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.

- If
*f''*> 0 on an interval I, then*f*is concave upward on I. - 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:

- the curve has a tangent line at P, and
- 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* = *x*^{2}*e*^{x} changes concavity.

Enter *y*1=*x*^2 * e^*x* and *y*2=*d*(*y*1(*x*), *x*, 2). The graph of *y*2, 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*) = *x*^{3} – 6*x*^{2} + 12*x* – 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*) = 3

*x*

^{2}– 12

*x*+12

*f*''(

*x*) = 6

*x*– 12

Step 2: Set *f* ''(*x*) = 0.

- 6

*x*– 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:

*f*(5) <*f*'(5) <*f*''(5)*f*''(5) <*f*'(5) <*f*(5)*f*'(5) <*f*(5) <*f*''(5)*f*'(5) <*f*''(5) <*f*(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

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