How do I tell the difference between a max of f and an inflection point when looking at the graph of f'?

Look at what the f' graph is doing. Where f' crosses zero and changes sign, f has an extremum. Where f' reaches a peak or valley (a local extremum of f' itself), f has an inflection point. Crossing the axis controls extrema; the slope of f' controls concavity.

If I'm given the graph of f'', can I find where f is increasing?

No. The graph of f'' only gives concavity and inflection points of f. Increasing and decreasing behavior depends on the sign of f', which you cannot recover from f'' alone without additional information such as a value of f'.

Why is my sketch of f not unique when I'm only given f'?

Any two functions that differ by a constant have identical derivatives, so f' determines the shape of f but not its vertical position. Unless the problem provides an initial condition like f(0)=5, you can only draw the correct shape, not a fixed height.

Does f''=0 always mean there is an inflection point?

No. f''=0 is necessary but not sufficient. You must verify that f'' actually changes sign there. If f'' touches zero but stays the same sign — such as behaving like x^2 — the concavity does not change and there is no inflection point.

AP-CALCBC-5.8-5.9

U5.8 Connecting f, f', f'' through Graphs

Master reading f, f', and f'' from graphs. Learn to sketch antiderivatives, find extrema, inflection points, and concavity from graphical data for AP Calc BC.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U5.8 Connecting f, f', f'' through Graphs, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

Start U5.8 Connecting f, f', f'' through Graphs free

What this lesson covers

On the AP exam, you'll often be handed the graph of

f'

and asked questions about

f

itself — where it increases, where it has a local max, where it's concave up. This is one of the most heavily tested skills in Unit 5 because it forces you to think about derivatives as information rather than formulas.

This lesson ties together everything you know about first and second derivatives, but now the input is a picture. You'll learn to translate graphical features of

f'

and

f''

into precise statements about the shape of

f

, and to sketch

f

when only its derivatives are given.

Translating Between the Graphs

The central skill is knowing which feature of one graph corresponds to which feature of another. Every statement about

f

increasing or decreasing is really a statement about the sign of

f'

, and every statement about concavity is a statement about the sign of

f''

.

When you look at the graph of

f'

, you read the

y

-values as slopes of

f

. Where

f'

is above the axis,

f

rises; where

f'

is below,

f

falls. Where

f'

crosses zero,

f

has a potential extremum. The slope of the

f'

graph itself equals

f''

, which controls concavity.

Feature of $f$	Condition on $f'$	Condition on $f''$
Increasing	$f' > 0$	—
Decreasing	$f' < 0$	—
Local max	$f'$ changes $+$ to $-$	$f'' < 0$
Local min	$f'$ changes $-$ to $+$	$f'' > 0$
Concave up	$f'$ increasing	$f'' > 0$
Concave down	$f'$ decreasing	$f'' < 0$
Inflection point	$f'$ has local extremum	$f''$ changes sign

Memorizing this table is not enough; you must practice reading each row directly off a picture.

Reading f from the Graph of f'

Suppose you are given only the graph of

f'

. The most common mistake is treating the

f'

graph as if it were

f

. When students see a peak on the

f'

graph, they wrongly announce a local max of

f

. In reality, a peak of

f'

means

f'

has a maximum slope-value, so

f''=0

there and

f

has an inflection point, not an extremum.

To find local extrema of

f

, look for where the

f'

graph crosses the

x

-axis, then check the sign change. A crossing from positive to negative gives a local maximum of

f

; negative to positive gives a local minimum. A touch that does not cross (like

f'

tangent to the axis) is not an extremum because the sign does not change.

To find concavity of

f

, ask whether the

f'

graph is rising or falling. Rising

f'

means

f

is concave up; falling

f'

means concave down. Inflection points of

f

occur where

f'

switches from rising to falling or vice versa — the local maxima and minima of the

f'

graph.

The exam loves to combine these: a single

f'

graph can generate questions about increasing intervals, extrema, concavity, and inflection points all at once. Work through each independently rather than trying to picture

f

all at once.

Reading f from the Graph of f''

Sometimes the AP exam gives you the graph of

f''

directly. Here the only information about

f

you can extract is concavity and inflection points — you cannot determine where

f

increases or decreases, because

f''

tells you nothing about the sign of

f'

.

Where

f'' > 0

(graph above the axis),

f

is concave up. Where

f'' < 0

f

is concave down. Inflection points of

f

occur where

f''

changes sign, i.e., where the

f''

graph crosses the axis (not merely touches it).

A classic trap: students see

f''

equal to zero and immediately claim an inflection point. But

f''=0

is necessary, not sufficient. You must confirm a sign change. If

f''

touches zero and stays the same sign — for example

f''=x^2

style behavior — there is no inflection point.

When the problem gives

f''

and asks about the second derivative test for an extremum you already located, evaluate the sign of

f''

at that critical point: positive means local min, negative means local max, and zero is inconclusive.

Sketching f from Its Derivatives

To sketch

f

from the graph of

f'

, proceed feature by feature. First mark all

x

-values where

f'=0

; these are critical points. Second, note the sign of

f'

on each interval to decide where

f

rises and falls. Third, mark where

f'

has local extrema — these become inflection points of

f

. Finally, use whether

f'

is increasing or decreasing to set concavity.

Your sketch of

f

is unique only up to a vertical shift, because any constant added to

f

has the same derivative. If the problem gives an initial condition like

f(0)=2

, use it to pin down the vertical position; otherwise draw a representative curve with correct shape.

A smooth checklist helps: at each local max of

f

the curve should have a horizontal tangent and turn from rising to falling; at each inflection point the concavity flips but the curve keeps moving in the same direction. Make sure your

f

is steepest where

|f'|

is largest and flattest where

f'

is near zero. Graders reward correct qualitative features — turning points, concavity, and inflection locations — far more than artistic precision.

Key terms

Critical point.: An $x$ -value where $f'(x)=0$ or $f'(x)$ is undefined; a candidate for a local extremum of $f$ .
Local maximum.: A point where $f$ changes from increasing to decreasing, i.e., $f'$ changes from positive to negative.
Local minimum.: A point where $f$ changes from decreasing to increasing, i.e., $f'$ changes from negative to positive.
Inflection point.: A point where the concavity of $f$ changes; occurs where $f''$ changes sign, equivalently where $f'$ has a local extremum.
Concave up.: A region where $f'' > 0$ and $f'$ is increasing; the graph of $f$ bends upward like a cup.
Concave down.: A region where $f'' < 0$ and $f'$ is decreasing; the graph of $f$ bends downward like a frown.
Second Derivative Test.: At a critical point where $f'=0$ : if $f''>0$ it's a local min, if $f''<0$ it's a local max, if $f''=0$ the test is inconclusive.

Worked example

The graph of

f'

(the derivative of

f

) consists of a line and is positive on

(-\infty,1)

, crosses zero at

x=1

going negative, reaches a minimum at

x=3

, then rises and crosses zero at

x=5

going positive. Identify where

f

has local extrema, and locate any inflection points.

Start with extrema of

f

, which come from sign changes of

f'

.

At

x=1

f'

changes from positive to negative. Since

f

switches from increasing to decreasing,

f

has a local maximum at

x=1

.

At

x=5

f'

changes from negative to positive, so

f

switches from decreasing to increasing, giving a local minimum at

x=5

.

Now find inflection points, which come from local extrema of

f'

(where

f''=0

and changes sign). The graph of

f'

reaches a minimum at

x=3

. To the left of

x=3

f'

is decreasing, so

f''<0

and

f

is concave down. To the right,

f'

is increasing, so

f''>0

and

f

is concave up. Because concavity changes at

x=3

, there is an inflection point at $x=3$ .

Note that

x=3

is neither a max nor a min of

f

, even though it is a special point of the

f'

graph — a common trap. The final answer: local max at

x=1

, local min at

x=5

, inflection point at

x=3

Practice questions

The graph of

f'

has a local maximum at

x=2

. What can you conclude about

f

x=2

$f$ has a local maximum at $x=2$
$f$ has a local minimum at $x=2$
$f$ has an inflection point at $x=2$
$f$ has a vertical tangent at $x=2$

Answer: $f$ has an inflection point at $x=2$

A local maximum of

f'

means

f'

stops increasing and starts decreasing, so

f''

changes from positive to negative. A sign change in

f''

is exactly an inflection point of

f

. It says nothing about extrema of

f

, which depend on the sign of

f'

, not its slope.

You are given only the graph of

f''

, which is negative on

(0,4)

and positive on

(4,7)

. Describe the concavity of

f

and state where any inflection point occurs. Explain why you cannot determine the intervals where

f

is increasing.

Answer: $f$ is concave down on $(0,4)$ , concave up on $(4,7)$ , with an inflection point at $x=4$ . Increasing/decreasing behavior cannot be determined.

The sign of

f''

gives concavity directly: negative means concave down, positive means concave up, and the sign change at

x=4

marks an inflection point. Whether

f

increases depends on the sign of

f'

, and

f''

only tells us how

f'

is changing, not its actual value — so without more information the direction of

f

is unknown.

The graph of

f'

crosses the

x

-axis at

x=-2

(going from negative to positive) and touches the axis at

x=3

without crossing. Classify each point for

f

Answer: $x=-2$ is a local minimum of $f$ ; $x=3$ is not an extremum but is likely an inflection point.

x=-2

f'

changes sign from negative to positive, so

f

changes from decreasing to increasing: a local minimum. At

x=3

f'

touches zero but keeps the same sign, so

f

has a horizontal tangent yet no sign change — not an extremum. Because

f'

reaches an extreme value there,

f''=0

and concavity typically changes, giving an inflection point with a momentarily flat tangent.

FAQ

How do I tell the difference between a max of f and an inflection point when looking at the graph of f'?: Look at what the $f'$ graph is doing. Where $f'$ crosses zero and changes sign, $f$ has an extremum. Where $f'$ reaches a peak or valley (a local extremum of $f'$ itself), $f$ has an inflection point. Crossing the axis controls extrema; the slope of $f'$ controls concavity.
If I'm given the graph of f'', can I find where f is increasing?: No. The graph of $f''$ only gives concavity and inflection points of $f$ . Increasing and decreasing behavior depends on the sign of $f'$ , which you cannot recover from $f''$ alone without additional information such as a value of $f'$ .
Why is my sketch of f not unique when I'm only given f'?: Any two functions that differ by a constant have identical derivatives, so $f'$ determines the shape of $f$ but not its vertical position. Unless the problem provides an initial condition like $f(0)=5$ , you can only draw the correct shape, not a fixed height.
Does f''=0 always mean there is an inflection point?: No. $f''=0$ is necessary but not sufficient. You must verify that $f''$ actually changes sign there. If $f''$ touches zero but stays the same sign — such as behaving like $x^2$ — the concavity does not change and there is no inflection point.

Learn this with a teacher, not a page

The Crimsora tutor teaches U5.8 Connecting f, f', f'' through Graphs live — explaining on a whiteboard, asking you questions, and adapting to where you get stuck.

Try a live session free