercepts and symmetry.
(b) Every asymptote, each as the equation of a line, with the one sided behaviour at each vertical one.
(c) Intervals of increase and decrease, and every local extremum with its value.
(d) Intervals of concavity and every inflection point.
(e) A labelled sketch consistent with (a) to (d).
Hint 1/4
Before differentiating, divide. The identity $x^{3}=x(x^{2}-4)+4x$ gives the slanted line and makes both derivatives shorter.
Hint 2/4
Checklist order: domain, intercepts, symmetry, asymptotes, sign of $f'$, sign of $f''$, plot. Odd symmetry means everything found for $x>0$ is repeated, rotated, for $x\lt 0$.
Hint 3/4
Given again: $f(x)=x^{3}/(x^{2}-4)=x+\dfrac{4x}{x^{2}-4}$. The derivatives are $f'(x)=\dfrac{x^{2}(x^{2}-12)}{(x^{2}-4)^{2}}$ and $f''(x)=\dfrac{8x(x^{2}+12)}{(x^{2}-4)^{3}}$.
Hint 4/4
Domain $x\ne\pm2$; odd; asymptotes $x=-2$, $x=2$ and $y=x$; local maximum $\left(-2\sqrt3,-3\sqrt3\right)$ and local minimum $\left(2\sqrt3,3\sqrt3\right)$, with $x=0$ a critical number that is not an extremum; inflection point at the origin only.
Show solutionDomain, intercepts, symmetry
$$x^{2}-4=0 \text{ at } x=\pm 2 \Rightarrow \text{three pieces}$$
the picture is cut twice before any calculus
$$f(x)=0 \iff x^{3}=0 \iff x=0$$
the origin is the only intercept
$$f(-x)=\frac{-x^{3}}{x^{2}-4}=-f(x)$$
odd, so the left half is the right half rotated through the origin
Independent check of the extremum values by a second route: at a critical number $x^{2}=12$, so $f=\dfrac{x\cdot x^{2}}{x^{2}-4}=\dfrac{12x}{8}=\dfrac{3x}{2}$, which at $x=2\sqrt3$ gives $3\sqrt3$ — the same number reached without cubing anything. And a scale check on the slant asymptote: $f(100)=\dfrac{10^{6}}{9996}\approx 100.04$ against $y=100$.
Two things here are worth carrying: a critical number with an even repeated factor in $f'$ never turns the graph, and a sign change of $f''$ at a number outside the domain is not an inflection point.
2§07.4 — cheapest open container of fixed volume●●●●○
A cost minimisation of the kind that appears on a midterm with the marks split between the setup and the justification. Two materials at different prices, so the objective is not symmetric in the two dimensions.
Given
an open topped rectangular container with a square base of side $x$ metres and height $h$ metres
the volume must be exactly $10\ \mathrm{m^{3}}$
the base material costs $10$ TL per square metre and the side material $6$ TL per square metre
Find
(a) Write the total cost as a function of $x$ alone and state the interval.
(b) Find the dimensions that minimise the cost.
(c) Give the minimum cost, and prove it is the absolute minimum on that interval.
Hint 1/4
Draw the container. It has one square base and four rectangular sides, and no lid, so count the faces before pricing them.
Hint 2/4
Cost equals price times area, face by face: $10x^{2}$ for the base and $6\cdot 4xh$ for the sides. The constraint is the volume equation $x^{2}h=10$, which eliminates $h$.
Hint 3/4
Given again: volume $10$, base $10$ TL per square metre, sides $6$ TL per square metre. Substituting $h=10/x^{2}$ into $C=10x^{2}+24xh$ gives $C(x)=10x^{2}+\dfrac{240}{x}$ for $x>0$.
Hint 4/4
$C'(x)=20x-\dfrac{240}{x^{2}}=0$ gives $x^{3}=12$, so $x=\sqrt[3]{12}\approx 2.2894$ m, $h=10/x^{2}\approx 1.9079$ m and $C=\dfrac{360}{x}\approx 157.24$ TL.
Independent check by pricing two nearby containers of the same volume: $x=2$ gives $h=2.5$ and $C=40+120=160$ TL; $x=2.6$ gives $h\approx 1.479$ and $C\approx 67.6+92.3=159.9$ TL. Both are above $157.24$, one on each side of the optimum.
The expensive base pushed the container squat: a base priced the same as the sides would have given a taller shape. Prices enter through the objective, never through the constraint.
3§07.5 — one step of Newton on a transcendental equation●●●○○
An equation with no algebraic solution, of the kind Newton's method exists for. Only the first step is asked, so the arithmetic is short and the sign is where the marks are.
Given
The equation is $e^{-x}=x$
It is written as $f(x)=e^{-x}-x=0$, so $f'(x)=-e^{-x}-1$
$x_{0}=0.5$, and $e^{-0.5}=0.6065307$
Find
Which number is $x_{1}$?
Hint 1/4
Write the two ingredients before dividing anything: the height $f(x_{0})$ and the slope $f'(x_{0})$, each with its sign.
Hint 2/4
$x_{1}=x_{0}-\dfrac{f(x_{0})}{f'(x_{0})}$, and both $f(x_{0})$ and $f'(x_{0})$ have to be evaluated at $x_{0}=0.5$ before the division.
Hint 3/4
Given again: $e^{-0.5}=0.6065307$, so $f(0.5)=0.6065307-0.5=0.1065307$ and $f'(0.5)=-0.6065307-1=-1.6065307$.
a positive height divided by a negative slope is a negative correction, and subtracting it moves right
$$x_{1}=0.5663110$$
seven decimal places kept, because the next step would otherwise inherit a rounding error
Answer $$x_{1}=0.5663110$$
Check
Independent check that the step went the right way: $f(0.5)=0.107>0$ and $f(1)=0.368-1=-0.632\lt 0$, so a root lies between $0.5$ and $1$, and the new iterate $0.5663$ is inside that bracket.
On a falling curve the iterate moves towards the root from whichever side it starts. Losing the sign of $f'$ moves it the other way, which is the most common single error in this material.
D · interleaved 3 questions
1mixed practice — decide the method yourself●●●○○
Deliberately not labelled. Part of the exercise is recognising which two pieces of the course this joins together before you start writing.
Given
$f$ is differentiable, with $f(a)$ and $f'(a)\ne 0$ known
for part (c): $f(x)=x^{2}-7$ and $a=2.6$
Find
(a) Write down the linear approximation to $f$ at $a$.
(b) Solve the equation "linear approximation $=0$" for $x$, and compare the result with Newton's iteration.
(c) Use the result of (b) once on $f(x)=x^{2}-7$ from $a=2.6$, and compare with $\sqrt7=2.6457513$.
Hint 1/4
Two of the things you have learned produce the same formula. Write both down before deciding they are different.
Hint 2/4
The linear approximation, also called the linearisation, of $f$ at $a$ is $L(x)=f(a)+f'(a)(x-a)$; Newton's iteration is $x_{1}=a-f(a)/f'(a)$. Setting $L(x)=0$ and solving is one line.
Hint 3/4
Given again: $L(x)=f(a)+f'(a)(x-a)$. Setting it to zero gives $f'(a)(x-a)=-f(a)$, so $x=a-f(a)/f'(a)$, which is exactly $x_{1}$. For part (c), $f(2.6)=6.76-7=-0.24$ and $f'(2.6)=5.2$.
Hint 4/4
The root of the linear approximation is Newton's first iterate; for (c) it gives $2.6+0.24/5.2=2.6461538$, which agrees with $\sqrt7=2.6457513$ to three decimal places.
Show solutionSolve the easy equation
$$L(x)=0 \iff f(a)+f'(a)(x-a)=0$$
the linear approximation is a line, and a line can be set to zero and solved
Independent check against the known value: $\sqrt7=2.6457513$, so the single step is off by $4.0\times 10^{-4}$, while the starting guess was off by $4.6\times 10^{-2}$ — a hundredfold improvement, which is the squaring of the error in action.
Two names, one formula. Whenever a section seems to introduce a new tool, check first whether it is an old tool applied repeatedly.
2mixed practice — decide the method yourself●●●●○
Not labelled on purpose. It looks like a routine sketch until the far left end, where the usual shortcut gives the wrong sign.
Given
$f(x)=\dfrac{x}{\sqrt{x^{2}+1}}$
Find
(a) State the domain and test for symmetry.
(b) Find $\lim_{x\to\infty}f(x)$ and $\lim_{x\to-\infty}f(x)$, and name every horizontal asymptote.
(c) Show that $f$ is increasing on the whole real line.
(d) Find the intervals of concavity and every inflection point.
(e) Sketch the graph, and say whether it ever meets its asymptotes.
Hint 1/4
The denominator is never zero, so nothing breaks the domain. The care needed is at the far left end, where a square root and a negative number meet.
Hint 2/4
For the limits, divide top and bottom by $\sqrt{x^{2}}=|x|$, and remember that $|x|=-x$ when $x$ is negative. For the derivatives, write $f(x)=x(x^{2}+1)^{-1/2}$ and use the product and chain rules.
Hint 3/4
Given again: $f(x)=x/\sqrt{x^{2}+1}$. Differentiating gives $f'(x)=(x^{2}+1)^{-3/2}$ and $f''(x)=-3x(x^{2}+1)^{-5/2}$.
Hint 4/4
Domain all of $\mathbb{R}$; odd; horizontal asymptotes $y=1$ at the right and $y=-1$ at the left; $f'>0$ everywhere so $f$ is increasing with no extremum; concave up on $(-\infty,0)$, concave down on $(0,\infty)$, inflection point $(0,0)$; and $|f(x)|\lt 1$ for every $x$, so the graph never meets either asymptote.
Show solutionDomain and symmetry
$$x^{2}+1>0 \text{ for every } x \Rightarrow \text{domain }\mathbb{R}$$
Independent check of the left hand limit by direct evaluation rather than by algebra: $f(-100)=-100/\sqrt{10001}\approx-0.99995$, which is near $-1$ and not near $+1$. That single arithmetic line catches the missing minus sign in one second.
Whenever a square root of $x^{2}$ appears, the two ends have to be done separately. It is the standard trap of any question mixing limits at infinity with roots.
3mixed practice — decide the method yourself●●●●○
Not labelled. A journey in two stages at two different speeds, and the only decision to make is where to change from one to the other.
Given
a straight river $3$ km wide, with a target $8$ km downstream on the opposite bank
rowing speed $6$ km per hour, running speed $8$ km per hour
let $x$ be the distance, in km, from the point directly opposite the start to the landing point
Find
(a) Write the total travel time as a function of $x$ and state the interval.
(b) Find the landing point that minimises the time.
(c) Give the shortest time exactly and to four decimal places, and compare it with rowing straight across and with rowing directly to the target.
Hint 1/4
Draw the two legs. One is a slanted crossing whose length comes from a right triangle, the other is a straight run along the bank.
Hint 2/4
Time is distance over speed, added leg by leg: the crossing has length $\sqrt{9+x^{2}}$ at $6$ km per hour, the run has length $8-x$ at $8$ km per hour. Differentiating the square root needs the chain rule.
Hint 3/4
Given again: width $3$, target $8$ downstream, speeds $6$ and $8$. So $T(x)=\dfrac{\sqrt{9+x^{2}}}{6}+\dfrac{8-x}{8}$ for $0\le x\le 8$.
Hint 4/4
$T'(x)=\dfrac{x}{6\sqrt{9+x^{2}}}-\dfrac18=0$ gives $8x=6\sqrt{9+x^{2}}$, hence $28x^{2}=324$ and $x=\dfrac{9}{\sqrt7}\approx 3.4017$ km, with $T=1+\dfrac{\sqrt7}{8}\approx 1.3307$ hours.
Independent check by testing a landing point on each side: $x=3$ gives $T=\sqrt{18}/6+5/8\approx 0.7071+0.6250=1.3321$ and $x=4$ gives $T=5/6+0.5\approx 1.3333$. Both are above $1.3307$, and only just, which is what a flat minimum looks like.
Because running is faster than rowing, the optimum lands short of the target and runs the rest. If the two speeds were equal the answer would be the straight line, $x=8$ — worth checking as a limiting case.
Shaped like the real papers 4 questions
1§07.2 — asymptotes of a radical function, proved●●●●●
The opening question of a Quiz 2 in this course, and about fifteen minutes of it. Both parts are settled by limits rather than by dividing one polynomial into another, and the instruction to prove the claim is not decoration: an asymptote written down without the limit that produces it earns nothing.
Given
$g(x)=x+\sqrt{x^{2}+4x}$
$h(x)=\dfrac{x^{4}-3x^{2}+6x-4}{x^{2}-4x+3}$
Find
(a) State the domain of $g$, then find every horizontal and every slant asymptote of $g$. Prove each claim with the limit that produces it.
(b) Find every vertical asymptote of $h$. Prove each claim, and for any zero of the denominator that is not an asymptote say what happens there instead.
Hint 1/4
An asymptote is a statement about a limit, so decide first which limits each part is asking you to take. In (a) the domain of $g$ is not a single interval, and its two infinite ends are two separate questions rather than one question answered twice. In (b), count the zeros of the denominator, then ask of each one whether the quotient really runs off to infinity there.
Hint 2/4
For a line $y=mx+b$ to be an asymptote as $x\to\infty$ you need two limits, in this order: $m=\lim_{x\to\infty}\dfrac{f(x)}{x}$, and then $b=\lim_{x\to\infty}\left(f(x)-mx\right)$. If $m=0$ the line is horizontal. Repeat the pair separately at $-\infty$, and remember that $\sqrt{x^{2}}=|x|=-x$ when $x<0$. A line $x=c$ is a vertical asymptote of a quotient only when the quotient itself becomes unbounded there, so a common zero of numerator and denominator must be cancelled first and then re-examined.
Hint 3/4
Given again: $g(x)=x+\sqrt{x^{2}+4x}$ and $h(x)=\dfrac{x^{4}-3x^{2}+6x-4}{x^{2}-4x+3}$. Two openings: $\left(\sqrt{x^{2}+4x}-x\right)\left(\sqrt{x^{2}+4x}+x\right)=4x$, and $x^{2}-4x+3=(x-1)(x-3)$ while the numerator of $h$ vanishes at $x=1$.
Hint 4/4
$g$ has domain $(-\infty,-4]\cup[0,\infty)$, the slant asymptote $y=2x+2$ as $x\to\infty$, and the horizontal asymptote $y=-2$ as $x\to-\infty$. $h$ has the single vertical asymptote $x=3$; at $x=1$ the limit is $-2$, so there is a hole there and no asymptote.
Show solutionThe domain of $g$, and why its two ends are two problems
the cancelled form is the one that decides the behaviour, and it is valid everywhere except at the cancelled point itself
$$\lim_{x\to 1}h(x)=\frac{1+1-2+4}{1-3}=-2$$
finite, so $x=1$ is a hole at $(1,-2)$ and not an asymptote, even though the original denominator is zero there
$$x=3:\ \ 27+9-6+4=34\ne 0$$
the numerator of the cancelled form does not vanish at $3$, which is exactly the condition for the quotient to blow up
$$\lim_{x\to 3^{-}}h(x)=-\infty,\qquad \lim_{x\to 3^{+}}h(x)=+\infty\ \Rightarrow\ x=3 \ \text{ is a vertical asymptote}$$
with a positive numerator, the sign of $h$ near $3$ is the sign of $x-3$
Check
Numerical spot check of both ends of $g$: at $x=100$, $g=100+\sqrt{10400}\approx 201.98$ against $2x+2=202$; at $x=-100$, $g=-100+\sqrt{9600}\approx -2.02$ against $-2$. For $h$, put $x=1.001$ into the original quotient: numerator $\approx 0.0040$ and denominator $\approx -0.0020$ are both tiny, but their ratio is about $-2.00$, not large. That is what a hole looks like from the outside, and it is how you catch the trap without doing the division.
2§07.1 — a sketch built backwards, from data to graph●●●●●
A whole Midterm 2 question in this course, and it runs the checklist in reverse: the formula is withheld and the derivatives are handed to you. The printed table is part of the answer, not scratch work, and the last part asks you to recover the function that produced the data.
Given
$f$ is defined and twice differentiable on $\{x:x\ne 2\}$
$f$ has a vertical asymptote at $x=2$, and the line $y=x+3$ is a slant asymptote as $x\to\pm\infty$
$f(-2)=f(1)=0$, $f(0)=1$, $f(4)=9$
$f'(x)=\dfrac{x(x-4)}{(x-2)^{2}}$ and $f''(x)=\dfrac{8}{(x-2)^{3}}$
Find
(a) Give the intervals of increase and decrease, every local maximum and minimum with its value, the intervals of concavity, and every inflection point. Then fill in a table whose rows are $x$, $f'(x)$, $f''(x)$, $f(x)$ and whose columns are the split points in increasing order.
(b) Sketch the graph, marking on it every asymptote and every point named in (a), and say on which side of the slant line each branch lies.
(c) Find constants $a$, $b$, $c$ for which $f(x)=\dfrac{x^{2}+ax+b}{x+c}$ satisfies all of the conditions above.
Hint 1/4
Every number that can split the real line into pieces is already printed in the two derivative formulas. Write those numbers down in increasing order before anything else and let them be the column headings; a table is only as good as its headings, and a missing column is a missing interval.
Hint 2/4
$f'>0$ means rising and $f'<0$ means falling; a sign change of $f'$ at a point of the graph gives a local extremum. $f''>0$ means concave up and $f''<0$ concave down; an inflection point needs both a sign change of $f''$ and a point of the graph where it happens. A squared factor never changes sign, an odd power does. For (c), a quadratic over a linear factor has a slant asymptote, and long division displays it as quotient plus remainder over that factor.
Hint 3/4
Given again: $f'(x)=\dfrac{x(x-4)}{(x-2)^{2}}$, $f''(x)=\dfrac{8}{(x-2)^{3}}$, vertical asymptote $x=2$, slant asymptote $y=x+3$, and $f(0)=1$, $f(4)=9$. For (c), start from the division $\dfrac{x^{2}+ax+b}{x-2}=x+(a+2)+\dfrac{b+2a+4}{x-2}$.
Hint 4/4
Increasing on $(-\infty,0)$ and $(4,\infty)$, decreasing on $(0,2)$ and on $(2,4)$; local maximum $(0,1)$ and local minimum $(4,9)$; concave down on $(-\infty,2)$, concave up on $(2,\infty)$, and no inflection point at all. The function is $f(x)=\dfrac{x^{2}+x-2}{x-2}$, that is $a=1$, $b=-2$, $c=-2$.
Show solutionRead the split points off the given formulas
$$f'(x)=0 \iff x=0 \ \text{ or }\ x=4; \qquad f' \ \text{ is undefined at } x=2$$
the numerator of $f'$ arrives already factored, and $x=2$ is outside the domain in any case
$$f''(x)\ne 0 \ \text{ for every } x \ \text{ in the domain}$$
its numerator is the constant $8$, so $f''$ can never be zero and only the denominator can flip its sign
$$\text{columns of the table: } \ 0,\ 2,\ 4$$
every number at which some sign can change, listed in increasing order
Sign of $f'$
$$(x-2)^{2}>0 \ \text{ for } x\ne 2 \ \Rightarrow\ \operatorname{sign}f'(x)=\operatorname{sign}\left(x(x-4)\right)$$
an even power cannot carry a sign, so the whole question is the sign of one product
$$f''\lt 0 \ \text{ on } (-\infty,2), \qquad f''>0 \ \text{ on } (2,\infty)$$
an odd power changes sign exactly once, at $2$
$$\text{no inflection point}$$
concavity does change, but only across $x=2$, and $x=2$ is not in the domain; an inflection point is a point of the graph, not a boundary between two branches
forced by what is already known: $f$ decreases on $(0,2)$ toward a vertical asymptote, and decreases on $(2,4)$ down to the value $9$, so it must come from above
$$f(x)-(x+3)\longrightarrow 0 \ \text{ as } x\to\pm\infty$$
that is exactly what the given slant asymptote asserts, and it fixes the far behaviour of both branches
the numerator does not vanish at $x=2$, so the vertical asymptote really is there and nothing cancelled
Check
Two checks on (c) that use data never spent in finding $a,b,c$. Values: $f(0)=\dfrac{-2}{-2}=1$ and $f(4)=\dfrac{16+4-2}{2}=9$, both as given. Zeros: $x^{2}+x-2=(x+2)(x-1)$ vanishes exactly at $-2$ and $1$, matching $f(-2)=f(1)=0$. One more thing worth noticing in (a): the local maximum value $1$ lies below the local minimum value $9$. Nothing is wrong; the two sit on branches separated by the vertical asymptote, and comparing them would only mean something on a connected piece.
3§07.3 — an optimization whose derivative has infinitely many roots●●●●●
A full Midterm 2 optimization question in this course, carrying close to a third of the paper. The marks are spread along the whole chain: seeing what has to be made small, reducing it to one variable, differentiating it correctly, picking the right root out of infinitely many, and giving the answer back in the unit the question used.
Given
A control tower stands at the origin and all distances are in kilometres.
A survey drone repeats a fixed loop. At $t$ minutes after 09:00 its position is $x(t)=1+2\cos\!\left(\dfrac{\pi t}{50}\right)$ and $y(t)=4\sin\!\left(\dfrac{\pi t}{50}\right)$.
The flight runs from 09:00 until 11:00, so $0\le t\le 120$.
The operator may land the drone at any moment strictly later than 09:30.
Find
(a) Write the quantity that has to be made small as a function of one variable, and say in one sentence why you are entitled to work with its square instead.
(b) Find the clock time at which the operator should land the drone, and its distance from the tower at that moment. Calculators are not allowed, so leave both exact.
(c) The derivative of your reduced function vanishes at an interior point of the range of that variable. In one sentence, say why that point is not the answer to (b).
Hint 1/4
The word closest is the only thing in this question that still has to be turned into mathematics; the rest is bookkeeping. Decide what single number measures closeness, and notice that the two coordinates are tied to each other through one and the same angle, so what looks like a problem in two coordinates is a problem in one.
Hint 2/4
Minimising a nonnegative quantity and minimising its square have the same minimisers, because $s\mapsto\sqrt{s}$ is increasing on $[0,\infty)$. If the objective turns out to depend only on $\cos\theta$, the honest variable is $u=\cos\theta$ and its range is the closed interval $[-1,1]$; on a closed interval the absolute extrema are found by comparing the values at interior critical points with the values at both endpoints. Finally, an equation such as $\cos\theta=-1$ has infinitely many solutions, and only those inside the stated window are answers.
Hint 3/4
Given again: $x=1+2\cos\theta$ and $y=4\sin\theta$ with $\theta=\dfrac{\pi t}{50}$, and the window $30<t\le 120$. Use $\sin^{2}\theta=1-\cos^{2}\theta$ to remove the sine entirely, then write $u=\cos\theta$.
Hint 4/4
The squared distance is $g(u)=-12u^{2}+4u+17$ on $[-1,1]$, with $g(-1)=1$, $g(1)=9$ and $g\!\left(\tfrac16\right)=\tfrac{52}{3}$. The smallest is $g(-1)=1$, reached when $\cos\theta=-1$, that is $\theta=\pi+2k\pi$ and $t=50+100k$. The only such $t$ in the window is $t=50$: land at 09:50, one kilometre from the tower.
closeness to the tower is distance to the origin; squaring removes the root without moving the minimiser, since $D\ge 0$ and $s\mapsto\sqrt{s}$ is increasing
the answer is returned in the unit the question used, which here is clock time, and the distance stays exact
Why the interior critical point is not the answer
$$g\!\left(\tfrac16\right)=\tfrac{52}{3}>9>1$$
the interior critical point is where the drone is farthest from the tower, about $4.16$ km out
$$\text{closest} \ \Rightarrow\ u=-1, \ \text{an endpoint of the range of } \cos$$
on a closed interval an absolute extremum may sit at an endpoint, and here the one being asked for does; solving $g'=0$ alone would have produced the wrong moment
Check
The same answer without the substitution. Writing $S(t)=17+4\cos\theta-12\cos^{2}\theta$ and differentiating in $t$ gives $S'(t)=\dfrac{4\pi}{50}\sin\theta\left(6\cos\theta-1\right)$, which vanishes when $\sin\theta=0$, that is at $t=50$ and $t=100$, and when $\cos\theta=\tfrac16$. The candidate values are $S(50)=1$, $S(100)=9$ and $\tfrac{52}{3}$ at the third, against $S\approx 14.6$ at $t=30$ and $S\approx 17.1$ at $t=120$. Smallest is $S(50)=1$, as before. Geometric check: the loop is the ellipse $\dfrac{(x-1)^{2}}{4}+\dfrac{y^{2}}{16}=1$, whose leftmost point is $(-1,0)$, at distance exactly $1$ from the origin.
4§07.4 — an optimization with no endpoints to lean on●●●●○
The shape that half of this course's official optimization problems take: the objective is not a box or a fence but a line, and the variable is the place where that line touches a curve. About twelve minutes of work, and the last part is a single sentence that carries marks on its own.
Given
The parabola $y=4-x^{2}$.
For $0\lt a\lt 2$, let $\ell_{a}$ be the tangent line to that parabola at the point $\left(a,\,4-a^{2}\right)$.
$\ell_{a}$ together with the two coordinate axes bounds a triangle in the first quadrant.
Find
(a) Write the area of that triangle as a function of $a$ alone, and state the interval on which it is to be minimised. Do not differentiate yet.
(b) Find the value of $a$ that makes the area least, and the least area itself. Leave both exact.
(c) In one sentence, say what makes your answer in (b) the absolute minimum on that interval rather than merely a local one.
Hint 1/4
The number being asked about is an area, and an area needs a base and a height. Here both of them are intercepts of one and the same line, and that line is pinned down by the single number $a$. So the whole of part (a) is: write those two intercepts.
Hint 2/4
The tangent to $y=f(x)$ at $x=a$ is $y=f(a)+f'(a)(x-a)$; its $y$-intercept is its value at $x=0$ and its $x$-intercept is the solution of $y=0$. On an open interval there are no endpoint values to compare, so an absolute extremum has to be established from the sign of the derivative alone: if $A'<0$ everywhere left of $c$ and $A'>0$ everywhere right of $c$, then $A(c)$ is the absolute minimum on the whole interval.
Hint 3/4
Given again: the tangent to $y=4-x^{2}$ at $\left(a,4-a^{2}\right)$, with $0<a<2$. The slope there is $-2a$, so the tangent is $y=-2ax+a^{2}+4$, and its intercepts are $\left(\dfrac{a^{2}+4}{2a},\,0\right)$ and $\left(0,\,a^{2}+4\right)$.
Hint 4/4
$A(a)=\dfrac{\left(a^{2}+4\right)^{2}}{4a}$ on $(0,2)$, with $A'(a)=\dfrac{\left(a^{2}+4\right)\left(3a^{2}-4\right)}{4a^{2}}$; hence $a=\dfrac{2}{\sqrt3}$ and the least area is $\dfrac{32\sqrt3}{9}$.
Show solutionThe tangent line and its intercepts
$$\frac{d}{dx}\left(4-x^{2}\right)=-2x \ \Rightarrow\ \text{slope at } x=a \ \text{ is } -2a$$
the slope of a tangent is the derivative evaluated at the point of tangency
half of base times height; with the legs on the axes there is nothing else to subtract
$$a\in(0,2)$$
the interval comes from the picture, not from the algebra: $a>0$ keeps the tangent sloping down across the first quadrant and $a<2$ keeps the point of tangency above the $x$-axis
exact throughout: no decimal is asked for, and with calculators barred none is available
Check
Sample the objective on either side of the critical number, which needs no calculus at all: $A(1)=\dfrac{25}{4}=6.25$ and $A\!\left(\tfrac32\right)=\dfrac{\left(25/4\right)^{2}}{6}=\dfrac{625}{96}\approx 6.51$, both above $\dfrac{32\sqrt3}{9}\approx 6.16$, so the critical number really is a dip. Behaviour check at the ends: $A(a)\to\infty$ as $a\to 0^{+}$, which matches a nearly horizontal tangent pushed far out along the $x$-axis, and $A(a)\to 8$ as $a\to 2^{-}$; both ends exceed $\dfrac{32\sqrt3}{9}$, consistent with an interior minimum.
Mistake ledger (14 entries)
⚠ Joining samples across a hole in the domain
The values on each side are perfectly good numbers and the eye wants a single curve through them.
wrong$$f(0)=0,\ f(2)=4 \Rightarrow \text{one rising curve from } (0,0) \text{ to } (2,4)$$
right$$1\notin \text{domain} \Rightarrow \text{two separate branches, } f(0.9)=-8.1,\ f(1.1)=12.1$$
⚠ Skipping the symmetry line
It is the one step that produces no equation, so it feels like it produces nothing.
wrong$$\text{six sign computations for } f(x)=\frac{x^{2}}{x^{2}+3} \text{ on all of } \mathbb{R}$$
right$$f(-x)=f(x) \Rightarrow \text{compute on } [0,\infty) \text{ and mirror}$$
⚠ Calling the whole quotient the asymptote
The division has just been written out and the leftover term is sitting right there on the page.
wrong$$y=x+1+\frac{1}{x-1}$$
right$$y=x+1$$
⚠ Reporting a horizontal asymptote for a top heavy quotient
The habit of comparing degrees stops at "numerator bigger, so no horizontal asymptote" and never asks what replaces it.
wrong$$\frac{x^{2}}{x-1}: \ \text{no asymptote as } x\to\pm\infty$$
$N$ and $D$ are polynomials with no common factor and $\deg N=\deg D+1$
Single variable reduction
$$\boxed{\ Q(x,y)\ \text{ with }\ g(x,y)=0 \ \Longrightarrow\ Q(x) \text{ on an interval } I \ \Longrightarrow\ Q'(x)=0\ }$$
the constraint can be solved for one of the variables
First Derivative Test for Absolute Extreme Values
$$\boxed{\ f'>0 \text{ left of } c,\ f'\lt 0 \text{ right of } c \text{ on } I \ \Longrightarrow\ f(c)=\max_{I}f\ \ (\text{signs reversed for } \min)\ }$$
$c$ is a critical number and $f$ is continuous on the interval $I$, which may be open or infinite
Valid for any $f$ defined on an interval $(a,\infty)$, with no polynomial division anywhere in sight: the quotient rule $\deg N=\deg D+1$ is only the special case in which $f$ happens to be a reduced quotient of polynomials. Use it whenever the function carries a root, a fractional power or a negative power. $m=0$ makes the candidate line horizontal, and then $b=\lim f$, which must itself come out finite. If $m$ is finite but $\lim(f-mx)=\pm\infty$ there is no asymptote at that end. Run the whole pair again, separately, as $x\to-\infty$: the two ends may give two different lines.
Turning a difference of roots into a quotient at infinity
Valid wherever the displayed denominator is nonzero. This is what makes the second limit $b=\lim(f-mx)$ computable when it arrives in the form $\infty-\infty$: the square version for a square root, the cube version for a cube root, where no conjugate exists. After the repair, divide numerator and denominator by the highest power of $x$. Pair it with $\sqrt{x^{2}}=|x|=-x$ for $x\lt 0$, which is exactly why the end at $-\infty$ is a separate calculation and not a copy of the one at $+\infty$.
Optimise a monotone transform instead
$$\boxed{\ \varphi \ \text{increasing on the range of}\ Q \ \Longrightarrow\ Q\ \text{and}\ \varphi\circ Q\ \text{are extreme at the same points};\quad \min D \iff \min D^{2},\quad \max\theta \iff \max\tan\theta\ }$$
$\varphi$ must be increasing on the set of values $Q$ actually takes. Two uses carry marks in this course: a distance $D\ge 0$ replaced by $D^{2}$ so that no square root has to be differentiated, and a viewing angle $\theta\in[0,\tfrac{\pi}{2})$ replaced by $\tan\theta$, which is how an angle is maximised before inverse trigonometric functions are available. The transform moves the value but never the place, so convert back and report $Q$ itself, with its unit.
One infinite one sided limit taken from inside the domain is enough, and the two sides need not agree or even both exist. Cancel every common factor before deciding: a zero of the original denominator that cancels leaves a finite limit, so it is a hole and not an asymptote. A number where the domain simply stops, such as $x=0$ for $\sqrt{x}$, is neither, as long as the value there is finite. Use whenever the question says find the vertical asymptotes and the function is not already a reduced quotient of polynomials.
Every solution of a trigonometric equation, not just the first
$\theta_{0}$ is any one solution, found by inspection. Needed whenever the critical equation $Q'=0$ turns out to be trigonometric, so that it has infinitely many roots: the family carrying $k$ is what earns full marks and a single principal value earns only part of them. Note the periods differ, $2\pi$ for sine and cosine but $\pi$ for tangent. The answer is then the one member of the family that lies inside the window the problem allows.
write the line with every term on one side first; the distance to a curve is not a formula but a minimisation
Check yourself
Close the page and write, from memory: the eight steps of the sketching checklist in order, the degree condition that produces a slanted asymptote, the two sentences that can close an optimization argument and when each is available, the Newton iteration, and the four ways a Newton run can fail.
Say which checklist step tells you how many separate pieces a graph has, and name two features a plotted picture cannot place exactly.
c-sketch-checklist
Divide $\dfrac{x^{2}+1}{x-1}$ in your head and name both the asymptote and the side of the line each branch is on.
c-slant-asymptote
Take a worded problem and write, in under a minute, the objective, the constraint and the interval, without differentiating anything.
c-optimization-recipe
State the one sentence that closes an optimization argument on $(0,\infty)$, where there are no endpoints to compare.
c-absolute-extremum
Derive the Newton iteration from the tangent line in two lines, without looking it up.
c-newton
Look at the iterates $2.0,\ 5.3,\ 14.7,\ 45.1$ and name the failure mode and the repair.
c-newton-failure
Glossary (11 terms)
curve sketchingeğri çizimi
Producing the graph of a function from its formula by a fixed sequence of checks rather than by plotting points.
objective functionamaç fonksiyonu
The quantity an optimization problem asks to be made largest or smallest, written as a formula.
constraintkısıt
An equation the variables of an optimization problem must satisfy; it is used to remove one variable from the objective.
polinom bölmesi
Writing a quotient of polynomials as a quotient polynomial plus a remainder over the divisor, with the remainder of lower degree.
Newton's methodNewton yöntemi
A root finder that replaces the curve by its tangent at the current guess and takes the intercept of that tangent as the next guess.
iterationyineleme
One application of a rule that turns the current number into the next one, repeated to build a sequence of guesses.
initial guessbaşlangıç tahmini
The starting number of an iterative method, usually chosen from a sketch or from a sign change of the function.
denklemin kökü
A number at which the function is zero, that is, a solution of the equation written in the form $f(x)=0$.
quadratic convergencekaresel yakınsama
Behaviour in which the error of the next step is bounded by a constant times the square of the current error.
divergenceıraksama
Failure of a sequence of iterates to approach any limit, typically with the values growing without bound.
durdurma ölçütü
The condition that ends an iterative run, here the step size falling below a stated tolerance.
What comes next
§08 · Antiderivatives, area, and the definite integral
Everything so far has started from a function and asked what its derivative says. Next week the question turns around: given the derivative, recover the function — and the same limits that built the derivative are used again, this time to add up infinitely many thin pieces of area.
Sources
James Stewart, Calculus, Ninth Edition — sections 3.6 and 3.7, and the section on Newton's method in the same chapter The graphing discussion, the optimization recipe and the iteration are stated as this book states them; every function worked here is a different one. The third number on the week's line, 3.9, is the antiderivative section, and it opens the next section instead of this one.
Course syllabus, week 7: Applications of Differentiation 3.6, 3.7, 3.9 The assessment weights quoted on the card come from the same syllabus.
The stopping rule and the four failure modes Standard numerical practice rather than a theorem; what carries the answer is the value of $f$ at the final iterate.