Test for Concavity and Points of Inflection for AP Calculus

Practice problems for these concepts can be found at:  Graphs of Functions and Derivatives Practice Problems for AP Calculus

Test for Concavity

Let f be a differentiable function.

1. If f'' > 0 on an interval I, then f is concave upward on I.
2. If f'' < 0 on an interval I, then f is concave downward on I.

See Figures 7.2-14 and 7.2-15.

Points of Inflection

A point P on a curve is a point of inflection if:

1. the curve has a tangent line at P, and
2. the curve changes concavity at P (from concave upward to downward or from concave downward to upward).

See Figures 7.2-16–7.2-18.

Note that if a point (a, f (a)) is a point of inflection, then f ''(c)=0 or f ''(c) does not exist. (The converse of the statement is not necessarily true.)

Note: There are some textbooks that define a point of inflection as a point where the concavity changes and do not require the existence of a tangent at the point of inflection. In that case, the point at the cusp in Figure 7.2-18 would be a point of inflection.

Example 1

The graph of f ', the derivative of a function f, is shown in Figure 7.2-19. Find the points of inflection of f and determine where the function f is concave upward and where it is concave downward on [–3, 5].

Solution: See Figure 7.2-20.

Thus, f is concave upward on [–3, 0) and (3, 5], and is concave downward on (0, 3). There are two points of inflection: one at x = 0 and the other at x = 3.

Example 2

Using a calculator, find the values of x at which the graph of y = x²e^x changes concavity.

Enter y1=x^2 * e^x and y2=d(y1(x), x, 2). The graph of y2, the second derivative of y, is shown in Figure 7.2-21. Using the [Zero] function, you obtain x = –3.41421 and x = –0.585786.

(See Figures 7.2-21 and 7.2-22.)

Thus, f changes concavity at x = –3.41421 and x = –0.585786.

Example 3

Find the points of inflection of f(x) = x³ – 6x² + 12x – 8 and determine the intervals where the function f is concave upward and where it is concave downward.

Step 1:   Find f '(x) and f ''(x).

f ''(x) = 3x² – 12x +12

f ''(x) = 6x – 12

Step 2:   Set f ''(x) = 0.

6x – 12 = 0

x = 2

Note that f ''(x) is defined for all real numbers.

Step 3:   Determine intervals.

The intervals are (–∞, 2) and (2, ∞).

Step 4:   Set up a table.

Since f(x) has change of concavity at x = 2, the point (2, f (2)) is a point of inflection. f(2) = (2)³ – 6(2)² +12(2) – 8 = 0.

Step 5:  Write a conclusion.

Thus f(x) is concave downward on (–∞, 2), concave upward on (2, ∞) and f(x) has a point of inflection at (2, 0). (See Figure 7.2-23.)

Example 4

Find the points of inflection of and determine the intervals where the function f is concave upward and where it is concave downward.

Step 1:   Find f '(x) and f ''(x).

Step 2:   Find all values of x where f ''(x) = 0 or f ''(x) is undefined.

Note that f ''(x) ≠ 0 and that f''(1) is undefined.

Step 3: Determine intervals.

The intervals are (–∞, 1), and (1, ∞).

Step 4:   Set up a table.

Note that since f(x) has no change of concavity at x = 1, f does not have a point of inflection.

Step 5:   Write a conclusion.

Therefore f (x) is concave downward on (–∞, ∞) and has no point of inflection. (See Figure 7.2-24.)

Example 5

The graph of f is shown in Figure 7.2-25 and f is twice differentiable. Which of the following statements is true:

1. f(5) < f '(5) < f ''(5)
2. f ''(5) < f '(5) < f (5)
3. f '(5) < f (5) < f ''(5)
4. f '(5) < f ''(5) < f (5)
5. f ''(5) < f (5) < f '(5)

The graph indicates that (1) f (5) = 0, (2) f '(5) < 0, since f is decreasing; and (3) f ''(5) > 0, since f is concave upward. Thus, f '(5) < f (5) < f ''(5), choice (C).

Practice problems for these concepts can be found at: 

Graphs of Functions and Derivatives Practice Problems for AP Calculus