MATH 203 Calculus I • Concordia University, Montreal

Revision sheet: concavity and curve sketching (MATH 203)

This sheet is not a summary of section 4.4 of Thomas' Calculus: you already have the course notes and the procedure for graphing. It answers one question only, what makes students lose marks on concavity and curve sketching in MATH 203 at Concordia University, and which precise gesture avoids each loss.

The rules of the chapter fit in four lines. The marks are lost in the algebra between the rule and the sign table: a numerator expanded to degree four when a common factor was waiting to be cancelled, a fractional power whose sign is misread, a factor divided out and lost. The sheet puts that algebra first, then the candidates, then the complete sketch.

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The thread of the chapter

The concavity is read on the SIGN of $f''$ , and that sign can only be read on a factored form: pull out the common factor of a quotient rule, the smallest power of a fractional exponent, the exponential that never vanishes. Then every zero or undefined point of $f''$ is only a candidate, kept if the sign changes at a point of the domain.

This chapter is part of MATH 203, Calculus I (Concordia)

What is on this sheet

1. The essentials
2. The rules in table form
3. The mistakes that cost marks
4. Which method to choose
5. How the answer is expected to be written
6. Check before you hand in
7. The typical problem, taken apart
8. Learn by heart
9. Frequently asked questions

Topics covered

Concavity
Concavity test
Point of inflection
Second derivative test
Critical number
Curve sketching
Cusp and vertical tangent
Factoring fractional powers
Point of diminishing returns
Logistic growth

The essentials

Factor first: the sign of f double prime is read on a product

• Quotient rule on $\frac{u}{v^n}$ : the numerator of the derivative contains $v^{n-1}$ . Pull it out and cancel ONE power before expanding anything: $\frac{d}{dx}\frac{12 - x^2}{(x^2 + 12)^2}$ gives $\frac{2x(x^2 - 36)}{(x^2 + 12)^3}$ .
• Fractional powers: factor out the SMALLEST power, the most negative exponent. $\frac{28}{9}x^{1/3} + \frac{14}{9}x^{-5/3} = \frac{14}{9}x^{-5/3}(2x^2 + 1)$ , because $x^{1/3} = x^{-5/3} \cdot x^2$ .
• Exponentials: pull $e^{-x}$ or $e^{-x^2/2}$ out of every term; it is positive and never zero, so the sign is that of the polynomial left.
• Never divide by a factor that can vanish: $x^3 - 6x^2 + 6x = x(x^2 - 6x + 6)$ has THREE roots, not two.
• Signs of the usual factors: $(x - a)^2 \ge 0$ never changes sign; $x^{5/3}$ has the sign of $x$ ; $x^{4/3} = \left(\sqrt[3]{x}\right)^4 \ge 0$ ; $\sqrt{x + 3} > 0$ on its domain.

\frac{x}{x^2 + 12}

has

f'' = \frac{2x(x^2 - 36)}{(x^2 + 12)^3}

: three sign changes, three points of inflection at

-6

0

and

6

, and the concavity alternates between them.

Write $f''$ in factored form on its own line before the table: a marker gives the concavity marks only when the table can be checked against that line.

Candidates, then decisions

• Candidates for inflection: every $c$ of the DOMAIN where $f''(c) = 0$ or $f''(c)$ does not exist.
• Decision: $(c, f(c))$ is a point of inflection if the sign of $f''$ changes at $c$ and the graph has a tangent line there (a vertical one counts).
• Second Derivative Test: $f'(c) = 0$ with $f''(c) < 0$ gives a local max, with $f''(c) > 0$ a local min, with $f''(c) = 0$ NOTHING: use the First Derivative Test.
• Procedure for graphing: domain, intercepts, symmetry, asymptotes, $f'$ factored and its sign, $f''$ factored and its sign, one table, the curve.

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 sign of a factor in f double prime

Read a line as: this factor, on this set, has this sign. The red lines are the signs students write and that are false.

Factor	On	Sign
$e^{-x}$	$\mathbb{R}$	$+$
Example: $e^{-5} \approx 0.0067 > 0$ and $e^{5} \approx 148 > 0$ : $(x - 3)e^{-x}$ has the sign of $x - 3$ .
$e^{-x}$	$x > 0$	$-$ false sign
Example: $e^{-2} \approx 0.1353$ : a small positive number, not a negative one. What to do: An exponential is always positive; the minus sign in the exponent makes it SMALL, never negative.
$(x - 2)^2$	$x \ne 2$	$+$
Example: $12(x - 2)^2$ at $1.9$ and $2.1$ : $0.12$ both times, no sign change, no inflection.
$x^{5/3}$	$x \ne 0$	sign of $x$
Example: $(-8)^{5/3} = (-2)^5 = -32$ and $8^{5/3} = 32$ .
$x^{4/3}$	$x < 0$	$-$ false sign
Example: $(-8)^{4/3} = (-2)^4 = 16 > 0$ . What to do: Write $x^{p/3} = \left(\sqrt[3]{x}\right)^p$ : an even $p$ gives a positive factor, an odd $p$ the sign of $x$ .
$(x^2 + 12)^3$	$\mathbb{R}$	$+$
Example: $x^2 + 12 \ge 12$ , so $(x^2 + 12)^3 \ge 1728 > 0$ : only the numerator decides.

Once each factor has its line, the sign of $f''$ is a product of signs, and the table writes itself.

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. Expanding the quotient rule numerator instead of cancelling the common factor

2 to 3 marks: the whole concavity study stops there

What not to write

“ $f'' = \frac{-2x(x^2 + 12)^2 - (12 - x^2) \cdot 4x(x^2 + 12)}{(x^2 + 12)^4} = \frac{2x^5 - 48x^3 - 864x}{(x^2 + 12)^4}$ , and I cannot factor the numerator.”

What to write

“Pull out $-2x(x^2 + 12)$ : $f''(x) = \frac{-2x\left[(x^2 + 12) + 2(12 - x^2)\right]}{(x^2 + 12)^3} = \frac{2x(x^2 - 36)}{(x^2 + 12)^3}$ .”

Why: Both terms of the numerator contain $x^2 + 12$ ; cancelling one power leaves a cubic that factors at sight. Expanding creates a quintic whose roots must be rediscovered.

2. Declaring a point of inflection where f double prime vanishes

1 mark, and an S drawn where the curve is a cup

What not to write

“ $f(x) = x^4 + 4x$ has $f''(x) = 12x^2$ , so $f''(0) = 0$ and $(0, 0)$ is a point of inflection.”

What to write

“ $f''(x) = 12x^2 \ge 0$ on both sides of $0$ : no change of sign, so no point of inflection; $f$ is concave up on $\mathbb{R}$ .”

Why: A zero of $f''$ only NOMINATES a point. What makes an inflection is a change of concavity, and a square does not change sign.

3. Concluding no inflection point because f double prime never vanishes

1 to 2 marks

What not to write

“ $f(x) = x + \sqrt[3]{x}$ has $f''(x) = -\frac{2}{9x^{5/3}}$ , which is never $0$ : no point of inflection.”

What to write

“ $f''$ is undefined at $0$ , a point of the domain: it is a candidate. $f'' > 0$ for $x < 0$ and $< 0$ for $x > 0$ , and the tangent at the origin is vertical ( $f'(x) = 1 + \frac{1}{3x^{2/3}} \to \infty$ ): $(0, 0)$ is a point of inflection.”

x + \sqrt[3]{x}

bends up on the left, down on the right, and passes through the origin with a vertical tangent: an inflection point where

f''

does not exist.

Why: The candidates are the zeros of $f''$ AND the points of the domain where $f''$ does not exist. Forgetting the second list loses every inflection with a vertical tangent.

4. Misreading the sign of a fractional power, and calling a cusp an inflection point

1 mark for the false point, and a wrong sketch near the origin

What not to write

“ $f(x) = x + 3x^{2/3}$ has $f''(x) = -\frac{2}{3x^{4/3}}$ , and $x^{4/3}$ changes sign at $0$ : $(0, 0)$ is a point of inflection.”

What to write

“ $x^{4/3} = \left(\sqrt[3]{x}\right)^4 > 0$ for $x \ne 0$ , so $f'' < 0$ on both sides of $0$ : concave down on $(-\infty, 0)$ and on $(0, \infty)$ , no point of inflection. At $0$ , $f'(x) = 1 + 2x^{-1/3}$ jumps from $-\infty$ to $+\infty$ : a cusp, and a local minimum.”

x + 3x^{2/3}

bends down on both sides of the origin: a cusp and a local minimum, not an inflection point; the top

(-8, 4)

is a local maximum.

Why: An even power of a cube root is never negative. The origin is a critical number where $f'$ jumps from $-\infty$ to $+\infty$ , not a change of concavity.

5. Reading a silent Second Derivative Test as no extremum

1 mark, and often the extremum the question was about

What not to write

“ $g(x) = 3 - x^6$ : $g'(0) = 0$ and $g''(0) = 0$ , so by the Second Derivative Test $g$ has no extremum at $0$ .”

What to write

“The test fails at $0$ . First Derivative Test: $g'(x) = -6x^5$ goes from $+$ to $-$ at $0$ , so $g(0) = 3$ is a local (in fact absolute) maximum.”

Why: When $f''(c) = 0$ the test gives NO information: a max, a min or neither are all possible. Silent is not negative.

6. Giving an exponential factor a sign

the whole concavity row of the table

What not to write

“ $f(x) = (x - 1)e^{-x}$ has $f''(x) = (x - 3)e^{-x}$ ; for $x > 0$ the exponent $-x$ is negative, so $e^{-x} < 0$ and $f'' > 0$ on $(0, 3)$ .”

What to write

“ $e^{-x} > 0$ for every $x$ , so $f''$ has the sign of $x - 3$ : concave down on $(-\infty, 3)$ , up on $(3, \infty)$ , point of inflection $(3, 2e^{-3})$ .”

Why: A negative exponent makes the power small, never negative: $e^{-3} = \frac{1}{e^3} \approx 0.0498$ . The exponential is pulled out and forgotten.

7. Dividing by x and losing a point of inflection

1 mark, and usually the most visible point of the graph

What not to write

“ $f''(x) = (x^3 - 6x^2 + 6x)e^{-x} = 0$ , divide by $x$ : $x^2 - 6x + 6 = 0$ , so the inflection points are at $3 \pm \sqrt 3$ .”

What to write

“ $f''(x) = x(x^2 - 6x + 6)e^{-x}$ : the candidates are $0$ and $3 \pm \sqrt 3$ , and $f''$ changes sign at all three.”

Why: Dividing an equation by $x$ assumes $x \ne 0$ and silently throws away the root $0$ . Factor it out instead.

8. Placing the point of diminishing returns at the top of the curve

the whole application question

What not to write

“The yield stops increasing at the point of diminishing returns, so I look for $Y'(x) = 0$ .”

What to write

“The point of diminishing returns is where $Y''$ changes from $+$ to $-$ : the yield still increases after it, but its RATE $Y'$ is largest there and decreases afterwards.”

Why: Diminishing returns is a statement about the marginal output $Y'$ , not about $Y$ . An inflection point of $Y$ is an extremum of $Y'$ .

Which method to choose

Which algebra makes the sign of f double prime readable, by the FORM of f prime

Look at the formula of the first derivative before differentiating again

If a quotient $\frac{u}{v^n}$ with $n \ge 2$ → quotient rule, then pull $v^{n-1}$ out of the numerator and cancel it

Example: $\frac{72x}{(x^2 + 27)^2}$ gives $f'' = \frac{216(9 - x^2)}{(x^2 + 27)^3}$
If a sum of powers $x^{p}$ with fractional exponents → factor out the smallest (most negative) power, then read the sign of each factor

Example: $\frac{4}{9}x^{-2/3} - \frac{8}{9}x^{-5/3} = \frac{4}{9}x^{-5/3}(x - 2)$
If a power of a composite, $(x - 1)(x^2 - 2x)^{-1/3}$ → product and chain rules, then factor out $(x^2 - 2x)^{-4/3}$

Example: $f'' = \frac{4(x^2 - 2x - 2)}{9(x^2 - 2x)^{4/3}}$ for $f = (x^2 - 2x)^{2/3}$
If a polynomial times $e^{\pm x}$ or $e^{-x^2/2}$ → pull the exponential out of every term, it never vanishes; study the polynomial

Example: $(x^3 - 6x^2 + 6x)e^{-x} = x(x^2 - 6x + 6)e^{-x}$
If a logistic $\frac{A}{1 + Be^{-kt}}$ → multiply by $e^{kt}$ to get $\frac{Ae^{kt}}{e^{kt} + B}$ , then quotient rule and cancel

Example: inflection when $e^{kt} = B$ , at $N = \frac{A}{2}$

If no branch applies, the numerator is a polynomial: factor it (common factor, then roots). Never expand a power of the denominator.

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.

Concavity and points of inflection

When to use it: “Find the intervals of concavity and the points of inflection of $f$ ”

1 State the domain of $f$ .
2 Compute $f''$ and write it FACTORED, on its own line.
3 List the candidates: zeros of $f''$ and points of the domain where $f''$ does not exist; reject those outside the domain.
4 Sign of each factor on each interval, then the sign of $f''$ , in a table.
5 Conclude interval by interval, then give each point of inflection as a POINT $(c, f(c))$ , checking that the graph has a tangent there.

Concluding sentence

“Since $f''$ changes from negative to positive at $x = -6$ , which is in the domain, the graph is concave down on $(-\infty, -6)$ , concave up on $(-6, 0)$ , and $\left(-6, -\frac{1}{8}\right)$ is a point of inflection.”

The trap: Stopping at the x-values: the question asks for points, with both coordinates.

Marking: Typically 1 mark for the factored second derivative, 1 for the candidates, 1 for the sign table, 1 for the intervals and points.

Check before you hand in

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

Expand back one value

Pick one easy x, say 1 or 2, and compute f double prime from the unfactored expression and from the factored one. The two numbers must agree.

For $\frac{x}{x^2 + 12}$ at $x = 2$ : the expanded form gives $\frac{64 - 384 - 1728}{16^4} = -\frac{2048}{65536} = -\frac{1}{32}$ , and $\frac{2 \cdot 2 \cdot (4 - 36)}{16^3} = -\frac{128}{4096} = -\frac{1}{32}$ . Two different numbers would mean a slip in the factoring.
Concavity agrees with the ends

A curve that rises toward a horizontal asymptote from below ends concave down; one that falls toward it from above ends concave up. Compare with the last column of the table.

$\frac{x^2 - 9}{x^2 + 27}$ rises toward $y = 1$ from below, and indeed $f'' < 0$ for $x > 3$ .
Each inflection point is in the domain

Go through the list of inflection points and check each x against the domain written on the first line.

$(x - 2)\sqrt{x + 1}$ has $f'' = 0$ at $x = -2$ , outside $[-1, \infty)$ : the list must be empty.
Radian mode

Before any decimal value of a trigonometric function, check that the calculator is in radian mode.

$\frac{\pi}{6} + 2\cos\frac{\pi}{6} \approx 2.2556$ in radians; in degree mode the calculator returns about $2.5235$ , a wrong answer that looks plausible.

The typical problem, taken apart

A logarithm squared: a one-sided domain, L'Hôpital twice, two extrema and one inflection

Sketch the graph of $f(x) = x(\ln x)^2$ , giving its extrema and its point of inflection exactly.

A scientific calculator is allowed for decimals, to four decimal places.

Step 1

Domain $(0, \infty)$ . $f \ge 0$ , and $f(x) = 0$ only at $x = 1$ . At $0^+$ the form is $0 \cdot \infty$ : write $\frac{(\ln x)^2}{1/x}$ , form $\frac{\infty}{\infty}$ ; L'Hôpital gives $\frac{2\ln x \cdot \frac{1}{x}}{-1/x^2} = -2x\ln x$ , again $0 \cdot \infty$ ; write $\frac{-2\ln x}{1/x}$ and apply it once more: $\frac{-2/x}{-1/x^2} = 2x \to 0$ . At $\infty$ , $f \to \infty$ .

Why

The limit at $0^+$ tells where the curve starts: at the origin, as a hole, since $0$ is not in the domain. The form is named before EACH use of the rule, which is where the marks are.

Step 2

$f'(x) = (\ln x)^2 + x \cdot 2\ln x \cdot \frac{1}{x} = (\ln x)^2 + 2\ln x = \ln x\,(\ln x + 2)$ . Zeros: $x = 1$ and $x = e^{-2}$ . With $u = \ln x$ , the sign is that of $u(u + 2)$ : positive on $(0, e^{-2})$ , negative on $(e^{-2}, 1)$ , positive after $1$ .

Why

Factoring out $\ln x$ turns the sign study into that of a quadratic in $\ln x$ , read at a glance because $\ln$ is increasing.

Step 3

Local maximum $f(e^{-2}) = e^{-2}(-2)^2 = \frac{4}{e^2} \approx 0.5413$ ; local minimum $f(1) = 0$ , absolute since $f \ge 0$ everywhere.

Why

The laws of logarithms do the evaluation: $\ln e^{-2} = -2$ . The local maximum is NOT absolute: $f \to \infty$ as $x \to \infty$ .

Step 4

$f''(x) = 2\ln x \cdot \frac{1}{x} + \frac{2}{x} = \frac{2(\ln x + 1)}{x}$ . On the domain $x > 0$ , so $f''$ has the sign of $\ln x + 1$ : negative on $(0, e^{-1})$ (concave down), positive after (concave up). Point of inflection $\left(\frac{1}{e}, \frac{1}{e}\right) \approx (0.3679, 0.3679)$ , since $f\left(\frac{1}{e}\right) = \frac{1}{e}(-1)^2$ .

Why

The candidate is in the domain and $f''$ changes sign there: both conditions are written, not assumed. The positive factor $\frac{1}{x}$ is set aside before reading the sign.

Step 5

Table and sketch: from the hole at the origin the curve rises, concave down, to $\left(e^{-2}, \frac{4}{e^2}\right)$ , falls, bends at $\left(\frac{1}{e}, \frac{1}{e}\right)$ , touches the axis at its minimum $(1, 0)$ and rises without bound, concave up. Range $[0, \infty)$ .

Why

The inflection point sits between the maximum and the minimum: a curve that turns from a cap into a cup must change its bending in between, which the table confirms.

The conclusion, written out

“ $f$ has a local maximum $\frac{4}{e^2}$ at $x = e^{-2}$ , an absolute minimum $0$ at $x = 1$ , is concave down on $\left(0, \frac{1}{e}\right)$ and concave up on $\left(\frac{1}{e}, \infty\right)$ , with point of inflection $\left(\frac{1}{e}, \frac{1}{e}\right)$ .”

The classic mistake on this problem: Writing $f'(x) = 2\ln x$ (differentiating only one factor, or treating $(\ln x)^2$ as $\ln(x^2)$ ), or calling $\frac{4}{e^2}$ the absolute maximum of a function that tends to infinity.

Learn by heart

• Concave up: $f'' > 0$ , $f'$ increasing. It does NOT mean $f$ increasing.
• Candidates for inflection: $f'' = 0$ OR $f''$ undefined, at a point of the DOMAIN; kept only if $f''$ changes sign.
• A point of inflection is a POINT $(c, f(c))$ , with a tangent line, possibly vertical.
• Second Derivative Test: $f''(c) < 0$ max, $f''(c) > 0$ min, $f''(c) = 0$ no information.
• Quotient rule twice: pull out the common factor, cancel one power, never expand the denominator.
• Fractional powers: factor out the smallest power; $x^{4/3} \ge 0$ , $x^{5/3}$ has the sign of $x$ .
• An inflection point of $f$ is an extremum of $f'$ : steepest point, point of diminishing returns, peak of new cases.

Frequently asked questions

How do you find the points of inflection of a function?

Compute the second derivative and factor it completely. List the candidates, the zeros of the second derivative and the points of the domain where it does not exist. Study the sign of each factor on each interval. A candidate is a point of inflection only if the sign changes there and the graph has a tangent line. Give the answer as a point, with both coordinates.

Is every zero of the second derivative a point of inflection?

No. A zero of the second derivative is only a candidate. For x to the fourth plus 4x, the second derivative 12x squared vanishes at 0 but stays positive on both sides, so the graph is a cup with no inflection. Conversely, x plus the cube root of x has an inflection at the origin where the second derivative does not even exist.

What does it mean when the second derivative test is inconclusive?

When the first and second derivatives are both zero at a critical number, the second derivative test gives no information: the function may have a maximum, a minimum or neither there. It does not mean there is no extremum. Go back to the first derivative test and read the sign of the first derivative on both sides of the point.

How do I simplify the second derivative of a quotient?

After the quotient rule, the numerator contains a factor of the denominator, because the derivative of a squared or cubed denominator keeps one copy of its base. Factor that base out of the whole numerator and cancel one power before expanding anything. For example the second derivative of x over x squared plus 12 becomes 2x times x squared minus 36, over x squared plus 12 cubed.

Practise it

Corrected exercises: Concavity and curve sketching, MATH 203 at Concordia

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.

10 corrected exercises
100 points
150 minutes

Do the exercises

Previous sheet Monotonic functions and the First Derivative Test Next sheet Applied optimization

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math203-concavity-curve-sketching. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: concavity and curve sketching (MATH 203)

The essentials

Factor first: the sign of f double prime is read on a product

Candidates, then decisions

The rules in table form

The sign of a factor in f double prime

The mistakes that cost marks

1. Expanding the quotient rule numerator instead of cancelling the common factor

2. Declaring a point of inflection where f double prime vanishes

3. Concluding no inflection point because f double prime never vanishes

4. Misreading the sign of a fractional power, and calling a cusp an inflection point

5. Reading a silent Second Derivative Test as no extremum

6. Giving an exponential factor a sign

7. Dividing by x and losing a point of inflection

8. Placing the point of diminishing returns at the top of the curve

Which method to choose

Which algebra makes the sign of f double prime readable, by the FORM of f prime

How the answer is expected to be written

Concavity and points of inflection

Check before you hand in

The typical problem, taken apart

A logarithm squared: a one-sided domain, L'Hôpital twice, two extrema and one inflection

Learn by heart

Frequently asked questions

How do you find the points of inflection of a function?

Is every zero of the second derivative a point of inflection?

What does it mean when the second derivative test is inconclusive?

How do I simplify the second derivative of a quotient?

Corrected exercises: Concavity and curve sketching, MATH 203 at Concordia

See also

Looking for a MATH 203 tutor in Montreal?