From the graph of f′ to the graph of f: what each feature means

By Marius, private math tutor in Montreal · 2026-10-05

The graph of $f^{'}$ shows the slope of $f$ at every $x$ , so each feature of it is a fact about $f$ : where $f^{'}$ is positive, $f$ rises; where it crosses zero, $f$ turns around; where it peaks or bottoms out, $f$ changes the way it bends. Here is the full translation, a worked example drawn both ways, and a self-check on a graph of $f^{'}$ .

The translation, feature by feature

Every reading uses one of two things: where the graph of $f^{'}$ is (above or below the axis) and which way it is going (uphill or downhill, which is what $f^{″}$ measures).

$f^{'}$ above the axis on an interval: $f$ is increasing there.
$f^{'}$ below the axis: $f$ is decreasing.
$f^{'}$ crosses the axis going down, from positive to negative: $f$ has a relative maximum.
$f^{'}$ crosses the axis going up, from negative to positive: $f$ has a relative minimum.
$f^{'}$ touches the axis and turns back without crossing: $f$ has a horizontal tangent there but no extremum. It levels off for a moment and keeps rising, or keeps falling, as before.
$f^{'}$ increasing, above the axis or below it: $f$ is concave up.
$f^{'}$ decreasing: $f$ is concave down.
A peak or a valley of $f^{'}$ , where it switches between increasing and decreasing: the concavity of $f$ changes, so $f$ has a point of inflection.
The height of $f^{'}$ : how steep $f$ is. On a stretch where $f$ rises, the highest point of $f^{'}$ marks where it rises fastest.
$f^{'}$ constant at a value $m$ on an interval: $f$ is a straight line of slope $m$ there.

Some features count twice. A touch point is also a peak or a valley of $f^{'}$ , so it is a horizontal tangent and a point of inflection of $f$ . And a peak of $f^{'}$ above the axis is a point of inflection where $f$ is still rising, at its steepest.

The peak of $f^{'}$ is not the peak of $f$ . $f$ has a relative maximum where $f^{'}$ crosses the axis from positive to negative. The highest point of the graph of $f^{'}$ is something else: the place where the slope of $f$ is greatest, and when it is a peak of $f^{'}$ it is a point of inflection of $f$ , not an extremum. The trap is reading the bumps of the graph you were given as bumps of $f$ .

Worked example: reading $f^{'} (x) = (x + 1)^{2} (2 - x)$

The top graph in the figure is $f^{'} (x) = (x + 1)^{2} (2 - x)$ . Here is what each of its features says about $f$ .

Sign of $f^{'}$ . $(x + 1)^{2}$ is never negative, so $f^{'}$ has the sign of $2 - x$ : positive for $x < 2$ , except at $x = - 1$ where it is $0$ , and negative for $x > 2$ . So $f$ is increasing on $(- \infty, 2)$ and decreasing on $(2, \infty)$ .
At $x = 2$ , a crossing going down. $f^{'}$ changes from positive to negative, so $f$ has a relative maximum at $x = 2$ .
At $x = - 1$ , a touch. $f^{'} (- 1) = 0$ , but $f^{'} > 0$ on both sides. There is no sign change, so there is no extremum: $f$ flattens and keeps rising.
Direction of $f^{'}$ . $f^{″} (x) = 3 (x + 1) (1 - x)$ is negative for $x < - 1$ , positive on $(- 1, 1)$ and negative for $x > 1$ . So $f$ is concave down on $(- \infty, - 1)$ , concave up on $(- 1, 1)$ and concave down on $(1, \infty)$ .
Points of inflection at $x = - 1$ and $x = 1$ , at the valley of $f^{'}$ (the touch point) and at its peak. For $x > - 1$ , the largest value of $f^{'}$ is $f^{'} (1) = 4$ , so that is where $f$ climbs most steeply.

The bottom graph is one function with this derivative, $f (x) = - \frac{x^{4}}{4} + \frac{3 x^{2}}{2} + 2 x$ , and every conclusion above can be checked on it.

Top:

f^{'} (x) = (x + 1)^{2} (2 - x)

. Bottom:

f (x) = - \frac{x^{4}}{4} + \frac{3 x^{2}}{2} + 2 x

, one function with that derivative. The dashed lines at

x = - 1

1

and

2

match each feature of

f^{'}

to what it does to

f

What the graph of $f^{'}$ cannot tell you

The graph of $f^{'}$ fixes the shape of $f$ , not its height. The functions $f$ and $f + 5$ have the same derivative, so their graphs are the same curve, one sitting five units above the other. To place $f$ you need one actual value of it; the bottom graph above uses $f (0) = 0$ . How much $f$ changes between two points can also be read from $f^{'}$ , as an area, but that is integration, the subject of Unit 6 of AP Calculus AB.

How the AP course frames it

Reading $f$ from a graph of $f^{'}$ is built into the course. The College Board's Course and Exam Description advises that “given a graph of the derivative, f′, of a function, f, it is safer and easier for students to make arguments about f based directly on the graph of the derivative,” and one of its sample free-response questions is built on a graph of $f^{'}$ . Its model sentence is the one to copy: “f is concave up on a < x < b because the graph of f′ is increasing on a < x < b.” The same passage expects you to call $f$ , $f^{'}$ and $f^{″}$ by name rather than “it”.

Check your understanding

Every question refers to this graph of $f^{'}$ , the derivative of a function $f$ . No calculator needed.

The graph of

f^{'}

[0, 20]

. It crosses the

x

-axis at

x = 1

and

x = 9

, touches it at

x = 19

, and has horizontal tangents at

x = 4

x = 13

and

x = 19

On what open interval is $f$ decreasing?
At what value of $x$ does $f$ have a relative maximum, and at what value a relative minimum? Give a reason for each.
$f^{'} (19) = 0$ . Does $f$ have a relative extremum at $x = 19$ ? Explain.
On what open intervals is the graph of $f$ concave up?
Find the $x$ -coordinate of every point of inflection of the graph of $f$ on $(0, 20)$ .
On the interval $(9, 19)$ , at what value of $x$ is $f$ increasing most rapidly?
Find the error: “On $(9, 19)$ the graph of $f^{'}$ is highest at $x = 13$ , so $f$ has a relative maximum at $x = 13$ .”
True or false: the graph of $f^{'}$ is enough to decide whether $f (0)$ is positive.

Show the answers

$(1, 9)$ , the only stretch where the graph of $f^{'}$ is below the axis.
A relative maximum at $x = 1$ , because $f^{'}$ changes from positive to negative there, and a relative minimum at $x = 9$ , because $f^{'}$ changes from negative to positive.
No. $f^{'} > 0$ just left and just right of $19$ , so $f$ is increasing on both sides. The graph of $f$ has a horizontal tangent there and keeps rising.
$(4, 13)$ and $(19, 20)$ , where the graph of $f^{'}$ is increasing, which is what concave up means.
$x = 4$ , $x = 13$ and $x = 19$ . At each one $f^{'}$ switches between decreasing and increasing, so the concavity of $f$ changes.
$x = 13$ . On that interval $f^{'}$ is largest at its peak, and the value of $f^{'}$ is the rate at which $f$ increases.
A relative maximum of $f$ needs $f^{'}$ to change from positive to negative. At $x = 13$ , $f^{'}$ is positive on both sides, so $f$ is still rising; the peak of $f^{'}$ is where $f$ rises fastest, a point of inflection.
False. $f$ and $f + C$ have the same derivative for every constant $C$ , so the graph fixes the shape of $f$ but not its height. You would need the value of $f$ at some point.

To practise writing these justifications in full, with sign charts for $f^{'}$ and $f^{″}$ , work through the free AP Calculus AB Unit 5 worksheet.

Want one-on-one help with this? I offer AP Calculus AB tutoring and exam preparation in Montreal, in person or online. Sessions start with a short written diagnostic, and a written recap goes to whoever books after every session.

Practise it on a free worksheet

Sources

AP Calculus AB and BC Course and Exam Description (Unit 5 overview, topic 5.9, and sample free-response question 2) — College Board. Checked 2026-10-05.

From the graph of f′ to the graph of f: what each feature means

The translation, feature by feature

Worked example: reading f′(x)=(x+1)2(2−x)

What the graph of f′ cannot tell you