The Second Derivative Test — AP® Calculus AB

Part 1 · Hook

Why this matters

A skateboarder at the bottom of a half-pipe is at a flat spot, and the ramp curves up on both sides. At the lip of a hill the board is also flat, but the ground curves down. Flatness plus the direction of the curve tells you which you have: that is the Second Derivative Test.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $f''(x) < 0$ on an interval, the graph of $f$ there is

concave down
concave up
decreasing
increasing

Show the answer

$f'' < 0$ means $f'$ is decreasing: the graph bends down.

Correct: concave down:
concave up:
decreasing:
increasing:

2. $f'$ changes from negative to positive at $x = c$ . Then $f$ has

a relative minimum at $c$
a relative maximum at $c$
a point of inflection at $c$
no extremum at $c$

Show the answer

The First Derivative Test.

Correct: a relative minimum at $c$ :
a relative maximum at $c$ :
a point of inflection at $c$ :
no extremum at $c$ :

Part 4 · See it

See it first

Three panels with a flat tangent line at x = c where f'(c) = 0. Left: concave up, f''(c) > 0, a relative minimum. Middle: concave down, f''(c) < 0, a relative maximum. Right: f''(c) = 0, and x to the fourth (a minimum), negative x to the fourth (a maximum) and x cubed (neither) show that the test says nothing. — At a critical point with $f'(c) = 0$ : concave up means a valley, concave down a peak. When $f''(c) = 0$ , anything can happen. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$f'(c) = 0$ and $f''(c) > 0$ the tangent line at $c$ is flat and the graph bends up away from it, so $f(c)$ is a relative minimum
$f'(c) = 0$ and $f''(c) < 0$ the graph bends down from a flat point, so $f(c)$ is a relative maximum
$f'(c) = 0$ and $f''(c) = 0$ (or $f''(c)$ does not exist)the test is inconclusive; go back to the First Derivative Test
A continuous $f$ has exactly one critical point on an interval, and it is a relative extremumthat relative extremum is also the absolute extremum on the interval

Part 6 · Key ideas

Key ideas

Second Derivative Test: if $f'(c) = 0$ and $f''(c) > 0$ , $f$ has a relative minimum at $c$ ; if $f'(c) = 0$ and $f''(c) < 0$ , a relative maximum.
It says nothing when $f''(c) = 0$ or when $f'(c)$ does not exist. Then use the First Derivative Test.
Justify with both facts: " $f'(1) = 0$ and $f''(1) > 0$ ."
One critical point on an interval, and it is a relative minimum (maximum)? Then it is the absolute minimum (maximum) there, even on an open interval.

Part 7 · Misconception

A common mistake

The wrong idea: $f''(c) > 0$ , so $f$ has a relative minimum at $c$ .

What actually happens: Concave up alone is not a minimum: $f(x) = x^2$ at $x = 1$ is concave up and rising. The test needs a critical point first, $f'(c) = 0$ , and then the sign of $f''(c)$ . And when $f''(c) = 0$ , the test says nothing at all.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Values of $f'$ and $f''$

f

is twice differentiable for all real numbers. The table gives values of its first and second derivatives.

Values of $f'(x)$ and $f''(x)$ at selected $x$
$x$	-2	0	1	3	4
$f'(x)$	0	-2	0	2	0
$f''(x)$	-3	2	4	-1	0

1. At which value of $x$ in the table does $f$ have a relative maximum?

$x = 1$
$x = -2$
$x = 4$
$x = 3$ , where $f''(3) < 0$

Show the answer

Second Derivative Test: $f'(-2) = 0$ and $f''(-2) < 0$ , so $f$ has a relative maximum at $x = -2$ .

$x = 1$ : $f'(1) = 0$ and $f''(1) > 0$ : a relative minimum.
Correct: $x = -2$ : $f'(-2) = 0$ and $f''(-2) = -3 < 0$ : concave down at a flat point.
$x = 4$ : $f''(4) = 0$ : the test gives no answer at $x = 4$ .
$x = 3$ , where $f''(3) < 0$ : $f'(3) = 2 \ne 0$ , so $x = 3$ is not a critical point.

2. Which justification earns credit for " $f$ has a relative minimum at $x = 1$ "?

" $f'(1) = 0$ and $f''(1) > 0$ ."
" $f''(1) > 0$ ."
" $f'(1) = 0$ ."
" $f'(0) < 0$ and $f'(3) > 0$ , so $f'$ changes sign."

Show the answer

The Second Derivative Test needs both: $f'(1) = 0$ (critical point) and $f''(1) = 4 > 0$ (concave up there), so $f$ has a relative minimum at $x = 1$ .

Correct: " $f'(1) = 0$ and $f''(1) > 0$ .": Both conditions of the Second Derivative Test.
" $f''(1) > 0$ .": Concave up alone is not a minimum; $f$ must also have a flat tangent line there.
" $f'(1) = 0$ .": A critical point alone is only a candidate.
" $f'(0) < 0$ and $f'(3) > 0$ , so $f'$ changes sign.": Values of $f'$ at two separate points do not show a sign change at $x = 1$ ; $f'$ might change sign more than once.

3. What does the Second Derivative Test say about $f$ at $x = 4$ ?

A relative maximum, because $f''(4) \le 0$
A point of inflection, because $f''(4) = 0$ and $f''$ is continuous
A relative minimum, because $f'(4) = 0$
Nothing, because $f''(4) = 0$ makes the test inconclusive

Show the answer

The Second Derivative Test needs $f''(c) \ne 0$ . Here $f''(4) = 0$ , so it gives no conclusion; check the sign of $f'$ on each side instead.

A relative maximum, because $f''(4) \le 0$ : The test needs $f''(4) < 0$ strictly.
A point of inflection, because $f''(4) = 0$ and $f''$ is continuous: $f''(4) = 0$ does not show a sign change of $f''$ .
A relative minimum, because $f'(4) = 0$ : $f'(4) = 0$ only makes 4 a critical point.
Correct: Nothing, because $f''(4) = 0$ makes the test inconclusive: With $f''(4) = 0$ , the point could be a maximum, a minimum or neither. Use the First Derivative Test.

4. Which describes the graph of $f$ at $x = 3$ ?

Increasing and concave up
Decreasing and concave down
Increasing and concave down
A relative maximum, because $f''(3) < 0$

Show the answer

$f'(3) = 2 > 0$ , so $f$ is increasing; $f''(3) = -1 < 0$ , so it is concave down.

Increasing and concave up: $f''(3) < 0$ means concave down.
Decreasing and concave down: $f'(3) > 0$ means increasing.
Correct: Increasing and concave down: $f'(3) = 2 > 0$ and $f''(3) = -1 < 0$ .
A relative maximum, because $f''(3) < 0$ : $x = 3$ is not a critical point, so the test does not apply.

5. Find the absolute minimum value of $f(x) = x + \frac{1}{x}$ on $(0, \infty)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$f'(x) = 1 - \frac{1}{x^2} = 0$ at $x = 1$ (the one critical point in $(0, \infty)$ ). $f''(x) = \frac{2}{x^3}$ , so $f''(1) = 2 > 0$ : a relative minimum, and as the only critical point it is the absolute minimum, $f(1) = 2$ .

Answer: 2

6. A company's profit is $P(x)$ dollars when it makes $x$ units. $P'(50) = 0$ and $P''(50) = -0.4$ . What can you conclude?

Making 50 units gives a relative maximum of profit
Making 50 units gives a relative minimum of profit
Profit is increasing at a decreasing rate at 50 units
At 50 units, profit is $-0.4$ dollars

Show the answer

$P'(50) = 0$ and $P''(50) < 0$ : by the Second Derivative Test, $P$ has a relative maximum at $x = 50$ .

Correct: Making 50 units gives a relative maximum of profit: A flat point where the graph is concave down: a peak.
Making 50 units gives a relative minimum of profit: $P'' < 0$ means concave down: a maximum, not a minimum.
Profit is increasing at a decreasing rate at 50 units: $P'(50) = 0$ , so profit is not increasing at 50 units.
At 50 units, profit is $-0.4$ dollars: $-0.4$ is the value of $P''$ , not of $P$ .

Part 9 · Summary

Summary

At a critical point where

f'(c) = 0

, the sign of

f''(c)

decides: positive gives a relative minimum, negative a relative maximum, zero gives no conclusion. State both conditions in a justification. With exactly one critical point on an interval, a relative extremum there is absolute.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Values of f′f' and f″f''

Summary

What comes next

Connections

Values of $f'$ and $f''$