5.6 Determining Concavity

Concavity describes how a curve bends: concave up when it opens upward and concave down when it opens downward. You find concavity by checking the sign of the second derivative, where $f''(x)>0$ means concave up and $f''(x)<0$ means concave down. For AP Calculus, justify concavity with the sign of $f''$ or with whether $f'$ is increasing or decreasing.

Why This Matters for the AP Calculus Exam

Concavity is one of the core ways AP Calculus asks you to analyze a function using its derivatives. You will use the second derivative to describe how a graph bends, find points of inflection, and justify those conclusions with clear reasoning. This shows up both in multiple-choice questions, where you read graphs and signs quickly, and in free-response questions, where you have to explain why a function is concave a certain way based on $f'$ or $f''$. Strong justification language here also sets you up for later topics like the second derivative test, sketching graphs, and connecting $f$, $f'$, and $f''$.

Key Takeaways

• A graph is concave up on an interval when $f'$ is increasing, and concave down when $f'$ is decreasing.
• $f''(x)>0$ means concave up; $f''(x)<0$ means concave down.
• A possible point of inflection occurs where $f''(x)=0$ or where $f''$ is undefined.
• A point is an actual point of inflection only if $f''$ changes sign there, meaning the concavity really switches.
• Build a sign chart for $f''$ to find concavity over intervals, just like you do with $f'$ for increasing/decreasing.
• Always justify concavity by referring to $f''$ or to $f'$ increasing/decreasing, and name $f$, $f'$, and $f''$ specifically.

Determining Concavity

A function is concave up when it opens upward and concave down when it opens downward. More precisely:

• If the slopes of the tangent lines are increasing (that is, $f'$ is increasing), the function is concave up.
• If the slopes of the tangent lines are decreasing (that is, $f'$ is decreasing), the function is concave down.

The Second Derivative

Since the second derivative is the derivative of the first derivative, it tells you whether $f'$ is increasing or decreasing. That connects directly to concavity:

• A positive second derivative means $f'$ is increasing, so the function is concave up.
• A negative second derivative means $f'$ is decreasing, so the function is concave down.

The point where a function switches concavity is called an inflection point. Because $f'$ changes from increasing to decreasing (or the reverse) there, $f''$ switches signs at that point. So if $f''(x)=0$ at a point, that point is a possible inflection point. You still have to confirm that the sign of $f''$ actually changes.

To sum up the key points:

1. A graph is concave up if $f'(x)$ is increasing, and concave down if $f'(x)$ is decreasing.
2. If $f''(x)>0$, the function is concave up. If $f''(x)<0$, the function is concave down.
3. A point of inflection is a point where $f$ changes concavity. If you only know $f''(x)=0$, you have a possible point of inflection, not a confirmed one.

Concavity Walkthrough

Consider the function:

$f(x)=x^3-3x^2+2x+1$

a) Determine the intervals where $f(x)$ is concave up and concave down.

b) Find where $f(x)$ has points of inflection.

Walkthrough of Part A

To determine concavity, find the intervals where $f''(x)$ is positive (concave up) or negative (concave down).

Find the first and second derivatives using the power rule.

$f'(x)=3x^2-6x+2$

$f''(x)=6x-6$

To understand the behavior of $f''(x)$, find a possible inflection point by setting $f''(x)=0$.

$0=6x-6$

$6=6x$

$x=1$

There is a possible point of inflection at $x=1$. This divides the domain into two intervals, $(-\infty,1)$ and $(1,\infty)$. Test a point in each interval to find the sign of $f''$.

Test $x=0$:

$f''(0)=6(0)-6=-6$

Since $f''(0)<0$, the interval to the left of $x=1$ is concave down.

Test $x=2$:

$f''(2)=6(2)-6=6$

Since $f''(2)>0$, the interval to the right of $x=1$ is concave up.

So $f(x)$ is concave down on $(-\infty,1)$ and concave up on $(1,\infty)$.

Walkthrough of Part B

Now check whether the possible point of inflection at $x=1$ is real. For that to be true, $f$ must change concavity at $x=1$.

Part A showed that it does, going from concave down to concave up. So $f(x)$ has a point of inflection at $x=1$, because $f$ changes concavity there and $f''(1)=0$.

How to Use This on the AP Calculus Exam

MCQ

For multiple-choice, you often read a graph of $f'$ or $f''$ and translate it into a statement about $f$. Remember that a graph of $f'$ that is increasing means $f$ is concave up, and a graph of $f''$ above the x-axis means $f$ is concave up. Watch for questions that hand you $f''$ and ask about concavity or inflection points directly, since you can answer those by checking signs.

Free Response

For free-response, you usually have to find concavity and justify it. State the second derivative, identify where $f''=0$ or is undefined, build a sign chart, and report the intervals. Then justify clearly, for example: "$f$ is concave up on $1<x<\infty$ because $f''(x)>0$ there." When the question gives you a graph of $f'$, it is often easier to argue from $f'$ directly: "$f$ is concave up because $f'$ is increasing."

Problem Solving

Use a clean sign chart for $f''$ the same way you do for $f'$. Find the candidates where $f''=0$ or is undefined, split the domain at those values, and test one point in each interval. An inflection point only counts if $f''$ changes sign, so always confirm the switch.

Concavity Practice Problems

Question 1: Let $h(x)=5x^3$. What is the concavity of $h$ at $x=5$?

Question 2: Let $h(x)=3x^4+2x^3$. What is the concavity of $h$ at $x=-\frac{1}{3}$?

Concavity Answers and Solutions

Question 1:

To evaluate concavity at a point, find the second derivative at that point.

$h'(x)=15x^2$

$h''(x)=30x$

$h''(5)=150$

Since $h''(5)=150$ is positive, $h$ is concave up at $x=5$.

Question 2:

To evaluate concavity at a point, find the second derivative at that point.

$h'(x)=12x^3+6x^2$

$h''(x)=36x^2+12x$

$h''\left(-\frac{1}{3}\right)=36\cdot \left(-\frac{1}{3}\right)^2+12\cdot \left(-\frac{1}{3}\right)=0$

Since $h''\left(-\frac{1}{3}\right)=0$, $x=-\frac{1}{3}$ might be a point of inflection. Check the second derivative on both sides to see if the sign changes. If it switches, it is a point of inflection.

Since $f''$ switches sign, the concavity changes, so $x=-\frac{1}{3}$ is a point of inflection.

Common Misconceptions

• Setting $f''(x)=0$ does not automatically give an inflection point. You must confirm that $f''$ actually changes sign there. A function can have $f''=0$ at a point and still keep the same concavity.
• Inflection points can also occur where $f''$ is undefined, not only where $f''=0$. Always include those candidates in your sign chart.
• Concave up does not mean increasing. A function can be decreasing while still concave up. Concavity is about how the slope changes, not whether the function is going up or down.
• Do not mix up the graph of $f'$ with the graph of $f$. If $f'$ is increasing, that tells you $f$ is concave up, not that $f$ is increasing.
• Be precise in your language. Say which function you mean by naming $f$, $f'$, or $f''$ instead of using "it," since a vague justification can lose meaning.

Related AP Calculus Guides

• Unit 5 Overview: Analytical Applications of Differentiation
• 5.1 Using the Mean Value Theorem
• 5.2 Extreme Value Theorem, Global vs Local Extrema, and Critical Points
• 5.3 Determining Intervals on Which a Function is Increasing or Decreasing
• 5.4 Using the First Derivative Test to Determine Relative (Local) Extrema
• 5.11 Solving Optimization Problems