MATH 140 Calculus 1 • McGill University, Montreal

Revision sheet: the shape of a graph (MATH 140)

Q: How do I find where a function is increasing or decreasing in MATH 140?

Compute the derivative and factor it completely. List the critical numbers, where the derivative is zero or does not exist. Then make a sign chart with one test value in every interval between them. Where the derivative is positive the function is increasing, where it is negative it is decreasing, stated interval by interval.

Q: What is the difference between the first and second derivative tests?

The first derivative test looks at the sign of f prime on each side of a critical number: a change from plus to minus is a maximum, from minus to plus a minimum, no change means no extremum. It always applies. The second derivative test only looks at the sign of f double prime at the point; it is faster, but when f double prime is zero or undefined it gives no answer.

Q: Is every point where the second derivative is zero an inflection point?

No. An inflection point needs a change of concavity, so the second derivative must change sign there. The function x to the fourth has a second derivative equal to zero at the origin, yet it is concave upward on both sides, and the origin is its minimum. A zero of the second derivative is only a candidate to be checked.

Q: How do I read f from the graph of f prime?

Where the graph of f prime is above the axis, f is increasing; below, f is decreasing. Where f prime crosses the axis, f has a local maximum or minimum; where it only touches the axis, f has none. Where the graph of f prime rises, f is concave upward, and the turning points of f prime are the inflection points of f.

Q: Can a function be increasing and concave down at the same time?

Yes. Increasing is about the sign of the first derivative, concave down about the sign of the second, and the two are independent. The square root function increases while bending down: it keeps rising, but more and more slowly. This is exactly what economists call diminishing returns.

This sheet is not a summary of section 4.3 of Stewart: you already have the course notes. It answers one question only, what makes students lose marks when they use $f'$ and $f''$ to describe a graph in MATH 140 at McGill University, and which precise gesture avoids each loss.

The chapter looks mechanical, differentiate, set equal to zero, solve, and that is exactly where the marks go: the equation $f'(c) = 0$ gives candidates, and the points are for what happens on each side. Every value below is exact and done by hand, as on the exam.

Mark this sheet as read or add it to your favourites: a free account, no password, keeps your read sheets and favourites from one visit to the next and tells you which chapter to tackle next. Create your space, an email is enough.

The thread of the chapter

A zero NOMINATES, a sign DECIDES: $f'(c) = 0$ , $f''(c) = 0$ or a derivative that does not exist only give candidates, and the sign of $f'$ (for an extremum) or of $f''$ (for an inflection point) on each side of the candidate is what the marker wants to see.

This chapter is part of MATH 140, Calculus 1 (McGill)

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

Increasing/Decreasing Test
Critical number
First Derivative Test
Local maximum and minimum
Concavity
Concavity Test
Inflection point
Second Derivative Test
Sign chart
Graph of the derivative
Point of diminishing returns

The essentials

Two derivatives, two independent questions

• The sign of $f'$ answers: is the graph going UP or DOWN? The sign of $f''$ answers: is it bending UP (a cup) or DOWN (a cap)? The four combinations all exist.
• Increasing does not mean concave upward: $\sqrt x$ increases while bending down, $e^{-x}$ decreases while bending up.
• Increasing/Decreasing Test and Concavity Test work on ONE interval at a time. A union of intervals is never an answer to “where is $f$ increasing” when the domain has a gap.
• On the graph of $f'$ : the SIGN (above or below the axis) gives the direction of $f$ , the SLOPE (rising or falling) gives the concavity of $f$ .

The four shapes: up and cup, up and cap, down and cup, down and cap. The sign of

f'

and the sign of

f''

are read separately.

Before reading any figure, write its name on it: “this is $y = f'(x)$ ”. Half of the errors on graph questions come from reading $f'$ as if it were $f$ .

Candidates, then verdicts

• Candidates for a local extremum: the critical numbers, where $f'(c) = 0$ OR $f'(c)$ does not exist ( $x^{1/3}$ , $|x|$ , $x^{2/3}$ at $0$ ).
• Candidates for an inflection point: where $f''(c) = 0$ OR $f''(c)$ does not exist, with $f$ continuous at $c$ .
• Verdict for an extremum: the sign of $f'$ changes (First Derivative Test), or $f''(c) \neq 0$ at a point where $f'(c) = 0$ (Second Derivative Test).
• Verdict for an inflection point: the sign of $f''$ changes. There is no shortcut test: $f''(c) = 0$ decides nothing.
• A factor squared, $(x - a)^2$ , never changes sign: a double root of $f'$ or $f''$ is the candidate that usually fails.

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.

What the Second Derivative Test says at a critical number

Read a line as: at a number $c$ with the value of $f'(c)$ of the first column and of $f''(c)$ of the second, the verdict is in the third. The red lines are not verdicts: the test is silent, and the First Derivative Test must decide.

$f'(c)$	$f''(c)$	Verdict at $c$
$0$	$> 0$	local minimum
Example: $f(x) = 2x^3 - 3x^2 - 36x + 1$ : $f'(3) = 0$ , $f''(3) = 30$ , local minimum $f(3) = -80$ .
$0$	$< 0$	local maximum
Example: Same $f$ : $f'(-2) = 0$ , $f''(-2) = -30$ , local maximum $f(-2) = 45$ .
$0$	$0$	no conclusion test is silent
Example: $p(x) = 3x^5 - 5x^3$ at $0$ : $p'(0) = p''(0) = 0$ , and $p' < 0$ on both sides, so no extremum. Same form, other result: $x^4$ , $-x^4$ and $x^3$ all have $f'(0) = f''(0) = 0$ : a minimum, a maximum, and neither. What to do: Go back to the sign of $f'$ on each side of $c$ : First Derivative Test.
$0$	undefined	no conclusion test cannot apply
Example: $x^{4/3}$ : $f'(0) = 0$ but $f''(x) = \frac{4}{9}x^{-2/3}$ is undefined at $0$ ; $f'$ goes from $-$ to $+$ , minimum $0$ . What to do: The hypotheses of the test fail: use the First Derivative Test.
undefined	undefined	no conclusion test cannot apply
Example: $x - 3x^{1/3}$ at $0$ : $f'(0)$ does not exist; $f' < 0$ on both sides, so no extremum. What to do: A critical number where $f'$ does not exist is ALWAYS classified by the sign of $f'$ .

The test is a shortcut for the easy cases. When it is silent it has said nothing, and a paper that writes “no extremum” at that point is wrong half the time.

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. Declaring an extremum wherever f' is zero

2 marks, the whole classification

What not to write

“ $g'(x) = 3(x - 2)^2$ , so $g'(2) = 0$ and $g$ has a local minimum at $2$ .”

What to write

“ $g'(2) = 0$ , but $3(x - 2)^2 > 0$ on both sides of $2$ : no sign change, so by the First Derivative Test $g$ has no local extremum at $2$ .”

Why: $f'(c) = 0$ gives a horizontal tangent, and a horizontal tangent can sit on a rising curve, like $x^3$ at the origin. Only a change of sign of $f'$ makes an extremum.

2. Declaring an inflection point wherever f'' is zero

1 to 2 marks, and a wrong sketch

What not to write

“ $f(x) = x^4$ : $f''(x) = 12x^2$ , $f''(0) = 0$ , so $(0, 0)$ is an inflection point.”

What to write

“ $f''(0) = 0$ , but $12x^2 > 0$ on both sides of $0$ : the concavity does not change, so $(0, 0)$ is not an inflection point.”

Both curves have

f''(0) = 0

. The cube

x^3

changes from cap to cup at the origin;

x^4

is a cup on both sides, so the origin is its minimum, not an inflection point.

Why: An inflection point is defined by a CHANGE of concavity. A double root of $f''$ , as $x^2$ in $g''(x) = 20x^2(x - 3)$ , is a candidate that fails; the root $3$ , simple, is a real inflection point.

3. Forgetting the critical numbers where f' does not exist

1 mark, more if the missing point was the extremum

What not to write

“ $h(x) = x - 3x^{1/3}$ , $h'(x) = 1 - x^{-2/3} = 0$ gives $x = \pm 1$ : the critical numbers are $-1$ and $1$ .”

What to write

“The critical numbers are $-1$ and $1$ ( $h' = 0$ ) AND $0$ ( $h'(0)$ does not exist, $0$ in the domain). The sign of $h'$ is then checked on each side of all three.”

Why: The definition has two halves. With $|x - 1|$ or $x^{2/3}$ , the forgotten point IS the minimum; here it is rejected, but only the sign of $h'$ is allowed to reject it.

4. Alternating the signs of a chart across a double root

2 marks, and a maximum that does not exist

What not to write

“ $f'(x) = 12x(x - 1)^2$ : the signs go $-$ , $+$ , $-$ around $0$ and $1$ , so a minimum at $0$ and a maximum at $1$ .”

What to write

“ $(x - 1)^2 \ge 0$ never changes sign: $f'$ is $-$ on $(-\infty, 0)$ , $+$ on $(0, 1)$ and $+$ on $(1, \infty)$ . Minimum at $0$ , no extremum at $1$ .”

Why: Signs alternate only across SIMPLE roots. Take one test value in EVERY interval of the chart, or read the sign of each factor, never a rhythm $+ - + -$ written in advance.

5. Reading the graph of f' as the graph of f

the whole question

What not to write

“The curve of $f'$ has its lowest point at $x = 0$ , so $f$ has a local minimum at $x = 0$ .”

What to write

“The zeros of $f'$ where it changes sign are the extrema of $f$ ; the lowest point of the curve of $f'$ is where $f$ decreases fastest, an inflection point of $f$ .”

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

has its extrema at

x = \pm 1

, where the dashed

f'

crosses the axis; the lowest point of

f'

, at

x = 0

, is the inflection point of

f

Why: The extrema of $f'$ are the inflection points of $f$ , and the zeros of $f'$ are its extrema. On the graph of $f'$ , look at the axis crossings for extrema and at the turning points for inflection points.

6. Reading a silent Second Derivative Test as a no

2 marks

What not to write

“ $f(x) = x^4$ : $f'(0) = 0$ and $f''(0) = 0$ , so by the Second Derivative Test $f$ has no extremum at $0$ .”

What to write

“ $f''(0) = 0$ : the Second Derivative Test gives no conclusion. $f'(x) = 4x^3$ goes from $-$ to $+$ at $0$ , so by the First Derivative Test $f$ has a local minimum at $0$ .”

Why: The test has only two outputs, maximum and minimum. When $f''(c) = 0$ it has not answered, and $x^4$ , $-x^4$ and $x^3$ show that all three answers are possible.

7. Claiming f is increasing on a union of intervals

1 mark

What not to write

“ $f(x) = -\frac{1}{x}$ , $f'(x) = \frac{1}{x^2} > 0$ , so $f$ is increasing on $(-\infty, 0) \cup (0, \infty)$ .”

What to write

“ $f' > 0$ on $(-\infty, 0)$ and on $(0, \infty)$ , so $f$ is increasing on $(-\infty, 0)$ and increasing on $(0, \infty)$ .”

Why: The test works on one interval. Across the gap it fails: $-1 < 1$ but $f(-1) = 1 > f(1) = -1$ . The union sign turns two true statements into one false one.

8. Giving where instead of how much

1 mark

What not to write

“The local maximum of $k(x) = x^2e^x$ is $x = -2$ .”

What to write

“ $k$ has a local maximum at $x = -2$ , and the local maximum VALUE is $k(-2) = 4e^{-2}$ .”

Why: “Find the local maximum values” asks for $f(c)$ ; “where does $f$ have a local maximum” asks for $c$ . Answer both when in doubt, and an inflection point is always a POINT, $(c, f(c))$ .

Which method to choose

Which test, by what you have in front of you

Look at the candidate and at the data given before choosing the test

If $f'(c) = 0$ and $f''(c)$ is quick to compute and NOT zero → Second Derivative Test: the sign of f'' at c gives max or min

Example: $\sin x + \cos x$ at $\frac{\pi}{4}$ : $f'' = -\sqrt 2 < 0$ , maximum $\sqrt 2$
If $f'(c) = 0$ and $f''(c) = 0$ → the second test is silent: First Derivative Test, sign of f' on each side

Example: $3x^5 - 5x^3$ at $0$ : $p' < 0$ on both sides, no extremum
If $f'(c)$ does not exist, or $f''(c)$ does not exist → First Derivative Test only

Example: $x^{4/3}$ at $0$ : $f'$ from $-$ to $+$ , minimum $0$
If $f'$ factors with a squared factor $(x - a)^2$ → sign chart, factor by factor: no sign change at $a$

Example: $12x(x - 1)^2$ : minimum at $0$ , nothing at $1$
If the graph of f' is given → axis crossings give extrema of f, turning points give inflection points of f

Example: $f' = \frac{(x + 2)(x - 1)^2}{4}$ : minimum at $-2$ , inflections at $-1$ and $1$
If the question asks for inflection points → list the zeros and undefined points of f'', then check the sign change of f'' at each

Example: $x^5 - 5x^4$ : $f'' = 20x^2(x - 3)$ , only $(3, -162)$

No branch uses a limit or an asymptote: those belong to the complete curve sketch. Here each question isolates what $f'$ or $f''$ says.

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.

Classifying the critical numbers with a sign chart

When to use it: Any question asking where $f$ is increasing or decreasing, or for its local maximum and minimum values

1 Give the domain, then compute $f'$ and FACTOR it completely.
2 List the critical numbers: the zeros of $f'$ AND the numbers of the domain where $f'$ does not exist.
3 Draw the sign chart: one row per factor, one column per critical number, one test value or factor sign in every interval.
4 Conclude the intervals of increase and decrease by naming the Increasing/Decreasing Test, one interval at a time.
5 At each critical number, name the First Derivative Test and give the VALUE $f(c)$ of each local extremum.

Concluding sentence

“Since $f'$ changes from negative to positive at $x = -4$ , $f$ has a local minimum there by the First Derivative Test, and the local minimum value is $f(-4) = -127$ .”

The trap: Stopping after solving $f'(x) = 0$ : the solutions are candidates, and the chart is the proof.

Marking: Typically 2 marks for f' factored, 1 for the complete list of critical numbers, 3 for the sign chart, 2 for the intervals, 2 for the classified extrema with their values.

Check before you hand in

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

Maxima and minima alternate

For a function that is continuous and differentiable on an interval, two local maxima always have a local minimum between them. A list max, max is a sign error.

$x^4 + 4x^3 - 8x^2 + 1$ : min at $-4$ , max at $0$ , min at $1$ ; the pattern alternates, as it must.
The values match the order

A local maximum value must be larger than the values at the neighbouring local minima. Compute the three values and compare.

$f(-4) = -127 < f(0) = 1 > f(1) = -2$ : consistent. A maximum smaller than a neighbouring minimum means a slip in the chart.
The ends of a polynomial

The sign of f' far to the left and far to the right is the sign of its leading term. The first and last cells of the chart must agree with it.

$f'(x) = 4x^3 + \dots$ : negative far left, positive far right, so the chart must start with $-$ and end with $+$ .
Inflection between extrema

Between two consecutive local extrema of a twice differentiable function, the concavity must change at least once. No inflection point there means a lost candidate.

$2x^3 - 3x^2 - 12x - 1$ : max at $-1$ , min at $2$ , inflection at $\frac{1}{2}$ , between them.

The typical problem, taken apart

A minimum, a flat inflection point, and a silent test

Let $f(x) = 3x^4 - 8x^3 + 6x^2 + 1$ . Find the intervals of increase and decrease, the local extreme values, the intervals of concavity and the inflection points of $f$ .

No calculator. Every conclusion must name the test that gives it, as on a MATH 140 final.

The graph drawn after the study: the only local minimum is

(0, 1)

, and at

(1, 2)

the curve flattens to a horizontal tangent without turning back, an inflection point.

Step 1

$f'(x) = 12x^3 - 24x^2 + 12x = 12x(x^2 - 2x + 1) = 12x(x - 1)^2$ . $f$ is a polynomial, so $f'$ exists everywhere: the critical numbers are $0$ and $1$ .

Why

Factoring completely is what makes the sign readable. The square $(x - 1)^2$ is the warning: at $1$ the sign of $f'$ will not change.

Step 2

Sign of $f'$ : $12x < 0$ and $(x - 1)^2 > 0$ on $(-\infty, 0)$ , so $f' < 0$ ; on $(0, 1)$ and $(1, \infty)$ , $f' > 0$ . $f$ is decreasing on $(-\infty, 0]$ and increasing on $[0, \infty)$ . Local minimum $f(0) = 1$ ; no extremum at $1$ .

Why

The increasing intervals $[0, 1]$ and $[1, \infty)$ share the point $1$ , so $f$ is increasing on $[0, \infty)$ as a whole. The candidate $1$ is rejected by the sign, not by a guess.

Step 3

$f''(x) = 36x^2 - 48x + 12 = 12(3x^2 - 4x + 1) = 12(3x - 1)(x - 1)$ . At $0$ : $f''(0) = 12 > 0$ , the Second Derivative Test confirms the minimum. At $1$ : $f''(1) = 0$ , the test is silent, and step 2 has already decided.

Why

Using both tests where they apply is a free check. At $1$ the silence of the second test is not a result: the result came from the sign chart.

Step 4

$12(3x - 1)(x - 1)$ is positive on $\left(-\infty, \frac{1}{3}\right)$ and $(1, \infty)$ , negative on $\left(\frac{1}{3}, 1\right)$ . The concavity changes at $\frac{1}{3}$ and at $1$ : inflection points $\left(\frac{1}{3}, \frac{38}{27}\right)$ and $(1, 2)$ .

Why

Both roots of $f''$ are SIMPLE, so both are real inflection points. The value $f\left(\frac{1}{3}\right) = \frac{1}{27} - \frac{8}{27} + \frac{18}{27} + \frac{27}{27} = \frac{38}{27}$ is computed over a common denominator.

Step 5

Check: $f(0) = 1 < f\left(\frac{1}{3}\right) = \frac{38}{27} < f(1) = 2$ , increasing as found in step 2; the point $(1, 2)$ has $f'(1) = 0$ and $f''$ changing sign: a horizontal tangent at an inflection point.

Why

The values must respect the monotonicity just proved. A value at $\frac{1}{3}$ below $1$ would reveal an arithmetic slip before the marker finds it.

The conclusion, written out

“ $f$ is decreasing on $(-\infty, 0]$ and increasing on $[0, \infty)$ , with a local minimum value $f(0) = 1$ and no extremum at $1$ . It is concave upward on $\left(-\infty, \frac{1}{3}\right)$ and $(1, \infty)$ , concave downward on $\left(\frac{1}{3}, 1\right)$ , with inflection points $\left(\frac{1}{3}, \frac{38}{27}\right)$ and $(1, 2)$ .”

The classic mistake on this problem: Declaring a maximum at $1$ because the chart was filled $-$ , $+$ , $-$ by habit, or skipping the inflection point $(1, 2)$ because it is also a critical number.

Learn by heart

• Critical number: $f'(c) = 0$ OR $f'(c)$ undefined, with $c$ in the domain. A candidate, never a verdict.
• Extremum: $f'$ CHANGES sign. Inflection point: $f''$ CHANGES sign, $f$ continuous at $c$ .
• Second Derivative Test: $f'(c) = 0$ , $f''(c) > 0$ minimum; $f''(c) < 0$ maximum; $f''(c) = 0$ or undefined, NO conclusion.
• A squared factor never changes sign: a double root of $f'$ or $f''$ is the candidate that fails.
• On the graph of $f'$ : crossings of the axis are extrema of $f$ , turning points are inflection points of $f$ .
• Increasing and decreasing are stated on INTERVALS, one at a time, never on a union across a gap.
• Local maximum at $c$ , local maximum VALUE $f(c)$ , inflection POINT $(c, f(c))$ .

Frequently asked questions

How do I find where a function is increasing or decreasing in MATH 140?

Compute the derivative and factor it completely. List the critical numbers, where the derivative is zero or does not exist. Then make a sign chart with one test value in every interval between them. Where the derivative is positive the function is increasing, where it is negative it is decreasing, stated interval by interval.

What is the difference between the first and second derivative tests?

The first derivative test looks at the sign of f prime on each side of a critical number: a change from plus to minus is a maximum, from minus to plus a minimum, no change means no extremum. It always applies. The second derivative test only looks at the sign of f double prime at the point; it is faster, but when f double prime is zero or undefined it gives no answer.

Is every point where the second derivative is zero an inflection point?

No. An inflection point needs a change of concavity, so the second derivative must change sign there. The function x to the fourth has a second derivative equal to zero at the origin, yet it is concave upward on both sides, and the origin is its minimum. A zero of the second derivative is only a candidate to be checked.

How do I read f from the graph of f prime?

Where the graph of f prime is above the axis, f is increasing; below, f is decreasing. Where f prime crosses the axis, f has a local maximum or minimum; where it only touches the axis, f has none. Where the graph of f prime rises, f is concave upward, and the turning points of f prime are the inflection points of f.

Can a function be increasing and concave down at the same time?

Yes. Increasing is about the sign of the first derivative, concave down about the sign of the second, and the two are independent. The square root function increases while bending down: it keeps rising, but more and more slowly. This is exactly what economists call diminishing returns.

Practise it

Corrected exercises: Shape of a graph, MATH 140 at McGill

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 The Mean Value Theorem Next sheet L'Hospital's rule

© Ahmed Squalli Houssaini. Revision sheet published at www.letuteurscientifique.ca/en/fiches/math140-shape-of-graph. Free for personal and classroom use; republishing it elsewhere requires written permission (legal notice).

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Revision sheet: the shape of a graph (MATH 140)

The essentials

Two derivatives, two independent questions

Candidates, then verdicts

The rules in table form

What the Second Derivative Test says at a critical number

The mistakes that cost marks

1. Declaring an extremum wherever f' is zero

2. Declaring an inflection point wherever f'' is zero

3. Forgetting the critical numbers where f' does not exist

4. Alternating the signs of a chart across a double root

5. Reading the graph of f' as the graph of f

6. Reading a silent Second Derivative Test as a no

7. Claiming f is increasing on a union of intervals

8. Giving where instead of how much

Which method to choose

Which test, by what you have in front of you

How the answer is expected to be written

Classifying the critical numbers with a sign chart

Check before you hand in

The typical problem, taken apart

A minimum, a flat inflection point, and a silent test

Learn by heart

Frequently asked questions

How do I find where a function is increasing or decreasing in MATH 140?

What is the difference between the first and second derivative tests?

Is every point where the second derivative is zero an inflection point?

How do I read f from the graph of f prime?

Can a function be increasing and concave down at the same time?

Corrected exercises: Shape of a graph, MATH 140 at McGill

See also

Looking for a MATH 140 tutor in Montreal?