Sketching graphs of f, f', f'' — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Connecting slope of f to f' output values, concavity of f to f'' output values, identifying critical points, inflection points, increasing/decreasing intervals, and matching/sketching graphs of f, f', f'' from any given function.

You should already know: How to compute first and second derivatives of common functions. How to find zeros of algebraic and transcendental functions. The formal definitions of critical points and inflection points.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Calculus BC style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Sketching graphs of f, f', f''?

This topic, part of Unit 5 (Analytical Applications of Differentiation) in the AP Calculus BC CED, makes up approximately 4-6% of the total exam weight, and appears regularly in both multiple-choice (MCQ) and free-response (FRQ) sections. The core goal of this topic is not just to plot points to draw a graph, but to use analytical relationships between a function, its first derivative, and its second derivative to sketch or match graphs without relying on a graphing calculator. Standard notation is consistent across the exam: $f(x)$ refers to the original function, $f^{\prime }(x)$ is its first derivative (representing the instantaneous slope of $f$ at any $x$), and $f^{\prime \prime }(x)$ is the second derivative (representing the instantaneous slope of $f^{\prime }$, and the rate of change of the slope of $f$). Common synonyms you may see on the exam include "analyzing curves" or "graph matching of derivatives". On the AP exam, you will commonly encounter problems that give you the graph of one function (e.g., $f^{\prime }$) and ask you to identify which of four options is the graph of $f$, or ask you to label key features on a sketch of $f$ given derivative information. This topic builds directly on derivative rules to connect algebraic derivative information to geometric graphical behavior.

2. Relationship Between the Graphs of f and f'

Every point on the graph of $f^{\prime }(x)$ equals the slope of the tangent line to $f(x)$ at that same $x$-value. This core relationship gives a set of consistent rules that let us connect the behavior of $f$ to $f^{\prime }$:

1. When $f$ is increasing on an interval, $f^{\prime }(x)>0$ for all $x$ in that interval, so the graph of $f^{\prime }$ lies above the $x$-axis.
2. When $f$ is decreasing on an interval, $f^{\prime }(x)<0$ for all $x$ in that interval, so the graph of $f^{\prime }$ lies below the $x$-axis.
3. At any critical point of $f$ where the slope changes sign (a local extremum), $f^{\prime }(x)=0$, so the graph of $f^{\prime }$ crosses or touches the $x$-axis at that $x$-value.
4. If $f$ is linear on an interval, its slope is constant, so $f^{\prime }$ is constant on that interval, meaning the graph of $f^{\prime }$ is a horizontal line.
5. If the slope of $f$ is increasing on an interval, then $f^{\prime }$ is increasing on that interval, so the graph of $f^{\prime }$ slopes upward.

These rules let you sketch or match $f^{\prime }$ directly from $f$, without fully computing the derivative if you only need key features.

Worked Example

Problem: Given $f(x)=x^3-\frac{3}{2}{x}^2-18x+10$, sketch the graph of $f^{\prime }(x)$ by analyzing the behavior of $f(x)$.

Solution:

1. First, find the critical points of $f$, which occur where $f^{\prime }(x)=0$. Compute the derivative: $f^{\prime }(x)=3x^2-3x-18=3(x-3)(x+2)$, so critical points of $f$ are at $x=-2$ and $x=3$. This means $f^{\prime }(x)$ crosses the $x$-axis at $(-2,0)$ and $(3,0)$.
2. Test the slope of $f$ on intervals separated by critical points: For $x<-2$, pick $x=-3$: $f^{\prime }(-3)=36>0$, so $f$ is increasing on $(-\infty ,-2)$, meaning $f^{\prime }(x)$ is positive (above the $x$-axis) here.
3. For $-2<x<3$, pick $x=0$: $f^{\prime }(0)=-18<0$, so $f$ is decreasing here, meaning $f^{\prime }(x)$ is negative (below the $x$-axis) here.
4. For $x>3$, pick $x=4$: $f^{\prime }(4)=18>0$, so $f$ is increasing here, meaning $f^{\prime }(x)$ is positive (above the $x$-axis) here.
5. Since $f$ is a cubic polynomial, its derivative is quadratic: $f^{\prime }(x)$ has a positive leading coefficient, so the final sketch is an upward-opening parabola crossing the $x$-axis at $x=-2$ and $x=3$, matching the sign intervals we found.

Exam tip: When matching $f$ from $f^{\prime }$ on MCQ, always eliminate options first by checking the sign of $f^{\prime }$: any option for $f$ that is increasing where $f^{\prime }$ is negative can be immediately crossed out, saving valuable exam time.

3. Relationship Between the Graphs of f and f''

Just as $f^{\prime }$ describes the slope of $f$, $f^{\prime \prime }$ describes the slope of $f^{\prime }$, which corresponds directly to the concavity of $f$. Concavity describes the direction the curve of $f$ bends: concave up curves bend upward (shaped like a cup) and concave down curves bend downward (shaped like a cap). The core rules connecting $f$ and $f^{\prime \prime }$ are:

1. When $f$ is concave up on an interval, $f^{\prime \prime }(x)>0$, so the graph of $f^{\prime \prime }$ lies above the $x$-axis on that interval.
2. When $f$ is concave down on an interval, $f^{\prime \prime }(x)<0$, so the graph of $f^{\prime \prime }$ lies below the $x$-axis on that interval.
3. At any inflection point of $f$ (where concavity changes), $f^{\prime \prime }(x)=0$ (for continuous $f$), so the graph of $f^{\prime \prime }$ crosses or touches the $x$-axis at that $x$-value.
4. If $f$ has constant concavity over an interval, $f^{\prime \prime }$ is constant over that interval, so the graph of $f^{\prime \prime }$ is a horizontal line.

This relationship also connects $f^{\prime }$ and $f^{\prime \prime }$ directly: when $f^{\prime }$ is increasing, $f^{\prime \prime }$ is positive; when $f^{\prime }$ is decreasing, $f^{\prime \prime }$ is negative.

Worked Example

Problem: The graph of $f(x)$ has an inflection point at $x=1$, is concave down on $(-\infty ,1)$, and concave up on $(1,\infty )$. If $f(x)$ is a cubic polynomial, what does the graph of $f^{\prime \prime }(x)$ look like?

Solution:

1. Concavity of $f$ directly translates to the sign of $f^{\prime \prime }$. Since $f$ is concave down for $x<1$, $f^{\prime \prime }(x)<0$ for all $x<1$, so $f^{\prime \prime }$ lies below the $x$-axis on $(-\infty ,1)$.
2. For $x>1$, $f$ is concave up, so $f^{\prime \prime }(x)>0$ for all $x>1$, meaning $f^{\prime \prime }$ lies above the $x$-axis on $(1,\infty )$.
3. $f$ has an inflection point at $x=1$, so $f^{\prime \prime }(1)=0$, so $f^{\prime \prime }$ crosses the $x$-axis at $(1,0)$.
4. Since $f$ is a cubic polynomial, its first derivative is quadratic, and its second derivative is linear. The sign of $f^{\prime \prime }$ goes from negative left of 1 to positive right of 1, so the slope of the line must be positive. The final graph is a straight line with positive slope crossing the $x$-axis at $x=1$.

Exam tip: On the AP exam, inflection points only occur where concavity of $f$ actually changes, not just where $f^{\prime \prime }(x)=0$. Always confirm the sign of $f^{\prime \prime }$ changes on either side of $x=a$ before marking an inflection point at $a$ on your sketch.

4. Matching and Sketching All Three Graphs From One Given Graph

Most AP exam problems for this topic give you the graph of one of $f$, $f^{\prime }$, $f^{\prime \prime }$ and ask you to identify or sketch either of the other two. The standardized step-by-step process for this is:

1. Label all key $x$-values on the given graph: any $x$ where the given graph crosses the $x$-axis, has a local extremum, or changes direction. These are the only key $x$-values you need for the target graph, as all changes in behavior occur at these points.
2. Split the $x$-axis into intervals separated by these key $x$-values.
3. For each interval, find the sign of the given graph (above or below the $x$-axis), which tells you whether the target function is increasing/decreasing (if given $f^{\prime }$) or concave up/concave down (if given $f$).
4. Connect the key points to get the full graph, checking that the slope of the current graph matches the value of the derivative graph.

Worked Example

Problem: Given the graph of $f^{\prime }(x)$ which is a quadratic crossing the $x$-axis at $x=-1$ and $x=2$, opening downward, so $f^{\prime }(x)>0$ between $x=-1$ and $x=2$, and $f^{\prime }(x)<0$ for $x<-1$ and $x>2$. Identify the key features of the graph of $f(x)$ and $f^{\prime \prime }(x)$.

Solution:

1. First analyze $f(x)$ from $f^{\prime }(x)$: The zeros of $f^{\prime }$ are at $x=-1$ and $x=2$, so $f$ has critical points at these $x$-values. For $x<-1$, $f^{\prime }(x)<0$, so $f$ is decreasing. For $-1<x<2$, $f^{\prime }(x)>0$, so $f$ is increasing. For $x>2$, $f^{\prime }(x)<0$, so $f$ is decreasing. So $f$ has a local minimum at $x=-1$ and a local maximum at $x=2$.
2. Next, find features of $f^{\prime \prime }(x)$ from $f^{\prime }(x)$: $f^{\prime }$ is a downward-opening quadratic, so its slope (which is $f^{\prime \prime }(x)$) is linear. The vertex of the quadratic (its maximum) is at the midpoint of $-1$ and $2$, which is $x=0.5$. For $x<0.5$, $f^{\prime }$ is increasing, so the slope of $f^{\prime }$ is positive, so $f^{\prime \prime }(x)>0$, meaning $f$ is concave up on $(-\infty ,0.5)$.
3. For $x>0.5$, $f^{\prime }$ is decreasing, so the slope of $f^{\prime }$ is negative, so $f^{\prime \prime }(x)<0$, meaning $f$ is concave down on $(0.5,\infty )$. $f^{\prime }$ has a maximum at $x=0.5$, so $f^{\prime \prime }(0.5)=0$, which means $f$ has an inflection point at $x=0.5$.
4. Final summary: $f^{\prime \prime }$ is a straight line with negative slope crossing the $x$-axis at $x=0.5$, and $f$ has key points at $x=-1$ (local min), $x=0.5$ (inflection point), and $x=2$ (local max) with the correct increasing/decreasing and concavity behavior in each interval.

Exam tip: When working from a given graph, write the sign of the given function above each interval first, then write the corresponding behavior of the target function below each interval. This simple step eliminates 90% of common sign mix-ups.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming that $f$ has a local extremum at $x=a$ just because $f^{\prime \prime }(a)=0$. Why: Students confuse inflection point rules with extremum rules, mixing up what $f^{\prime }$ and $f^{\prime \prime }$ tell you. Correct move: Only conclude $f$ has a local extremum at $a$ if $f^{\prime }$ changes sign at $a$; $f^{\prime \prime }(a)=0$ only tells you of a possible inflection point for $f$.
• Wrong move: Drawing $f$ as crossing the $x$-axis at the same $x$ where $f^{\prime }$ crosses the $x$-axis. Why: Students confuse x-intercepts of $f^{\prime }$ with x-intercepts of $f$, matching all key points to the same $x$ across graphs regardless of meaning. Correct move: Remember that x-intercepts of $f^{\prime }$ correspond to local extrema (critical points) of $f$, not x-intercepts of $f$.
• Wrong move: Stating that $f$ is concave up where $f^{\prime }$ is positive. Why: Students mix up the meaning of first vs second derivative signs, conflating increasing/decreasing with concavity. Correct move: Always associate $f^{\prime }$ sign with $f$ increasing/decreasing, and $f^{\prime \prime }$ sign with $f$ concavity, and write this association down on scratch paper for every problem.
• Wrong move: Drawing $f^{\prime \prime }$ as touching (not crossing) the $x$-axis at an inflection point of $f$ when concavity changes. Why: Students memorize that crossing means sign change, but forget that a sign change of $f^{\prime \prime }$ is required for an inflection point. Correct move: If concavity changes at $x=a$, $f^{\prime \prime }$ must change sign at $a$, so $f^{\prime \prime }$ crosses the $x$-axis at $a$; only a tangent that doesn't cross means no sign change, hence no inflection point.
• Wrong move: When given $f^{\prime }$ and asked for $f$'s inflection points, looking for where $f^{\prime }$ crosses the $x$-axis. Why: Students confuse where $f^{\prime }$ has zero value with where $f^{\prime }$ has zero slope. Correct move: Inflection points of $f$ occur at extrema of $f^{\prime }$, where $f^{\prime }$ changes slope, i.e., where $f^{\prime }$ crosses its own extrema, not its x-intercepts.
• Wrong move: Assuming that if $f$ is increasing everywhere, then $f^{\prime }$ is also increasing everywhere. Why: Students assume increasing $f$ means positive $f^{\prime }$, so positive $f^{\prime }$ must be increasing. Correct move: An increasing $f$ can have a decreasing positive slope (e.g., $f(x)=\ln x$ for $x>0$), so always check the slope of $f^{\prime }$ to get behavior of $f^{\prime \prime }$.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

The graph of $y=f^{\prime \prime }(x)$ crosses the $x$-axis at $x=-2$, $x=1$, and $x=3$. It is negative on $(-\infty ,-2)$, positive on $(-2,1)$, negative on $(1,3)$, and positive on $(3,\infty )$. For the original function $f(x)$, how many inflection points does $f$ have? A) 1 B) 2 C) 3 D) 0

Worked Solution: Inflection points of $f$ occur where $f^{\prime \prime }$ changes sign, which corresponds to where the graph of $f^{\prime \prime }$ crosses the $x$-axis (changing from positive to negative or vice versa). The given $f^{\prime \prime }$ crosses the $x$-axis at all three points, and changes sign at every crossing: it goes from negative to positive at $x=-2$, positive to negative at $x=1$, and negative to positive at $x=3$. All three points have a sign change of $f^{\prime \prime }$, so $f$ has three inflection points. Correct answer: C.

Question 2 (Free Response)

Let $f^{\prime }(x)=2\sin x$ on the interval $0\le x\le 2\pi$. (a) Identify the intervals where $f(x)$ is increasing and decreasing, and find the $x$-coordinates of all local extrema of $f(x)$ on $[0,2\pi ]$. (b) Identify the intervals where $f(x)$ is concave up and concave down, and find the $x$-coordinates of all inflection points of $f(x)$ on $[0,2\pi ]$. (c) Given that $f(0)=2$, find the coordinates of all key points (local extrema, inflection points) of $f(x)$ to prepare for sketching.

Worked Solution: (a) First, find zeros of $f^{\prime }(x)$: $2\sin x=0$ at $x=0$, $x=\pi$, $x=2\pi$. Testing signs: $f^{\prime }(x)>0$ on $(0,\pi )$, so $f$ is increasing on $(0,\pi )$; $f^{\prime }(x)<0$ on $(\pi ,2\pi )$, so $f$ is decreasing on $(\pi ,2\pi )$. By the First Derivative Test, $f$ has a local maximum at $x=\pi$ (endpoints are not local extrema).

(b) $f^{\prime \prime }(x)=\frac{d}{dx}[2\sin x]=2\cos x$. Zeros of $f^{\prime \prime }(x)$ are at $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$. Testing signs: $f^{\prime \prime }(x)>0$ on $\left(0,\frac{\pi }{2}\right)\cup \left(\frac{3\pi }{2},2\pi \right)$, so $f$ is concave up on these intervals; $f^{\prime \prime }(x)<0$ on $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$, so $f$ is concave down here. $f^{\prime \prime }$ changes sign at both $\frac{\pi }{2}$ and $\frac{3\pi }{2}$, so inflection points are at $x=\frac{\pi }{2}$ and $x=\frac{3\pi }{2}$.

(c) The antiderivative of $f^{\prime }(x)=2\sin x$ is $f(x)=-2\cos x+C$. Substitute $f(0)=2$: $-2(1)+C=2⟹C=4$, so $f(x)=-2\cos x+4$. Calculate key points: $(0,2)$ (start), $\left(\frac{\pi }{2},4\right)$ (inflection), $(\pi ,6)$ (local maximum), $\left(\frac{3\pi }{2},4\right)$ (inflection), $(2\pi ,2)$ (end).

Question 3 (Application / Real-World Style)

The velocity of a rocket moving straight upward is given by $v(t)=s^{\prime }(t)$, where $t\ge 0$ is time in seconds after launch, and $s(t)$ is the height of the rocket in meters. Acceleration $a(t)=v^{\prime }(t)=s^{\prime \prime }(t)$. The graph of $a(t)$ is constant negative ($-5{\text{ m/s}}^2$) for $0\le t<10$, then jumps to constant positive ($2{\text{ m/s}}^2$) for $t>10$ after the rocket ignites its second stage. Given $v(0)=50\text{ m/s}$, identify the key features of $v(t)$ and $s(t)$ and interpret them in context.

Worked Solution: For $0\le t<10$: $a(t)=v^{\prime }(t)=-5<0$, so $v(t)$ is linear decreasing, and $s^{\prime \prime }(t)=-5<0$, so $s(t)$ is concave down. At $t=0$, $v(t)$ starts at $(0,50)$. For $t>10$: $a(t)=v^{\prime }(t)=2>0$, so $v(t)$ is linear increasing, and $s^{\prime \prime }(t)=2>0$, so $s(t)$ is concave up. At $t=10$, $a(t)$ changes from negative to positive, so $v(t)$ has a local minimum at $t=10$, and $s(t)$ has an inflection point at $t=10$. Calculating $v(10)=50+(-5)(10)=0$, so the rocket slows to a stop at $t=10$ before the second stage ignites, then begins accelerating upward again. The height $s(t)$ is always increasing (the rocket moves upward the entire time), with the rate of height increase slowing for the first 10 seconds, then speeding up after the second stage ignites.

7. Quick Reference Cheatsheet

8. What's Next

This topic is the foundational prerequisite for the next major topics in Unit 5: Optimization, the First and Second Derivative Tests, and the Mean Value Theorem, and later for related rates and area under curve problems. Without being able to reliably connect the behavior of $f$ to its derivatives, you will struggle to confirm that a critical point you found in an optimization problem is actually a maximum or minimum, and you will not be able to interpret the meaning of derivative graphs in applied FRQ problems. This topic also feeds directly into the study of differential equations in Unit 7, where you will sketch slope fields and solution curves using the same relationships between a function and its derivative you practiced here.

• Mean Value Theorem
• First and Second Derivative Tests
• Optimization
• Slope Fields and Differential Equations