10 worked examples24 exercises6 figuresbölüm 7 / 14
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 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.
The picture on the screen is not the graph
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
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.
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$f(-x)=f(x) \Rightarrow \text{compute on } [0,\infty) \text{ and mirror}$
The slanted line the graph settles onto
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 $\deg N=\deg D+1$
equivalently: the division leaves a quotient of degree one
$$\boxed{\ f(x)=mx+b+\frac{R(x)}{D(x)}\ \text{ with }\ \frac{R(x)}{D(x)}\xrightarrow[x\to\pm\infty]{}0 \ \Longrightarrow\ \lim_{x\to\pm\infty}\bigl[f(x)-(mx+b)\bigr]=0\ }$$
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$\frac{x^{2}}{x-1}: \ \text{no asymptote as } x\to\pm\infty$
Turning a sentence into one function of one variable
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
$$\boxed{\ \underbrace{Q(x,y)}_{\text{objective}}\ \text{ with }\ \underbrace{g(x,y)=0}_{\text{constraint}}\ \Longrightarrow\ Q(x)\ \text{ on }\ I\ \Longrightarrow\ Q'(x)=0\ }$$
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
$y=13-2(3.25)=6.5,\qquad A=3.25\cdot 6.5=21.125$
translate back into the words of the question: two sides of $3.25$ m and one of $6.5$ m
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.
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$0\lt x\lt 8 \Rightarrow \text{only } x=\tfrac{10}{3} \text{ is admissible}$
Proving it is the best and not merely a candidate
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.
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$S'\lt 0 \text{ on } (0,r_{c}),\ S'>0 \text{ on } (r_{c},\infty) \Rightarrow S(r_{c})=\min$
Replacing the curve by its tangent
Every optimization above ended with an equation that factored; change the cost slightly and $f'(x)=0$ becomes $x^{3}-x-1=0$, which no amount of factoring will open.
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 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.
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.
Most area for 36 m of fence
Fixed perimeter, and the area is what we are allowed to move.
Given
a rectangle with sides $x$ and $y$
perimeter fixed: $2x+2y=36$
Find
the largest possible area
SolutionReduce and differentiate
$y=18-x,\qquad A(x)=x(18-x),\qquad 0\lt x\lt 18$
the constraint is the perimeter equation, so the area is the objective
$A'(x)=18-2x=0 \iff x=9$
the only critical number in the interval
$A'>0 \text{ on } (0,9),\ A'\lt 0 \text{ on } (9,18) \Rightarrow \text{absolute maximum}$
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.
Mistake ledger (12 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
$f''$ continuous near $x^{*}$, $f'(x^{*})\ne 0$, and $x_{0}$ close enough to $x^{*}$
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)
eğri 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.