Connecting f, f', f'' qualitatively

AP Calculus AB· AP Calculus AB CED — Analytical Applications of Differentiation· 14 min read

1. Relating f and f': Increasing/Decreasing and Critical Points★★☆☆☆⏱ 3 min

The core relationship between $f$ and $f^{'}$ comes from the definition of the derivative as the instantaneous slope of $f$ at any point $x$ .

If $f^{'} (x) > 0$ for all $x$ in an interval, $f$ is increasing: as $x$ increases, $f (x)$ increases
If $f^{'} (x) < 0$ for all $x$ in an interval, $f$ is decreasing: as $x$ increases, $f (x)$ decreases

📘 Definition

Critical Point

Any point $x = c$ in the domain of $f$ where $f^{'} (c) = 0$ or $f^{'} (c)$ is undefined. Critical points are the only locations where $f$ can change from increasing to decreasing (or vice versa).

📐 Worked Example

The graph of $f^{'} (x)$ is negative for $x < - 2$ , positive for $- 2 < x < 1$ , negative for $1 < x < 4$ , and positive for $x > 4$ , with $f^{'} (x)$ crossing the x-axis exactly at $x = - 2$ , $x = 1$ , and $x = 4$ , and no discontinuities. What intervals is $f$ increasing on? What are the critical points of $f$ ?

1
By definition, $f$ is increasing when $f^{'} (x) > 0$ .
2
The intervals where $f^{'} (x)$ is positive are $(- 2, 1)$ and $(4, \infty)$ , so these are the intervals where $f$ is increasing.
3
Critical points occur where $f^{'} (x) = 0$ or $f^{'} (x)$ is undefined. Here, $f^{'} (x) = 0$ at $x = - 2$ , $x = 1$ , $x = 4$ , and there are no points where $f^{'} (x)$ is undefined.
4
Final answer: $f$ increases on $(- 2, 1) \cup (4, \infty)$ , critical points at $x = - 2$ , $x = 1$ , $x = 4$ .

Exam tip:

On all AP exam questions asking for intervals of increase/decrease, always use open intervals. AP graders never accept closed intervals for this question type.

2. Relating f and f'': Concavity and Inflection Points★★☆☆☆⏱ 3 min

Concavity describes how the slope of $f$ changes as $x$ increases, so it is determined by the derivative of $f^{'}$ , which is $f^{''}$ .

If $f^{''} (x) > 0$ , $f^{'}$ is increasing on the interval, so $f$ is concave up (shaped like a cup $\cup$ )
If $f^{''} (x) < 0$ , $f^{'}$ is decreasing on the interval, so $f$ is concave down (shaped like a cap $\cap$ )

📘 Definition

Inflection Point

A point where the concavity of $f$ changes from up to down (or down to up). For twice-differentiable $f$ , this requires $f^{''} (x)$ changes sign at the point, which can only happen if $f^{''} (c) = 0$ or $f^{''} (c)$ is undefined. Not all points where $f^{''} (c) = 0$ are inflection points.

Example:

A useful shortcut: inflection points of $f$ occur at the local extrema of $f^{'}$

📐 Worked Example

$f$ is twice differentiable for all real $x$ , and the graph of $f^{'}$ has a local maximum at $x = 2$ and a local minimum at $x = - 3$ . Where does $f$ have inflection points? Justify your answer.

1
Inflection points of $f$ require a sign change in $f^{''}$ , which corresponds to a change in whether $f^{'}$ is increasing or decreasing.
2
At a local maximum of $f^{'}$ , $f^{'}$ changes from increasing (so $f^{''} > 0$ ) to decreasing (so $f^{''} < 0$ ). This means $f^{''}$ changes sign at $x = 2$ , so $x = 2$ is an inflection point of $f$ .
3
At a local minimum of $f^{'}$ , $f^{'}$ changes from decreasing (so $f^{''} < 0$ ) to increasing (so $f^{''} > 0$ ). This means $f^{''}$ also changes sign at $x = - 3$ , so $x = - 3$ is also an inflection point of $f$ .
4
Final answer: $f$ has inflection points at $x = - 3$ and $x = 2$ .

Exam tip:

Always justify inflection points by explicitly stating that concavity (or the sign of $f^{''}$ ) changes at the point. AP graders will deduct points if you only state that $f^{''} (c) = 0$ with no mention of a sign change.

3. Classifying Local Extrema Qualitatively★★★☆☆⏱ 3 min

Once you have identified the critical points of $f$ , you can classify them as local maxima, local minima, or neither using two qualitative tests:

First Derivative Test: Works for all critical points: $f^{'}$ changes from positive to negative = local maximum; negative to positive = local minimum; no sign change = no extremum
Second Derivative Test: Only for critical points where $f^{'} (c) = 0$ : $f^{''} (c) < 0$ = local maximum; $f^{''} (c) > 0$ = local minimum; $f^{''} (c) = 0$ = test is inconclusive, use the first derivative test

📐 Worked Example

$f$ is twice differentiable, and has a critical point at $x = 3$ where $f^{'} (3) = 0$ and $f^{''} (3) = - 4$ . $f^{'} (2.9) = 2$ and $f^{'} (3.1) = - 1$ . Classify the critical point at $x = 3$ .

1
First apply the second derivative test: we have $f^{'} (3) = 0$ and $f^{''} (3) = - 4 < 0$ . By the second derivative test, this means $f$ has a local maximum at $x = 3$ .
2
Confirm with the first derivative test: check the sign of $f^{'}$ on either side of $x = 3$ .
3
Left of $x = 3$ , $f^{'} (2.9) = 2 > 0$ , so $f$ is increasing before $x = 3$ . Right of $x = 3$ , $f^{'} (3.1) = - 1 < 0$ , so $f$ is decreasing after $x = 3$ .
4
$f^{'}$ changes from positive to negative at $x = 3$ , so the first derivative test confirms the result: $x = 3$ is a local maximum.

Exam tip:

If an FRQ asks you to justify a local extremum, you must explicitly reference the test you use (e.g., "by the second derivative test, $f^{''} (3) < 0$ so $x = 3$ is a local maximum"). A bare conclusion earns zero points.

4. Sketching One Graph From Another★★★☆☆⏱ 3 min

A common AP question asks you to sketch the graph of $f$ given the graph of $f^{'}$ (or vice versa), using only qualitative relationships. The process follows three simple steps: mark all key points, divide the x-axis into intervals between key points, then assign the correct increasing/decreasing and concavity to each interval.

📐 Worked Example

The graph of $f^{'} (x)$ is a parabola opening upward with roots at $x = 0$ and $x = 4$ , and $f (0) = 2$ . Identify all key features of $f$ (extrema, inflection points) to prepare a sketch.

1
First, find intervals of increase/decrease for $f$ : since $f^{'} (x)$ is an upward opening parabola, it is negative between its roots ( $0 < x < 4$ ) and positive outside ( $x < 0$ and $x > 4$ ). So $f$ increases on $(- \infty, 0)$ , decreases on $(0, 4)$ , and increases on $(4, \infty)$ .
2
Find extrema of $f$ : $f^{'}$ changes from positive to negative at $x = 0$ , so $f$ has a local maximum at $(0, 2)$ . $f^{'}$ changes from negative to positive at $x = 4$ , so $f$ has a local minimum at $x = 4$ .
3
Find concavity and inflection points of $f$ : the vertex of the parabola $f^{'} (x)$ is at $x = 2$ , so $f^{'} (x)$ is decreasing for $x < 2$ and increasing for $x > 2$ . This means $f^{''} (x) < 0$ for $x < 2$ and $f^{''} (x) > 0$ for $x > 2$ , so $f$ changes concavity at $x = 2$ , which is an inflection point.
4
Final key features: local maximum at $(0, 2)$ , inflection point at $x = 2$ , local minimum at $x = 4$ .

Exam tip:

When asked to sketch a graph on AP FRQ, you only need to correctly plot and label all required key features and get the general shape right. You do not need to plot every point to earn full credit.

5. Concept Check★★★★☆⏱ 2 min

✓ Quick check

Test your understanding with this AP-style multiple choice question:

The graph of the second derivative $f^{''}$ of a function $f$ is negative for $x < 1$ , crosses the x-axis at $x = 1$ , is positive for $1 < x < 3$ , touches the x-axis (does not cross) at $x = 3$ , and remains positive for all $x > 3$ . For what values of $x$ does $f$ have an inflection point?
- A) $x = 1$ only
- B) $x = 3$ only
- C) $x = 1$ and $x = 3$
- D) $x < 1$ and $1 < x < 3$

📐 Worked Example

The function $f$ is twice differentiable for all real $x$ , with first derivative $f^{'} (x) = x (x - 2)^{2}$ . (a) Identify all critical points of $f$ . (b) On what intervals is $f$ decreasing? Justify your answer. (c) Classify each critical point as a local maximum, local minimum, or neither. (d) Identify all inflection points of $f$ . Justify your answer.

1
(a) Critical points occur where $f^{'} (x) = 0$ or $f^{'} (x)$ is undefined. $f^{'} (x)$ is a polynomial, so defined everywhere. Set $x (x - 2)^{2} = 0$ , giving critical points at $x = 0$ and $x = 2$ .
2
(b) $f$ is decreasing when $f^{'} (x) < 0$ . $(x - 2)^{2}$ is non-negative for all real $x$ , so the sign of $f^{'} (x)$ matches the sign of $x$ . $f^{'} (x) < 0$ when $x < 0$ , so $f$ is decreasing on $(- \infty, 0)$ .
3
(c) For $x = 0$ : $f^{'} (x) < 0$ when $x < 0$ , and $f^{'} (x) > 0$ when $0 < x < 2$ . $f^{'}$ changes from negative to positive, so $x = 0$ is a local minimum. For $x = 2$ : $f^{'} (x) > 0$ when $0 < x < 2$ and $f^{'} (x) > 0$ when $x > 2$ , so there is no sign change. $x = 2$ is neither a local maximum nor minimum.
4
(d) To find inflection points, first compute $f^{''} (x)$ :
5
$f'(x) = x^3 - 4x^2 + 4x \implies f''(x) = 3x^2 - 8x + 4 = (3x - 2)(x - 2)$
6
Set $f^{''} (x) = 0$ to get $x = \frac{2}{3}$ and $x = 2$ . Testing sign: $f^{''} (x) > 0$ for $x < \frac{2}{3}$ , $f^{''} (x) < 0$ for $\frac{2}{3} < x < 2$ , $f^{''} (x) > 0$ for $x > 2$ . $f^{''}$ changes sign at both points, so inflection points at $x = \frac{2}{3}$ and $x = 2$ .

6. Common Pitfalls

Wrong move:

Calling $x = c$ an inflection point of $f$ just because $f^{''} (c) = 0$ , with no check for sign change.

Why:

Students memorize that inflection points occur where $f^{''} = 0$ , so they assume all such points qualify, forgetting the concavity change requirement.

Correct move:

Always check the sign of $f^{''}$ on either side of $c$ ; only label it an inflection point if the sign changes.

Wrong move:

Stating $f$ is increasing on $[a, b]$ when asked for intervals of increase.

Why:

Students incorrectly assume closed intervals are acceptable because monotonicity can extend to endpoints.

Correct move:

Always write intervals of increase/decrease as open intervals, per AP exam convention.

Wrong move:

Confusing the y-value of a graph of $f^{'}$ with the slope of the graph.

Why:

When given a graph of $f^{'}$ , students mix up what tells you about increase/decrease of $f$ versus concavity of $f$ .

Correct move:

Label the graph immediately: $y = f^{'} (x)$ , so $y > 0 = f$ increasing; slope of this graph = $f^{''} (x)$ , so positive slope = $f$ concave up.

Wrong move:

Classifying every critical point as a local maximum or minimum.

Why:

Students assume all critical points are extrema by definition.

Correct move:

Always check for a sign change of $f^{'}$ around the critical point; if no sign change, it is not an extremum.

Wrong move:

Concluding there is no extremum at $x = c$ when the second derivative test gives $f^{''} (c) = 0$ .

Why:

Students forget the test is inconclusive, not negative, when $f^{''} (c) = 0$ .

Correct move:

If $f^{''} (c) = 0$ , fall back to the first derivative test to check for a sign change of $f^{'}$ .

Wrong move:

Stating inflection points of $f$ are the same as critical points of $f$ .

Why:

Students confuse the location of concavity changes with slope changes.

Correct move:

Remember inflection points of $f$ correspond to extrema of $f^{'}$ , not critical points of $f$ .

7. Quick Reference Cheatsheet

Category	Rule	Notes
$f$ increasing on interval	$f^{'} (x) > 0$ for all $x$ in interval	Always use open intervals for AP; endpoints are not required.
$f$ decreasing on interval	$f^{'} (x) < 0$ for all $x$ in interval	AP does not accept closed intervals for this question type.
Critical point of $f$	$f^{'} (c) = 0$ or $f^{'} (c)$ undefined	Only points in the domain of $f$ count; not all are extrema.
$f$ concave up on interval	$f^{''} (x) > 0 ⟺ f^{'}$ increasing	Shaped like a cup $\cup$ ; slope of $f$ increases as $x$ increases.
$f$ concave down on interval	$f^{''} (x) < 0 ⟺ f^{'}$ decreasing	Shaped like a cap $\cap$ ; slope of $f$ decreases as $x$ increases.
Inflection point of $f$	Concavity of $f$ changes at $x = c$	$f^{''} (c) = 0$ is necessary but not sufficient; must confirm sign change.
First Derivative Test for Extrema	$f^{'}$ : $+$ to $-$ = local max; $-$ to $+$ = local min; no change = no extremum	Works for all critical points, even when $f^{''}$ does not exist.
Second Derivative Test for Extrema	$f^{'} (c) = 0$ , $f^{''} (c) < 0$ = local max; $f^{'} (c) = 0$ , $f^{''} (c) > 0$ = local min	Inconclusive if $f^{''} (c) = 0$ ; use first derivative test in that case.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2022 · MCQ
Inflection point identification
2023 · FRQ
Extrema classification and justification

What's Next

This topic is the foundational conceptual framework for all further applications of differentiation in AP Calculus AB, including full curve sketching, optimization problems, and related rates, all of which are heavily tested on the AP exam. Mastering the qualitative relationships between $f$ , $f^{'}$ , and $f^{''}$ will make it much easier to interpret results from numerical derivative calculations and solve real-world application problems that require you to describe the behavior of a function over time. Next, you will apply these concepts to full curve sketching, then to solving optimization problems that use the same rules for identifying extrema we covered here, continuing your work through Unit 5 of the AP Calculus AB syllabus.