Determining Concavity of Functions over Their Domains - AP Calculus AB Study Guide

The Core Idea: Determining Concavity of Functions over Their Domains

The first derivative, $f^{\prime }(x)$, tells us about the rate of change, or slope, of a function $f(x)$. The second derivative, $f^{\prime \prime }(x)$, tells us about the rate of change of the slope. This concept is visualized as the concavity of the function's graph. Concavity describes the way the graph of a function bends. A graph that is "cupped up" (like a smile) is concave up, while a graph that is "cupped down" (like a frown) is concave down.

By analyzing the sign of the second derivative, we can determine the intervals over which a function is concave up or concave down. The points on the graph where the concavity changes are called points of inflection. This analysis provides a deeper understanding of a function's behavior and is essential for creating an accurate sketch of its graph.

Key Definitions

The relationship between a function $f$, its first derivative $f^{\prime }$, and its second derivative $f^{\prime \prime }$ is central to understanding concavity.

• Concave Up: The graph of a function $f$ is concave up on an interval if its first derivative, $f^{\prime }(x)$, is increasing on that interval. This corresponds to the second derivative being positive, i.e., $f^{\prime \prime }(x)>0$. Visually, the graph of $f$ lies above its tangent lines on this interval.

• Concave Down: The graph of a function $f$ is concave down on an interval if its first derivative, $f^{\prime }(x)$, is decreasing on that interval. This corresponds to the second derivative being negative, i.e., $f^{\prime \prime }(x)<0$. Visually, the graph of $f$ lies below its tangent lines on this interval.

• Point of Inflection: A point of inflection is a point on the graph of $f$ at which the concavity changes (from up to down, or from down to up).

  • These changes can occur where $f^{\prime \prime }(x)=0$ or where $f^{\prime \prime }(x)$ is undefined.

  • A point $(c,f(c))$ is a point of inflection only if $f^{\prime \prime }$ changes sign at $x=c$.

Understanding the Connection Between $f$, $f^{\prime }$, and $f^{\prime \prime }$

The key nuance of this topic is understanding the hierarchical relationship between a function and its derivatives. Just as the sign of the first derivative determines if the original function is increasing or decreasing, the sign of the second derivative determines the behavior of the first derivative, which in turn defines the concavity of the original function.

• If $f^{\prime \prime }(x)>0$ on an interval:

  • This means the slope of $f^{\prime }(x)$ is positive.

  • Therefore, $f^{\prime }(x)$ is an increasing function.

  • This means the slopes of the tangent lines to $f(x)$ are increasing.

  • Therefore, the graph of $f(x)$ is concave up.

• If $f^{\prime \prime }(x)<0$ on an interval:

  • This means the slope of $f^{\prime }(x)$ is negative.

  • Therefore, $f^{\prime }(x)$ is a decreasing function.

  • This means the slopes of the tangent lines to $f(x)$ are decreasing.

  • Therefore, the graph of $f(x)$ is concave down.

A point of inflection on the graph of $f$ corresponds to a point where $f^{\prime \prime }$ changes sign. This also means that at a point of inflection, the first derivative $f^{\prime }$ must have a local extremum (a local maximum or minimum), because it is at these points that $f^{\prime }$ changes from increasing to decreasing or vice versa.

Core Concepts & Rules

• The sign of the second derivative, $f^{\prime \prime }(x)$, determines the concavity of the function $f(x)$.

• If $f^{\prime \prime }(x)>0$ on an interval, then $f(x)$ is concave up on that interval.

• If $f^{\prime \prime }(x)<0$ on an interval, then $f(x)$ is concave down on that interval.

• A point of inflection is a point on the graph of $f(x)$ where the concavity changes.

• To find potential points of inflection, identify all x-values where $f^{\prime \prime }(x)=0$ or $f^{\prime \prime }(x)$ is undefined. These are the candidates.

• To confirm a point of inflection at a candidate point $x=c$, you must verify that the sign of $f^{\prime \prime }(x)$ changes at $x=c$.

Step-by-Step Example 1: Finding Intervals of Concavity from an Equation

Problem: Find the intervals on which the function $f(x)=x^4-4x^3+10$ is concave up and concave down, and identify any points of inflection.

Step 1: Find the second derivative of $f(x)$.

First, find the first derivative:

$$f^{\prime }(x)=4x^3-12x^2$$

Next, find the second derivative:

$$f^{\prime \prime }(x)=12x^2-24x$$

Step 2: Find the potential points of inflection.

Set $f^{\prime \prime }(x)=0$ and find where $f^{\prime \prime }(x)$ is undefined. Since $f^{\prime \prime }(x)$ is a polynomial, it is defined for all real numbers.

$$12x^2-24x=0$$

$$12x(x-2)=0$$

The potential points of inflection occur at $x=0$ and $x=2$.

Step 3: Create a sign chart for $f^{\prime \prime }(x)$.

Use the potential points of inflection to create test intervals: $(-\infty ,0)$, $(0,2)$, and $(2,\infty )$. Choose a test value within each interval and evaluate the sign of $f^{\prime \prime }(x)$.

• Interval $(-\infty ,0)$: Let's test $x=-1$.

  $f^{\prime \prime }(-1)=12(-1{)}^2-24(-1)=12+24=36$. The sign is positive ($+$).

• Interval $(0,2)$: Let's test $x=1$.

  $f^{\prime \prime }(1)=12(1{)}^2-24(1)=12-24=-12$. The sign is negative ($-$).

• Interval $(2,\infty )$: Let's test $x=3$.

  $f^{\prime \prime }(3)=12(3{)}^2-24(3)=108-72=36$. The sign is positive ($+$).

Sign Chart:

      f''(x)   + + +   0   - - -   0   + + +

<----------------|-----------|---------------->

                 0           2

Step 4: Write the conclusion with justification.

• The function $f(x)$ is concave up on the intervals $(-\infty ,0)$ and $(2,\infty )$ because $f^{\prime \prime }(x)>0$ on these intervals.

• The function $f(x)$ is concave down on the interval $(0,2)$ because $f^{\prime \prime }(x)<0$ on this interval.

• Since the concavity changes at both $x=0$ and $x=2$, there are points of inflection at these x-values. To find the points, evaluate $f(x)$ at these values:

  • $f(0)=(0{)}^4-4(0{)}^3+10=10$. Point of inflection at $(0,10)$.

  • $f(2)=(2{)}^4-4(2{)}^3+10=16-32+10=-6$. Point of inflection at $(2,-6)$.

Step-by-Step Example 2: Determining Concavity from a Graph of $f^{\prime }(x)$

Problem: The graph of $f^{\prime }(x)$, the derivative of a function $f(x)$, is shown below. On what intervals is the graph of $f(x)$ concave up? Justify your answer.

(Imagine a graph of $f^{\prime }(x)$ that is a parabola opening down with its vertex at $(1,3)$ and x-intercepts at $x=-1$ and $x=3$)

Step 1: Recall the relationship between the concavity of $f(x)$ and the behavior of $f^{\prime }(x)$.

• The graph of $f(x)$ is concave up when $f^{\prime }(x)$ is increasing.

• The graph of $f(x)$ is concave down when $f^{\prime }(x)$ is decreasing.

• A point of inflection on $f(x)$ occurs where $f^{\prime }(x)$ changes from increasing to decreasing or vice versa (i.e., at local extrema of $f^{\prime }(x)$).

Step 2: Analyze the provided graph of $f^{\prime }(x)$ for increasing and decreasing behavior.

• Look at the graph of $f^{\prime }(x)$. Reading from left to right, the y-values of the graph are increasing until the graph reaches its peak.

• The graph of $f^{\prime }(x)$ is increasing on the interval $(-\infty ,1)$.

• The graph of $f^{\prime }(x)$ is decreasing on the interval $(1,\infty )$.

• The graph of $f^{\prime }(x)$ has a local maximum at $x=1$.

Step 3: Connect the behavior of $f^{\prime }(x)$ to the concavity of $f(x)$.

• Because $f^{\prime }(x)$ is increasing on $(-\infty ,1)$, the graph of $f(x)$ is concave up on this interval.

• Because $f^{\prime }(x)$ is decreasing on $(1,\infty )$, the graph of $f(x)$ is concave down on this interval.

• Because the behavior of $f^{\prime }(x)$ changes from increasing to decreasing at $x=1$, the graph of $f(x)$ has a point of inflection at $x=1$.

Step 4: Write the final answer with justification.

The graph of $f(x)$ is concave up on the interval $(-\infty ,1)$ because the graph of $f^{\prime }(x)$ is increasing on that interval.

Using Your Calculator

For problems where you are given a function rule, a graphing calculator can be used to verify your analytical work by visualizing the second derivative.

Problem: Use a calculator to find the point(s) of inflection for $f(x)=e^x\cos (x)$ on the interval $[0,2\pi ]$.

Method:

1. Graph the Second Derivative: You can do this by finding the second derivative by hand and graphing it, or by using the calculator's numerical derivative function twice. The latter can be slow and less accurate. Let's find $f^{\prime \prime }(x)$ by hand first:

   • $f^{\prime }(x)=e^x\cos (x)-e^x\sin (x)$

   • $f^{\prime \prime }(x)=(e^x\cos (x)-e^x\sin (x))-(e^x\sin (x)+e^x\cos (x))=-2e^x\sin (x)$

2. Enter f''(x) into the calculator:

   • Press Y=.

   • In Y1, enter -2e^(X)sin(X).

3. Set the Window:

   • Press WINDOW. Set $Xmin=0$, $Xmax=2\pi$ (approx. 6.28), and adjust $Ymin$ and $Ymax$ to see the graph (e.g., $Ymin=-30$, Ymax = 30`). 4. **Graph and Find Zeros:** * Press `GRAPH`. You are looking for where the graph of $f''(x) crosses the x-axis. This indicates a sign change in $f^{\prime \prime }$, and therefore a point of inflection on $f$.

   • Press 2nd then TRACE (CALC). Select 2:zero.

   • The calculator will ask for a "Left Bound," "Right Bound," and "Guess."

   • Move the cursor to the left of the first x-intercept and press ENTER. Move it to the right and press ENTER. Press ENTER again for the guess. The calculator will find the zero at $x=\pi$.

   • Repeat the process for the next x-intercept, which is at $x=2\pi$.

4. Interpret the Results:

   • The graph of $f^{\prime \prime }(x)$ is below the x-axis ($f^{\prime \prime }<0$) on $(0,\pi )$ and above the x-axis ($f^{\prime \prime }>0$) on $(\pi ,2\pi )$.

   • The sign of $f^{\prime \prime }$ changes at $x=\pi$. Therefore, $f(x)$ has a point of inflection at $x=\pi$.

AP Exam Quick Hit

Common Question Types

• Analytical from an Equation: Given $f(x)=2x^3-9x^2+12x-1$, find the intervals where $f$ is concave down and identify the x-coordinate of its point of inflection.

• Graphical from $f^{\prime }(x)$: Given the graph of the derivative, $f^{\prime }(x)$, identify the intervals where the original function $f(x)$ is concave up. This requires you to find where the provided graph of $f^{\prime }(x)$ is increasing.

• Tabular from $f^{\prime }(x)$: Given a table of selected values of a differentiable function $f^{\prime }(x)$, determine on which interval $f(x)$ must be concave down. You would look for an interval where the values of $f^{\prime }(x)$ are consistently decreasing.

Common Mistakes

• Confusing Concavity with Slope: A common error is to state that $f$ is concave up because $f^{\prime \prime }(x)$ is increasing. This is incorrect. Concavity depends on the sign of $f^{\prime \prime }(x)$, not whether $f^{\prime \prime }(x)$ is increasing or decreasing.

• Stopping at Candidates: Students find where $f^{\prime \prime }(x)=0$ but fail to test for a sign change. For $f(x)=x^4$, $f^{\prime \prime }(0)=0$, but $f^{\prime \prime }(x)=12x^2$ is positive on both sides of $x=0$. Thus, $x=0$ is not a point of inflection. You must confirm the change in concavity.

• Insufficient Justification: On free-response questions, writing "The function has a point of inflection at $x=c$ because $f^{\prime \prime }(c)=0$" will not earn full credit. A correct justification must state that $f^{\prime \prime }(x)$ changes sign at $x=c$, or that the concavity of $f(x)$ changes at $x=c$.

• Misinterpreting the Graph of $f^{\prime }(x)$: When given the graph of $f^{\prime }(x)$ and asked about the concavity of $f(x)$, students often mistakenly analyze the concavity of the graph they are looking at. You must instead analyze the increasing/decreasing behavior (i.e., the slope) of the given $f^{\prime }(x)$ graph.