Differential Calculus for Social Science 201-103 • CEGEP, Montreal

Revision sheet: complete curve sketching (201-103)

Curve sketching is the long question of the final exam: it is often worth fifteen marks on its own and it is decided by the order of the steps more than by the calculations. A correct derivative set down in the wrong order loses half the marks, because each step feeds the next.

This sheet gives that order, the table of the four possible shapes with a calculation under each, the nine mistakes that come back every term, and the complete write-up of a tax revenue curve, the typical exercise of the Social Science program.

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The thread of the chapter

$f'$ says the DIRECTION, $f''$ says the CURVATURE, and the two readings are independent: a curve that goes down can very well be concave up, and the variation table says nothing until both lines are filled in.

This chapter is part of 201-103, Differential Calculus for Social Science

The essentials

The order of a study, seven steps

• The DOMAIN, first, with the constraint of the context when there is one.
• The LIMITS at the edges of the domain and at infinity.
• The ASYMPTOTES that follow, vertical, horizontal or oblique.
• $f'$ , its zeros and its SIGN: that is the direction of variation.
• $f''$ , its zeros and its SIGN: that is the concavity and the inflection points.
• The TABLE, which gathers the two sign lines and the notable values.
• The SKETCH, last, deduced from the table and never from a table of values.

For

f(x)=x^{3}-3x

: the table gathers the limits, the sign of

f'(x)=3x^{2}-3

and the values

f(-1)=2

f(1)=-2

. The sketch is then read off the arrows, with no table of values at all.

Ten points calculated on a machine show neither an asymptote, nor an inflection, nor an exact extremum. The sketch is deduced from the study, otherwise the study served no purpose.

Two independent readings

• $f'>0$ : the function RISES. $f'<0$ : it FALLS. That is all $f'$ says.
• $f''>0$ : the curve is concave UP, above its tangent lines. $f''<0$ : concave down.
• The direction of variation is NEVER read from $f''$ , and the curvature is never read from $f'$ .
• All four combinations exist: rising and speeding up, rising and slowing down, falling and slowing down, falling and speeding up.
• In context, “the increase is slowing” is a sentence about $f''$ , “the quantity is falling” a sentence about $f'$ .

The first two rise and the last two fall, but their curvatures alternate: direction and curvature combine freely, so they have to be read separately.

Half of the wrong comments in Social Science come from mixing up these two readings, especially on curves that rise while flattening out.

The three asymptotes

• VERTICAL at $a$ if $\lim_{x\to a}f=\pm\infty$ . It marks an edge of the domain.
• HORIZONTAL $y=L$ if $\lim_{x\to\pm\infty}f=L$ . It describes the long run.
• OBLIQUE $y=mx+p$ if $f(x)-(mx+p)\to 0$ . It appears when the degree of the numerator exceeds that of the denominator by exactly one.
• A curve has at most ONE horizontal or oblique asymptote on each side, never both on the same side.
• An average cost of the form $\dfrac{F}{d}+ad+b$ has a vertical asymptote at $0$ , an oblique asymptote $y=ad+b$ , and a minimum where the two variable terms balance.

Vertical: the curve shoots off to infinity near a forbidden value. Horizontal: it levels off in the long run. Oblique: it ends up running alongside a rising line, without ever touching it.

Announcing a horizontal asymptote AND an oblique asymptote on the same side is a contradiction: the curve cannot tend toward two things at once.

Local extrema and absolute extrema

• $f'(a)=0$ makes $a$ a CANDIDATE, not an extremum. The cubic $x^{3}$ is the reminder at $0$ .
• First derivative test: a local extremum occurs where $f'$ CHANGES sign.
• Second derivative test: $f''(a)<0$ gives a maximum, $f''(a)>0$ a minimum. If $f''(a)=0$ , the test is SILENT.
• ABSOLUTE extrema on a closed bounded interval: three kinds of candidates, the critical points, the points of non-differentiability, and the ENDPOINTS.
• You then compare the VALUES of the function at these candidates, with no sign chart. On an open interval, an absolute extremum may not exist.

[0, 5.4]

, the local maximum at

x=2

5

, but the right endpoint gives more, about

6.16

: the ABSOLUTE maximum is at the endpoint, where

f'

does not vanish.

A local maximum is absolute only if no endpoint exceeds it. Forgetting to evaluate the function at both endpoints is the most common mistake on this question.

The rules in table form

Each row reads from left to right: the assumptions, then the result. A red cell is not an answer, it is the finding that the form settles nothing and the instruction to rewrite it. Every case is followed by a worked example.

The four shapes, and the two cases where the test stays silent

The first two columns are the two lines of the variation table. The last one says what the curve does. The two red boxes are not conclusions: they are orders to keep going.

Signs	What we write	What the curve does
$f'>0$ and $f''>0$	increasing, concave up	rises and speeds up
Example: $f(x)=0.25x^{2}$ at $x=2$ : $f'=1>0$ and $f''=0.5>0$ .
$f'>0$ and $f''<0$	increasing, concave down	rises and slows down
Example: $f(x)=6\sqrt{x}$ at $x=4$ : $f'=1.5>0$ and $f''=-0.1875<0$ .
$f'<0$ and $f''>0$	decreasing, concave up	falls and slows down
Example: $f(x)=\dfrac{4}{x}$ at $x=2$ : $f'=-1<0$ and $f''=1>0$ .
$f'<0$ and $f''<0$	decreasing, concave down	falls and speeds up
Example: $f(x)=-0.25x^{2}$ at $x=2$ : $f'=-1<0$ and $f''=-0.5<0$ .
$f'(a)=0$ and $f''(a)=0$	the second derivative test	silent test silent
Example: $x^{3}$ and $x^{4}$ both have $f'(0)=0$ and $f''(0)=0$ , and yet one has an inflection point and the other a minimum. What to do: Go back to the FIRST derivative test: look at the sign of $f'$ just before and just after.
$f''(a)=0$ gives an inflection point	vanishing alone	does not exist no such rule
Example: For $x^{4}$ , $f''(x)=12x^{2}$ vanishes at $0$ and stays positive on both sides: no inflection point. What to do: Require that $f''$ CHANGES sign, and not only that it vanishes.

The first four rows can be recited in thirty seconds and are enough to write any comment on a curve. The last two are the reasoning questions of the exam.

The mistakes that cost marks

These are the errors I correct most often in session. Each one costs marks on a paper, even when the reasoning behind it is right.

1. Reading the direction of variation from the second derivative

2 marks, and the whole variation line of the table

What not to write

“ $f''<0$ , so the function decreases on this interval.”

What to write

“ $f''<0$ says the curve is concave down. Since $f'>0$ here, the function INCREASES, while slowing down.”

Why: The two lines of the table are independent. A curve that rises while flattening out has precisely $f'>0$ and $f''<0$ , and it is the most common situation in Social Science.

2. Concluding there is an extremum as soon as the derivative vanishes

2 marks, and a wrong variation table

What not to write

“ $f'(0)=0$ , so $f$ has an extremum at $0$ .”

What to write

“ $f'(0)=0$ makes $0$ a CANDIDATE. For $f(x)=x^{3}$ , $f'(x)=3x^{2}$ stays positive on both sides: there is no extremum, but an inflection point with a horizontal tangent.”

Both dotted tangent lines are horizontal, so

f'

vanishes in both cases: on the left the curve keeps rising, on the right it turns back up after having gone down.

Why: An extremum requires a CHANGE of sign of $f'$ , not a vanishing. The cubic is the official counterexample, and it comes up every term.

3. Declaring an inflection point as soon as $f''$ vanishes

1.5 marks, and an invented inflection point on the sketch

What not to write

“ $f''(0)=0$ , so $0$ is an inflection point.”

What to write

“For $f(x)=x^{4}$ , $f''(x)=12x^{2}$ vanishes at $0$ but stays POSITIVE on both sides: the concavity does not change, there is no inflection point.”

In both cases

f''(0)=0

. For

x^{3}

f''=6x

changes sign: the curvature reverses at

0

. For

x^{4}

f''=12x^{2}

stays positive: the curve stays turned upward, no inflection point.

Why: It is exactly the same logic as the previous pitfall, one level up: vanishing is not enough, there has to be a sign change.

4. Forcing a conclusion when the second derivative test is silent

2 marks, and a conclusion that has a one in two chance of being wrong

What not to write

“ $f''(a)=0$ , so there is neither a maximum nor a minimum at $a$ .”

What to write

“ $f''(a)=0$ : the second derivative test is inconclusive. I go back to the FIRST derivative test and study the sign of $f'$ on either side of $a$ .”

x^{3}

x^{4}

and

-x^{4}

all have

f'(0)=0

and

f''(0)=0

, and yet three conclusions: neither a maximum nor a minimum, a minimum, a maximum. The silent test sends you back to the sign of

f'

Why: Silent does not mean negative. $x^{4}$ does have a minimum at $0$ even though $f''(0)=0$ : the test says nothing, it does not say no.

5. Forgetting the endpoints in a search for absolute extrema

2 marks out of 5 on this question, and sometimes the wrong answer

What not to write

“The only critical point is $t=33.3$ , so the absolute maximum is reached there.”

What to write

“The candidates are the critical points AND the endpoints: $t=0$ , $t=33.3$ and $t=100$ . We compare $R(0)=0$ , $R(33.3)\approx 1{,}481.5$ and $R(100)=0$ .”

Three candidates on

[0, 100]

: the critical point

t=33.3

and the two ENDPOINTS. We compare their values:

0

, about

1{,}481.5

and

0

. Here the maximum is interior, but only the comparison proves it.

Why: On a closed bounded interval, an absolute extremum may very well lie at an endpoint, where the derivative does not vanish at all.

6. Confusing a local maximum with an absolute maximum

1 mark, and the next question becomes wrong

What not to write

“ $f''(a)<0$ , so $a$ is the maximum of the function.”

What to write

“ $f''(a)<0$ gives a LOCAL maximum. It is absolute only if no endpoint of the domain gives a larger value, which has to be checked.”

Why: The second derivative test is information about a neighbourhood. It looks at one point and its immediate surroundings, never at the whole interval.

7. Announcing two asymptotes on the same side

1.5 marks, and a visible contradiction on the sketch

What not to write

“The curve has the asymptotes $y=2$ and $y=0.5x+3$ at $+\infty$ .”

What to write

“A curve tends toward only one thing at $+\infty$ : either a horizontal line or an oblique one. Here the degree of the numerator exceeds that of the denominator by one, so it is the oblique one.”

+\infty

, the curve runs alongside the oblique line

y=0.5x+3

and moves farther and farther from the line

y=2

(red): a limit is unique, the curve cannot follow both.

Why: A limit is unique. Announcing two behaviours at infinity amounts to saying that the function tends toward two different values at the same time.

8. Sketching from a table of values

3 marks on the sketching question, where the marking scheme pays for the elements, not for a pretty curve

What not to write

“I calculated ten points on the calculator and joined them.”

What to write

“The sketch is deduced from the variation table: asymptotes dashed, extrema marked, concavity respected, then the curve is drawn between these constraints.”

The two lines get closer without ever touching: the gap between the curve and the line is

\dfrac{40}{d}

, which tends to

0

without vanishing.

Why: Ten points show neither an asymptote, nor an inflection, nor an exact extremum. They can even jump over a vertical asymptote without anything signalling it.

9. Keeping a critical point that is not in the domain

2 marks, and a variation table built on nonexistent points

What not to write

“ $f'$ vanishes at $t=-4$ and at $t=12$ , so there are two extrema.”

What to write

“The domain is $[0, 10]$ : neither $-4$ nor $12$ belongs to it, so neither is a candidate. The only candidates are the endpoints.”

f'

vanishes at

-4

and at

12

(red), two points OUTSIDE the domain

[0, 10]

. On the domain (green), the function only goes down: the only candidates are the endpoints.

Why: Solving $f'(x)=0$ gives the roots of the derivative, not the critical points of the function being studied. They then have to be filtered by the domain, context included.

Which method to choose

How to decide on a critical point

We have solved $f'(a)=0$ . It remains to say what happens at $a$ , and there are four cases, settled in this order.

If $a$ does not belong to the domain → discard it immediately, it is not a candidate

Example: $t=-4$ for a rate that lives on $[0, 100]$
If $f''(a)<0$ → LOCAL maximum, to be confirmed as absolute by comparing with the endpoints

Example: $R''(33.3)=-2<0$
If $f''(a)>0$ → LOCAL minimum, same check

Example: $f''(2)=1>0$ for $\dfrac{4}{x}$
If $f''(a)=0$ → the test is silent: go back to the sign of $f'$ on either side

Example: $x^{3}$ and $x^{4}$ have the same diagnosis and opposite conclusions

this is the only branch where you really have to work, and it is the one that comes up on exams
If $f'$ changes sign at $a$ → local extremum, of the type indicated by the direction of the change

Example: from $+$ to $-$ : maximum
If $f'$ keeps the same sign around $a$ → no extremum, but an inflection point with a horizontal tangent

Example: $x^{3}$ at $0$

If the question asks for an ABSOLUTE extremum, none of these branches is enough to conclude on its own: you still have to evaluate the function at the endpoints and compare the values.

Which asymptote to look for, according to the form

We look at the form of the expression, then at the degrees. Three questions, three answers, and none requires a long calculation.

If a denominator vanishes while the numerator does not → VERTICAL asymptote, and the sign on each side decides the two infinities

Example: $\dfrac{40}{d}$ at $d=0$
If degree of the numerator less than or equal to that of the denominator → HORIZONTAL asymptote, $0$ or the ratio of the leading coefficients

Example: $\dfrac{6x^{3}+x}{2x^{3}-5}$ tends to $3$
If degree of the numerator exactly ONE more than that of the denominator → OBLIQUE asymptote, obtained by division

Example: $\dfrac{40}{d}+0.5d+2$ has the oblique asymptote $y=0.5d+2$
If degree of the numerator two or more above that of the denominator → no asymptote on that side, the curve heads off to infinity with no limiting direction

Example: $\dfrac{x^{3}}{x}=x^{2}$
If a decaying exponential or a logistic model → horizontal asymptote, and the logistic has TWO, one on each side

Example: $y=0$ and $y=M$

One curve can have a horizontal asymptote on one side and an oblique one on the other. What it cannot have is two asymptotes on the same side.

How the answer is expected to be written

A marker ticks steps. Here they are in order, with the concluding sentence expected word for word.

Writing up a complete study

When to use it: On the long question of the final exam, and whenever the question asks you to study and then sketch.

1 Give the domain, crossing the algebraic restrictions with the constraint of the context.
2 Calculate the limits at the edges and at infinity, then NAME the asymptotes that follow.
3 Calculate $f'$ , solve $f'(x)=0$ , FILTER the roots by the domain, then study the sign.
4 Calculate $f''$ , solve $f''(x)=0$ , study the sign and spot the changes.
5 Draw up the table with the two sign lines and the notable values.
6 Sketch from the table, placing the asymptotes and the notable points first.
7 Conclude with a sentence of interpretation in the context of the question.

Concluding sentence

“On $[0, 100]$ , $R'(t)=100-4t+0.03t^{2}$ vanishes at $t=\dfrac{100}{3}$ and at $t=100$ . Since $R''\left(\dfrac{100}{3}\right)=-2<0$ , the revenue is maximal for a rate of about $33.3$ percent, where it is about $1{,}481.5$ .”

The trap: Starting with the derivative. Without the domain and the limits, you do not know which roots to keep or how to close the table, and the marks lost cannot be recovered.

Marking: 1 mark for the domain, 2 for the limits and asymptotes, 3 for $f'$ and its sign, 2 for $f''$ , 2 for the table, 3 for the sketch, 2 for the interpretation.

Writing up a search for absolute extrema

When to use it: Whenever the question says “on the interval”, “at most”, “the largest value”, with a closed bounded interval.

1 Check that the function is CONTINUOUS on the closed interval, and say so: that is what guarantees that the extrema exist.
2 List the three kinds of candidates: critical points, points where $f$ is not differentiable, and the TWO endpoints.
3 Discard the critical points outside the interval.
4 Evaluate the function at each candidate kept, in a small table of values.
5 Compare the values and conclude: the largest is the absolute maximum, the smallest the absolute minimum.

Concluding sentence

“ $R$ is a polynomial, hence continuous on $[0, 100]$ . The candidates are $0$ , $\dfrac{100}{3}$ and $100$ , with $R(0)=0$ , $R\left(\dfrac{100}{3}\right)\approx 1{,}481.5$ and $R(100)=0$ . The absolute maximum is therefore about $1{,}481.5$ , reached at a rate of $33.3$ percent, and the absolute minimum is $0$ , reached at both endpoints.”

The trap: Drawing up a sign chart instead of comparing VALUES. For an absolute extremum, it is the list of values that decides, and it is much faster.

Marking: 1 mark for continuity, 1 mark for the complete list of candidates, 2 marks for the values, 1 mark for the conclusion.

Check before you hand in

Five minutes of checking recover more marks than one more problem started in a hurry.

The table must be consistent with itself

Reread the arrows: where $f'$ is positive, the arrow goes up. Where $f''$ changes sign, an inflection point is marked. An inconsistency jumps out in five seconds.

An arrow that goes down under a box $f'>0$ signals a sign error in the factorization of the derivative.
The sketch on the calculator

Graph the function in a suitable window and compare the shape with the one deduced from the table. The two must agree, extrema and asymptotes included.

If the machine shows two bumps and the table predicts only one, a root of $f'$ has been lost.
The values at the endpoints

Evaluate the function at both endpoints, even when the question does not explicitly ask. It is free and it settles the question of absolute extrema.

$R(0)=0$ and $R(100)=0$ : both endpoints give the same value, which confirms the bell shape of the revenue curve.
The number of inflection points

A polynomial function of degree $n$ has at most $n-2$ inflection points. Counting before concluding catches the vanishings with no sign change.

A cubic has at most one inflection point. Announcing two signals that $f''$ , which is linear, was solved wrongly.
The oblique asymptote, tested numerically

Calculate $f(x)-(mx+p)$ for a large value of $x$ . The result must be very small, and all the smaller as $x$ is larger.

For $\dfrac{40}{d}+0.5d+2$ and $y=0.5d+2$ : the gap is $0.4$ at $d=100$ and $0.04$ at $d=1000$ .

The typical problem, taken apart

The tax revenue curve: a complete study on a closed interval

The annual tax revenue of a territory, in millions of dollars, is modelled by $R(t)=100t-2t^{2}+0.01t^{3}$ , where $t$ is the tax rate as a percentage.

Study $R$ on $[0, 100]$ : critical points, nature, concavity, inflection, absolute extrema, then interpret.

Step 1

Domain: $R$ is a polynomial, so it is defined on $\mathbb{R}$ . The CONTEXT requires $0\le t\le 100$ , since $t$ is a rate as a percentage.

Why

The mathematical domain and the domain of the context are two different things, and it is the second one that is used here. Without it, we would keep negative roots.

Step 2

Values at the endpoints: $R(0)=0$ and $R(100)=10{,}000-20{,}000+10{,}000=0$ .

Why

Both extremes give zero, and that makes sense: with no tax there is no revenue, and at $100$ percent the tax base disappears. The model is therefore plausible.

Step 3

$R'(t)=100-4t+0.03t^{2}$ . We solve $0.03t^{2}-4t+100=0$ : the discriminant is $16-12=4$ , so $t=\dfrac{4\pm 2}{0.06}$ , that is, $t=\dfrac{100}{3}\approx 33.3$ or $t=100$ .

Why

Both roots are in the interval, so neither is discarded. Checking that they belong to the domain is the step that students skip.

Step 4

Sign of $R'$ : the coefficient of $t^{2}$ is positive, so $R'>0$ on $\left[0, \dfrac{100}{3}\right)$ and $R'<0$ on $\left(\dfrac{100}{3}, 100\right)$ . The revenue increases then decreases.

Why

The sign is read off the parabola $R'$ , with no test values: between its roots it is negative, outside them positive.

Step 5

$R''(t)=-4+0.06t$ , so $R''\left(\dfrac{100}{3}\right)=-4+2=-2<0$ : the critical point $t=\dfrac{100}{3}$ is a local MAXIMUM.

Why

The second derivative test is quicker here than the first derivative test, since $R''$ is linear and can be calculated in your head.

Step 6

$R''$ vanishes at $t=\dfrac{200}{3}\approx 66.7$ and changes sign, from negative to positive: it is an INFLECTION point.

Why

The sign change is what makes the inflection point. Since $R''$ is linear with nonzero slope, the change is guaranteed as soon as it vanishes.

Step 7

Absolute extrema on $[0, 100]$ : candidates $0$ , $\dfrac{100}{3}$ and $100$ . Values: $0$ , about $1{,}481.5$ and $0$ .

Why

The endpoints are candidates in their own right. Here they do not win, but they had to be evaluated before we could say so.

Step 8

Interpretation: beyond $33.3$ percent, raising the rate makes the revenue FALL. From $66.7$ percent on, the drop stops speeding up and starts slowing down.

Why

The two sentences come from two different lines of the table, one from $R'$ and the other from $R''$ . Giving both is what the interpretation marks pay for.

The conclusion, written out

“On $[0, 100]$ , the revenue increases up to $t=\dfrac{100}{3}\approx 33.3$ percent, where it reaches its absolute maximum of about $1{,}481.5$ million dollars, then decreases until it vanishes at $t=100$ . The inflection point at $t\approx 66.7$ percent marks the moment when the drop stops speeding up. The absolute minimum, $0$ , is reached at both endpoints.”

The classic mistake on this problem: Concluding that the revenue is maximal at $t=100$ because it is the highest rate. The reasoning confuses the variable and the function: it is $R$ that we maximize, not $t$ .

Learn by heart

• Order of a study: domain, limits, asymptotes, $f'$ and its sign, $f''$ and its sign, table, sketch.
• $f'$ says the DIRECTION, $f''$ says the CURVATURE. The two readings are independent.
• $f'(a)=0$ makes a CANDIDATE. An extremum requires a CHANGE of sign of $f'$ .
• $f''(a)=0$ does not make an inflection point: $f''$ must CHANGE sign.
• Second derivative test: $f''(a)<0$ maximum, $f''(a)>0$ minimum, $f''(a)=0$ SILENT.
• Absolute extrema on a closed bounded interval: critical points, points of non-differentiability, and the ENDPOINTS.
• A curve has at most ONE horizontal or oblique asymptote on each side.
• The sketch is deduced from the table, never from a table of values.

Frequently asked questions

In what order should you study a function?

Domain, limits at the edges and at infinity, asymptotes, first derivative with its sign, second derivative with its sign, variation table, and the sketch last. Each step feeds the next: without the domain you do not know which roots to keep, and without the limits you do not know how to close the table.

Does a zero derivative always mean an extremum?

No. A zero derivative only makes that point a candidate. For there to be an extremum, the derivative must change sign on either side. The cube function is the classic counterexample: its derivative vanishes at the origin but stays positive on both sides, and the curve keeps rising.

What do you do when the second derivative is zero at the critical point?

The second derivative test is inconclusive, and above all you must not deduce that there is no extremum. You go back to the first derivative test: look at the sign of the derivative just before and just after the point. The fourth power function does have a minimum even though its second derivative vanishes there.

How do you recognize an inflection point?

The second derivative must vanish and change sign. Vanishing alone is not enough: for the fourth power function, the second derivative vanishes at the origin but stays positive on both sides, so the concavity does not change and there is no inflection point. It is the change that counts, not the vanishing.

Where should you look for absolute extrema on a closed interval?

In three places only: the points where the derivative vanishes, the points where the function is not differentiable, and the two endpoints of the interval. You evaluate the function at each of these candidates and compare the values. The endpoints are the most common oversight, and they win more often than you might think.

When does a curve have an oblique asymptote?

When the degree of the numerator exceeds that of the denominator by exactly one. You then get the equation of the asymptote by division, and the gap between the curve and the line tends to zero without ever vanishing. A curve cannot have both a horizontal and an oblique asymptote on the same side.

Practise it

Corrected exercises: Curve sketching, 201-103

A method is proved on a paper, not on a sheet. The set for the same chapter takes each of these traps into a problem, with the solution written out step by step.

11 corrected exercises
110 points
150 minutes

Do the exercises

Something missing from this sheet? A method, a trap, an exercise to practise it, a mistake you spotted: tell me, I read every request.

Sending goes through your free space: that is how I can answer you and tell you when the exercise is added. An email is enough, no password. Create your space. Rather not create an account? Write to me directly.

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© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/cegep-103-curve-sketching. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: complete curve sketching (201-103)

The essentials

The order of a study, seven steps

Two independent readings

The three asymptotes

Local extrema and absolute extrema

The rules in table form

The four shapes, and the two cases where the test stays silent

The mistakes that cost marks

1. Reading the direction of variation from the second derivative

2. Concluding there is an extremum as soon as the derivative vanishes

3. Declaring an inflection point as soon as f′′f''f′′ vanishes

4. Forcing a conclusion when the second derivative test is silent

5. Forgetting the endpoints in a search for absolute extrema

6. Confusing a local maximum with an absolute maximum

7. Announcing two asymptotes on the same side

8. Sketching from a table of values

9. Keeping a critical point that is not in the domain

Which method to choose

How to decide on a critical point

Which asymptote to look for, according to the form

How the answer is expected to be written

Writing up a complete study

Writing up a search for absolute extrema

Check before you hand in

The typical problem, taken apart

The tax revenue curve: a complete study on a closed interval

Learn by heart

Frequently asked questions

In what order should you study a function?

Does a zero derivative always mean an extremum?

What do you do when the second derivative is zero at the critical point?

How do you recognize an inflection point?

Where should you look for absolute extrema on a closed interval?

When does a curve have an oblique asymptote?

Corrected exercises: Curve sketching, 201-103

See also

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3. Declaring an inflection point as soon as $f''$ vanishes