Determining concavity — AP Calculus BC Study Guide

For: AP Calculus BC candidates sitting AP Calculus BC.

Covers: Formal definition of concave up/down, the second derivative test for concavity, process for finding intervals of concavity, identifying candidate inflection points, and confirming inflection points on twice-differentiable functions.

You should already know: How to compute first and second derivatives of implicit and explicit functions. How to solve polynomial and rational inequalities to test sign changes. How to find the domain of a function and evaluate points on a function.

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 Determining concavity?

Concavity describes the direction of curvature of a function's graph, relative to its tangent lines across an interval. This sub-topic makes up roughly 4-6% of the total AP Calculus BC exam weight per the College Board CED, and it appears in both multiple choice (MCQ) and free response (FRQ) sections. It is almost always paired with other analytical differentiation topics: curve sketching, classifying extrema, or interpreting rates of change in applied problems.

By definition, a function concave up on an interval lies above all its tangent lines on that interval, while a concave down function lies below all its tangent lines. AP exclusively uses the terms "concave up" and "concave down" (avoid confusing alternate terminology like "convex" that may appear in other sources). Determining concavity relies on the relationship between the second derivative of a function and the behavior of its first derivative, making it a foundational step for almost all advanced curve analysis.

2. Concavity Definition and the Second Derivative Rule

The core definition of concavity is tied to the behavior of the first derivative, not the original function: a function $f$ is concave up on an open interval $I$ if and only if $f^{\prime }$ (the slope of the tangent to $f$) is increasing on $I$. Similarly, $f$ is concave down on $I$ if and only if $f^{\prime }$ is decreasing on $I$.

This definition directly leads to the second derivative rule for concavity, because the derivative of $f^{\prime }$ is $f^{\prime \prime }$, the second derivative. A function is increasing when its derivative is positive, so if $f^{\prime }$ is increasing, its derivative $f^{\prime \prime }$ must be positive. The formal rule is: $$ \begin{align*} f \text{ is concave up on } I &\iff f''(x) > 0 \text{ for all } x \in I \ f \text{ is concave down on } I &\iff f''(x) < 0 \text{ for all } x \in I \end{align*} $$ Intuition confirms this: for the upward-opening parabola $f(x)=x^2$, $f^{\prime \prime }(x)=2>0$ everywhere, and slopes increase from negative to positive, matching concave up curvature. For a downward-opening parabola $f(x)=-x^2$, $f^{\prime \prime }(x)=-2<0$, slopes decrease from positive to negative, matching concave down curvature.

Worked Example

Use the second derivative rule to confirm that $f(x)=2e^{2x}-12x$ is concave up over its entire domain.

1. Compute the first derivative: $f^{\prime }(x)=4e^{2x}-12$
2. Compute the second derivative: $f^{\prime \prime }(x)=8e^{2x}$
3. Check the sign of $f^{\prime \prime }(x)$: For all real $x$, $e^{2x}>0$, so $8e^{2x}>0$ for all $x$ in the domain of $f$.
4. By the second derivative rule, since $f^{\prime \prime }(x)$ is positive everywhere, $f(x)$ is concave up over its entire domain.

Exam tip: On AP FRQs, you must explicitly reference the sign of the second derivative to justify concavity. A conclusion of "concave up" without referencing the sign of $f^{\prime \prime }$ will lose points.

3. Finding Intervals of Concavity

To find all intervals where a function is concave up or down, we use a systematic process that mirrors the process for finding intervals of increase/decrease, but uses the second derivative instead of the first.

The key property here is that $f^{\prime \prime }(x)$ can only change sign at points where $f^{\prime \prime }(x)=0$ or $f^{\prime \prime }(x)$ is undefined (and $x$ is in the domain of $f$). These points are called candidate points, because they are the only points where concavity can change. The full process is:

1. Compute $f^{\prime \prime }(x)$ fully
2. Find all candidate points: all $x$ in the domain of $f$ where $f^{\prime \prime }(x)=0$ or $f^{\prime \prime }(x)$ is undefined
3. Candidate points split the domain of $f$ into open test intervals
4. Test the sign of $f^{\prime \prime }(x)$ in each test interval, and assign concavity based on the rule: positive = concave up, negative = concave down

Worked Example

Find all intervals of concavity for $f(x)=\frac{x^3}{3}-x^2-3x+2$.

1. Compute derivatives: $f^{\prime }(x)=x^2-2x-3$, so $f^{\prime \prime }(x)=2x-2=2(x-1)$
2. Find candidate points: $f^{\prime \prime }(x)$ is a polynomial, so defined everywhere on the domain (all real numbers). Set $f^{\prime \prime }(x)=0$: $2(x-1)=0⟹x=1$. This splits the domain into two test intervals: $(-\infty ,1)$ and $(1,\infty )$.
3. Test the sign of $f^{\prime \prime }$: For $(-\infty ,1)$, test $x=0$: $f^{\prime \prime }(0)=-2<0$. For $(1,\infty )$, test $x=2$: $f^{\prime \prime }(2)=2>0$.
4. Assign concavity: $f(x)$ is concave down on $(-\infty ,1)$ and concave up on $(1,\infty )$.

Exam tip: Never forget to check for points where $f^{\prime \prime }(x)$ is undefined (but $x$ is in the domain of $f$). Most students only check for $f^{\prime \prime }(x)=0$ and miss concavity changes at corners, cusps, or vertical tangents on $f$.

4. Identifying Inflection Points

An inflection point is a point on the graph of $f$ where the concavity changes from up to down, or down to up. For a point to be an inflection point, two conditions must be satisfied:

1. The point $(c,f(c))$ lies on the graph of $f$ (meaning $c$ is in the domain of $f$)
2. The concavity of $f$ changes across $x=c$ (which means the sign of $f^{\prime \prime }$ changes across $x=c$)

A common misconception is that any point where $f^{\prime \prime }(c)=0$ is an inflection point. This is not true: $f^{\prime \prime }(c)=0$ only makes $x=c$ a candidate point. If the sign of $f^{\prime \prime }$ does not change across $x=c$, there is no concavity change, so no inflection point. Similarly, a point where $f^{\prime \prime }(c)$ is undefined can be an inflection point if concavity changes across it and $c$ is in the domain of $f$.

The process for finding inflection points is: after finding intervals of concavity, check each candidate point for a sign change of $f^{\prime \prime }$. If the sign changes, compute the $y$-coordinate to get the full inflection point as an ordered pair.

Worked Example

Find all inflection points of $f(x)=x^4-8x^3+18x^2$.

1. Compute the second derivative: $f^{\prime }(x)=4x^3-24x^2+36x$, so $f^{\prime \prime }(x)=12x^2-48x+36=12(x^2-4x+3)=12(x-1)(x-3)$.
2. Find candidate points: $f^{\prime \prime }$ is defined everywhere, set to zero: $x=1$ and $x=3$, both in the domain of $f$.
3. Check for sign change: Test intervals: $(-\infty ,1)$: $f^{\prime \prime }(-1)=12(-2)(-4)=96>0$ (concave up). $(1,3)$: $f^{\prime \prime }(2)=12(1)(-1)=-12<0$ (concave down). $(3,\infty )$: $f^{\prime \prime }(4)=12(3)(1)=36>0$ (concave up). Sign changes at both $x=1$ and $x=3$, so both are inflection points.
4. Compute $y$-coordinates: $f(1)=1-8+18=11$, $f(3)=81-216+162=27$. Inflection points are $(1,11)$ and $(3,27)$.

Exam tip: AP FRQs require inflection points to be written as ordered pairs $(x,y)$, not just $x$-coordinates. Always compute the $y$-value to earn full credit.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Claiming that $f^{\prime \prime }(c)=0$ means $(c,f(c))$ must be an inflection point. Why: Students confuse a common property of inflection points with a requirement, and forget that sign change is mandatory. Correct move: Always test whether the sign of $f^{\prime \prime }$ changes across $x=c$ before concluding it is an inflection point.
• Wrong move: Ignoring points where $f^{\prime \prime }(x)$ is undefined (but $x$ is in the domain of $f$) when searching for inflection points. Why: Students only search for roots of $f^{\prime \prime }(x)=0$ and forget that $f^{\prime \prime }$ can be undefined at points on $f$ where concavity changes. Correct move: Always list all $x$ in the domain of $f$ where $f^{\prime \prime }$ is zero or undefined before dividing into test intervals.
• Wrong move: Justifying concavity by referencing the sign of the first derivative, not the second. Why: Students confuse the test for increasing/decrease (first derivative) with the test for concavity (second derivative). Correct move: Explicitly reference the sign of the second derivative in all FRQ justifications for concavity.
• Wrong move: Reporting an inflection point at an $x$-value not in the domain of the original function. Why: Students find a root of $f^{\prime \prime }$ but forget to check if the original function is defined there. Correct move: Always confirm that $x=c$ is in the domain of $f$ before checking for an inflection point.
• Wrong move: Closing intervals of concavity by including inflection point $x$-values. Why: Concavity is defined for open intervals, not individual points. Correct move: Always use open intervals when reporting intervals of concavity, as required by the AP exam.

6. Practice Questions (AP Calculus BC Style)

Question 1 (Multiple Choice)

How many distinct intervals of concave up does the function $f(x)=x^4-6x^2+8x+1$ have over the domain of all real numbers? A) $1$ B) $2$ C) $3$ D) $4$

Worked Solution: First, compute the second derivative of $f(x)$. The first derivative is $f^{\prime }(x)=4x^3-12x+8$, so the second derivative simplifies to $f^{\prime \prime }(x)=12x^2-12=12(x-1)(x+1)$. Candidate points for concavity change are $x=-1$ and $x=1$, splitting the domain into three intervals. Testing the sign of $f^{\prime \prime }(x)$: $f^{\prime \prime }(x)>0$ (concave up) on $(-\infty ,-1)$ and $(1,\infty )$, and negative (concave down) on $(-1,1)$. There are 2 distinct intervals of concave up. The correct answer is B.

Question 2 (Free Response)

Let $h(x)=e^{-x^2/2}$ for all real $x$. (a) Find all intervals where $h(x)$ is concave up and concave down. Justify your answer. (b) Find all inflection points of $h(x)$, showing your work. (c) Given $h^{\prime }(x)=-x{e}^{-x^2/2}$, use your result from (a) to identify intervals where $h^{\prime }(x)$ is increasing and decreasing. Explain the connection.

Worked Solution: (a) By product rule, $h^{\prime \prime }(x)=\frac{d}{dx}\left(-x{e}^{-x^2/2}\right)=-e^{-x^2/2}+(-x)(-x{e}^{-x^2/2})=e^{-x^2/2}(x^2-1)$. Since $e^{-x^2/2}>0$ for all $x$, the sign of $h^{\prime \prime }(x)$ matches the sign of $x^2-1$. $h^{\prime \prime }(x)>0$ when $|x|>1$, so $h(x)$ is concave up on $(-\infty ,-1)\cup (1,\infty )$. $h^{\prime \prime }(x)<0$ when $|x|<1$, so $h(x)$ is concave down on $(-1,1)$. Justification: Concavity is determined by the sign of the second derivative, as shown above. (b) Candidate points are $x=-1$ and $x=1$, both in the domain of $h$. The sign of $h^{\prime \prime }$ changes at both points, so both are inflection points. $h(-1)=e^{-1/2}=\frac{1}{\sqrt{e}}$ and $h(1)=\frac{1}{\sqrt{e}}$, so inflection points are $\left(-1,\frac{1}{\sqrt{e}}\right)$ and $\left(1,\frac{1}{\sqrt{e}}\right)$. (c) By definition, $f$ is concave up on an interval if and only if $f^{\prime }$ is increasing on that interval, and concave down if and only if $f^{\prime }$ is decreasing. Thus, $h^{\prime }(x)$ is increasing on $(-\infty ,-1)\cup (1,\infty )$ and decreasing on $(-1,1)$.

Question 3 (Application / Real-World Style)

The population of a town $t$ years after 2010 is modeled by the logistic function $P(t)=\frac{100000}{1+9e^{-0.2t}}$, where $P(t)$ is measured in people. Is the rate of population growth increasing or decreasing at $t=10$? Justify your answer, and interpret the result in context.

Worked Solution: The rate of population growth is $P^{\prime }(t)$. To test if the rate is increasing or decreasing at $t=10$, we check the sign of $P^{\prime \prime }(10)$ (positive = increasing growth, negative = decreasing growth). Using the derivative rule for logistic functions, $P^{\prime }(t)=0.2P(t)\left(1-\frac{P(t)}{100000}\right)$, so $P^{\prime \prime }(t)=0.2P^{\prime }(t)\left(1-\frac{2P(t)}{100000}\right)$. At $t=10$, $P(10)=\frac{100000}{1+9e^{-2}}\approx 45000<50000$. $P^{\prime }(t)$ is always positive for this growth model, so $P^{\prime \prime }(10)$ has the same sign as $1-\frac{2(45000)}{100000}=0.1>0$. Thus, the rate of population growth is increasing at $t=10$. Interpretation: In 2020, 10 years after 2010, the population of the town is growing, and the speed of that growth is still increasing.

7. Quick Reference Cheatsheet

8. What's Next

Determining concavity is a direct prerequisite for the second derivative test for local extrema, which you will apply next to classify critical points as local minima or maxima. Without a solid understanding of how to compute the second derivative and test its sign, you cannot correctly apply this test, a common topic across both MCQ and FRQ sections of the AP exam. Beyond extrema classification, concavity is a core tool for full analytic curve sketching, a frequent multi-part FRQ task. It also appears in applied problems, from kinematics (where it describes how acceleration changes) to economics (where it describes marginal utility changes), so mastery is critical for all applied derivative questions.

Next topics to study:

• Second derivative test for local extrema
• Analytic curve sketching
• Optimization of differentiable functions