Connecting f, f′, and f″

On the AP exam you’ll often be shown the graph of $f'$ , not $f$ , and asked about $f$ : where it increases, where it has a maximum, where it’s concave up. This page puts all the connections between $f$ , $f'$ , and $f''$ in one place, so you can move between the three graphs confidently. The trick is to keep asking: “which graph am I looking at, and what does its height tell me?”

Key ideas

The big table

If this is true…	…then this is true about $f$
$f' \gt 0$ (graph of $f'$ above the $x$ -axis)	$f$ is increasing
$f' \lt 0$ (graph of $f'$ below the $x$ -axis)	$f$ is decreasing
$f'$ changes from $+$ to $-$ (crosses the axis going down)	$f$ has a relative maximum
$f'$ changes from $-$ to $+$ (crosses the axis going up)	$f$ has a relative minimum
$f'$ is increasing, or $f'' \gt 0$	$f$ is concave up
$f'$ is decreasing, or $f'' \lt 0$	$f$ is concave down
$f'$ switches between increasing and decreasing, or $f''$ changes sign	$f$ has a point of inflection

The pattern: each graph’s sign tells you whether the graph above it rises or falls, and each graph’s zeros (with a sign change) mark the turning points of the graph above it.

Follow the dashed lines: the extrema of

f

line up with the zeros of

f'

, and the inflection point of

f

lines up with the minimum of

f'

and the zero of

f''

Reading a graph of f′

When the graph you’re given is $f'$ :

Its height (above or below the axis) is the slope of $f$ .
Its zeros are the critical points of $f$ . Check whether the graph crosses (extremum of $f$ ) or just touches (no extremum).
Its own slope (rising or falling) is $f''$ , so it tells you the concavity of $f$ .
Where it switches between rising and falling (its peaks and valleys, including corners), $f$ has inflection points.

What a graph of $f'$ does not tell you is the height of $f$ . Many functions have the same derivative (they differ by a constant), so you can’t read $f(2)$ off a graph of $f'$ without extra information. That’s what integration is for, in the next unit.

Sketching f′ from a graph of f

Go left to right across the graph of $f$ :

Wherever $f$ has a horizontal tangent, the graph of $f'$ touches or crosses the $x$ -axis.
Where $f$ rises, draw $f'$ above the axis; where it falls, below. Steeper means farther from the axis.
Where $f$ has a point of inflection (steepest rise or fall), $f'$ has a peak or valley.
Corners or cusps on $f$ become breaks in $f'$ ( $f'$ is undefined there).

Justify with the graph you’re given

If a question gives the graph of $f'$ , your reasons should talk about $f'$ : ” $f'$ changes from positive to negative at $x = 0$ ”, or ” $f'$ is decreasing on $(-2, 2)$ , so $f$ is concave down there.” Saying ” $f$ goes up then down” describes a graph you weren’t shown and doesn’t earn the point.

Worked examples

The figure below is used for Examples 1 and 2 and some practice questions. It shows the graph of $f'$ , the derivative of a function $f$ , on $-4 \le x \le 6$ . It is made of straight segments joining $(-4, -2)$ , $(-2, 2)$ , $(2, -2)$ , $(4, 0)$ , and $(6, -1)$ .

The graph of

f'

, not

f

. It crosses the

x

-axis at

x = -3

and

x = 0

, and touches it at

x = 4

Example 1: Increasing, decreasing, and extrema from f′

Using the graph of $f'$ , find the intervals where $f$ is increasing and decreasing, and the $x$ -values of the relative extrema of $f$ . Justify your answers.

Solution. Read the sign of $f'$ :

$f'(x) \gt 0$ on $(-3, 0)$ , so $f$ is increasing on $(-3, 0)$ .
$f'(x) \lt 0$ on $(-4, -3)$ , $(0, 4)$ , and $(4, 6)$ , so $f$ is decreasing on those intervals. (Since $f'$ is only $0$ at the single point $x = 4$ , you can also say $f$ is decreasing on $(0, 6)$ .)

Extrema:

$f'$ changes from negative to positive at $x = -3$ , so $f$ has a relative minimum at $x = -3$ .
$f'$ changes from positive to negative at $x = 0$ , so $f$ has a relative maximum at $x = 0$ .
At $x = 4$ , $f' = 0$ but $f'$ doesn’t change sign (negative on both sides), so $f$ has no relative extremum there.

Example 2: Concavity and inflection points from f′

Using the same graph, find the intervals where $f$ is concave up and concave down, and the $x$ -values of the points of inflection of $f$ .

Solution. Now read whether $f'$ is rising or falling (the slope of the graph of $f'$ is $f''$ ):

$f'$ is increasing on $(-4, -2)$ and $(2, 4)$ , so $f$ is concave up there.
$f'$ is decreasing on $(-2, 2)$ and $(4, 6)$ , so $f$ is concave down there.

$f'$ switches between increasing and decreasing at $x = -2$ , $x = 2$ , and $x = 4$ , so $f$ has points of inflection at all three. (At the corners of the graph of $f'$ , $f''$ doesn’t exist, but the concavity of $f$ still changes.)

Notice that $x = 4$ is both a critical point of $f$ and an inflection point: the graph of $f$ flattens out while switching from concave up to concave down, then keeps falling.

Example 3: From f to f′

The top graph in the stacked figure is $f(x) = \dfrac{x^3}{3} - x$ . Describe the graph of $f'$ without differentiating, then check.

Solution. Reading the graph of $f$ left to right:

$f$ rises until $x = -1$ , so $f' \gt 0$ for $x \lt -1$ .
$f$ has horizontal tangents at $x = -1$ and $x = 1$ , so $f'(-1) = f'(1) = 0$ .
$f$ falls between them, so $f' \lt 0$ on $(-1, 1)$ .
$f$ is steepest going down at its inflection point $x = 0$ , so $f'$ has its minimum there.
$f$ rises after $x = 1$ , more and more steeply, so $f' \gt 0$ and growing.

That describes an upward parabola with zeros at $\pm 1$ and its vertex at $x = 0$ . Check: $f'(x) = x^2 - 1$ . ✓

Example 4: From sign information to a sketch

A continuous function $f$ has these properties:

Interval	$x \lt -1$	$-1 \lt x \lt 1$	$1 \lt x \lt 3$	$x \gt 3$
$f'$	$+$	$-$	$-$	$+$
$f''$	$-$	$-$	$+$	$+$

Describe the shape of the graph of $f$ and name its key features.

Solution. Go interval by interval:

$x \lt -1$ : increasing and concave down (rising, but flattening).
$-1 \lt x \lt 1$ : decreasing and concave down (falling more and more steeply).
$1 \lt x \lt 3$ : decreasing and concave up (still falling, but levelling off).
$x \gt 3$ : increasing and concave up (rising more and more steeply).

Features: $f'$ changes from $+$ to $-$ at $x = -1$ , so a relative maximum there. $f''$ changes sign at $x = 1$ , so a point of inflection there. $f'$ changes from $-$ to $+$ at $x = 3$ , so a relative minimum there. The graph looks like a stretched “N” shape, similar to a cubic.

Common mistakes

Reading the graph of f′ as if it were f. The highest point on a graph of $f'$ is not a maximum of $f$ . It’s where $f$ is increasing fastest: an inflection point. Write “graph of $f'$ ” in the margin to remind yourself.

Thinking “f′ is decreasing” means “f is decreasing”. $f$ decreasing needs $f' \lt 0$ (below the axis). $f'$ decreasing means $f$ is concave down, which can happen while $f$ is still rising, like on $(-2, 0)$ in Example 1.

Calling every zero of f′ an extremum. At $x = 4$ in Example 1, the graph of $f'$ touches the axis without crossing, so there’s no extremum. Look for a sign change.

Missing inflection points at corners. On a piecewise-linear graph of $f'$ , the corners where $f'$ switches from rising to falling are inflection points of $f$ , even though $f''$ is undefined there.

Trying to read values of f from a graph of f′. The graph of $f'$ tells you about the slope and shape of $f$ , not its height. You need an initial value and integration to find $f(x)$ .

Practice

1. (Warm-up) If $f''(x) \lt 0$ on an interval, what is the graph of $f'$ doing there? What is the graph of $f$ doing?

Solution

$f''$ is the derivative of $f'$ , so $f'' \lt 0$ means $f'$ is decreasing. The graph of $f$ is concave down.

2. (Warm-up) On the interval $(2, 5)$ , $f'$ is positive and decreasing. Describe the graph of $f$ there.

Solution

$f' \gt 0$ , so $f$ is increasing. $f'$ is decreasing, so $f$ is concave down. The graph of $f$ is rising, but more and more slowly.

3. (Warm-up) The graph of $f'$ crosses the $x$ -axis from above to below at $x = 3$ . What does $f$ have at $x = 3$ ? Write the justification.

Solution

A relative maximum. ” $f'$ changes from positive to negative at $x = 3$ , so $f$ has a relative maximum at $x = 3$ .”

4. (Core) The graph of $f'$ is the parabola $y = x^2 - 2x - 3$ , with $x$ -intercepts $-1$ and $3$ and vertex $(1, -4)$ .

(a) Where is $f$ increasing and decreasing?
(b) Where does $f$ have relative extrema?
(c) Where is $f$ concave up and down, and where is its point of inflection?

Solution

(a) The parabola is above the axis for $x \lt -1$ and $x \gt 3$ , and below on $(-1, 3)$ . So $f$ is increasing on $(-\infty, -1)$ and $(3, \infty)$ , and decreasing on $(-1, 3)$ .

(b) $f'$ changes from $+$ to $-$ at $x = -1$ : relative maximum. $f'$ changes from $-$ to $+$ at $x = 3$ : relative minimum.

(c) The parabola is falling for $x \lt 1$ and rising for $x \gt 1$ . So $f$ is concave down on $(-\infty, 1)$ and concave up on $(1, \infty)$ , with a point of inflection at $x = 1$ (the vertex of the graph of $f'$ ).

5. (Core) Suppose $f'(x) = \sin x$ on $(0, 2\pi)$ (radians). Find where $f$ is increasing and decreasing, its relative extrema, its intervals of concavity, and the $x$ -values of its points of inflection.

Solution

$\sin x \gt 0$ on $(0, \pi)$ and $\lt 0$ on $(\pi, 2\pi)$ . So $f$ is increasing on $(0, \pi)$ and decreasing on $(\pi, 2\pi)$ , with a relative maximum at $x = \pi$ ( $f'$ changes from $+$ to $-$ ).

$f''(x) = \cos x$ , which is positive on $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3\pi}{2}, 2\pi\right)$ , and negative on $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$ . So $f$ is concave up on $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3\pi}{2}, 2\pi\right)$ , and concave down on $\left(\frac{\pi}{2}, \frac{3\pi}{2}\right)$ .

Points of inflection at $x = \dfrac{\pi}{2}$ and $x = \dfrac{3\pi}{2}$ , where $f'' = \cos x$ changes sign (and where $f' = \sin x$ has its maximum and minimum).

6. (Core) A differentiable function $f$ has a relative maximum at $(-1, 4)$ , a relative minimum at $(2, -1)$ , and a point of inflection at $(0.5, 1.5)$ , and no other turning points or inflection points. Describe the graph of $f'$ .

Solution

$f'(-1) = 0$ and $f'(2) = 0$ : the graph of $f'$ has zeros at $x = -1$ and $x = 2$ .
$f$ rises before $-1$ , falls between $-1$ and $2$ , and rises after $2$ : so $f' \gt 0$ for $x \lt -1$ , $f' \lt 0$ on $(-1, 2)$ , and $f' \gt 0$ for $x \gt 2$ .
$f$ is falling most steeply at the inflection point, so $f'$ has its minimum at $x = 0.5$ .

The graph of $f'$ looks like an upward-opening, U-shaped curve with $x$ -intercepts $-1$ and $2$ and its lowest point at $x = 0.5$ . (You can’t tell the exact $y$ -value of that lowest point.)

7. (Core) Using the graph of $f'$ from the worked examples, on which intervals is $f$ both decreasing and concave up? Explain.

Solution

$f$ is decreasing where $f' \lt 0$ and concave up where $f'$ is increasing. Both are true on $(-4, -3)$ and $(2, 4)$ : the graph of $f'$ is below the axis and rising.

8. (Core) Using the graph of $f'$ from the worked examples, find $f''(1)$ , $f''(3)$ , and $f''(5)$ . Is $f''(-2)$ defined?

Solution

$f''$ is the slope of the graph of $f'$ .

From $(-2, 2)$ to $(2, -2)$ , the slope is $\dfrac{-2 - 2}{2 - (-2)} = -1$ , so $f''(1) = -1$ .
From $(2, -2)$ to $(4, 0)$ , the slope is $\dfrac{0 - (-2)}{4 - 2} = 1$ , so $f''(3) = 1$ .
From $(4, 0)$ to $(6, -1)$ , the slope is $\dfrac{-1 - 0}{6 - 4} = -\dfrac{1}{2}$ , so $f''(5) = -\dfrac{1}{2}$ .

$f''(-2)$ is not defined: the graph of $f'$ has a corner there (slope $2$ on the left, $-1$ on the right).

9. (Challenge) True or false? Explain each.

(a) If $f''(c) = 0$ , then the graph of $f$ has a point of inflection at $x = c$ .
(b) If $f$ is continuous at $x = c$ and $f'$ changes from increasing to decreasing at $x = c$ , then the graph of $f$ has a point of inflection at $x = c$ .

Solution

(a) False. $f(x) = x^4$ has $f''(0) = 0$ , but $f''(x) = 12x^2$ doesn’t change sign, so there’s no inflection point.

(b) True. Where $f'$ is increasing, $f$ is concave up; where $f'$ is decreasing, $f$ is concave down. So $f$ changes from concave up to concave down at $x = c$ , and since $f$ is continuous there, that’s a point of inflection.

Be careful with the wording, though. ” $f'$ has a relative maximum at $c$ ” on its own isn’t quite enough: if $f(x) = x$ , then $f'(x) = 1$ is constant, and every point counts as a (non-strict) relative maximum of $f'$ , yet a straight line has no inflection points. What matters is that $f'$ actually switches from increasing to decreasing.

10. (Challenge) Using the graph of $f'$ from the worked examples, let $g(x) = f(x) - x$ . Find the intervals where $g$ is increasing, and the $x$ -values of the relative extrema of $g$ .

Solution

$g'(x) = f'(x) - 1$ , so $g$ is increasing where $f'(x) \gt 1$ : where the graph of $f'$ is above the line $y = 1$ .

On $[-4, -2]$ , the segment through $(-4, -2)$ and $(-2, 2)$ is $f'(x) = 2x + 6$ . $2x + 6 \gt 1$ when $x \gt -2.5$ .
On $[-2, 2]$ , the segment is $f'(x) = -x$ . $-x \gt 1$ when $x \lt -1$ .
For $x \ge 2$ , $f'(x) \le 0 \lt 1$ .

So $g$ is increasing on $(-2.5, -1)$ and decreasing on $(-4, -2.5)$ and $(-1, 6)$ .

$g'$ changes from negative to positive at $x = -2.5$ : relative minimum. $g'$ changes from positive to negative at $x = -1$ : relative maximum.