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# Graphs of Derivatives for AP Calculus

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By — McGraw-Hill Professional
Updated on Mar 10, 2014

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

The functions f, f ', and f" are interrelated, and so are their graphs. Therefore, you can usually infer from the graph of one of the three functions ( f, f ', or f ") and obtain information about the other two. Here are some examples.

### Example 1

The graph of a function f is shown in Figure 7.4-1. Which of the following is true for f on (a, b)?

1. f ' ≥ 0 on (a, b)
2. f " > 0 on (a, b)

Solution:

1. Since f is strictly increasing, f ' ≥ 0 on (a, b) is true.
2. The graph is concave downward on (a, 0) and upward on (0, b). Thus f " > 0 on (0, b) only. Therefore only statement I is true.

### Example 2

Given the graph of f ' in Figure 7.4-2, find where the function f (a) has its relative maximum(s) or relative minimums, (b) is increasing or decreasing, (c) has its point(s) of inflection, (d) is concave upward or downward, and (e) if f(–2)= f(2)=1 and f(0)= – 3, draw a sketch of f.

The function f has a relative maximum at x = –4 and at x = 4, and a relative minimum at x = 0.

A change of concavity occurs at x = – 2 and at x = 2 and f ' exists at x = – 2 and at x =2, which implies that there is a tangent line to the graph of f at x = – 2 and at x =2. Therefore, f has a point of inflection at x = – 2 and at x =2.

1. Summarize the information of f ' on a number line:
2. The function f is increasing on interval (–∞, –4] and [0, 4], and f is decreasing on [–4, 0] and [4,∞).
3. Summarize the information of f " on a number line:
4. The graph of f is concave upward on the interval (–2, 2) and concave downward on (–∞, –2) and (2,∞).
5. A sketch of the graph of f is shown in Figure 7.4-3.

### Example 3

Given the graph of f ' in Figure 7.4-4, find where the function f (a) has a horizontal tangent, (b) has its relative extrema, (c) is increasing or decreasing, (d) has a point of inflection, and (e) is concave upward or downward.

The First Derivative Test indicates that f has relative maximums at x = – 4 and 4; and f has relative minimums at x =2 and 8.

A change of concavity occurs at x = – 1, 3, and 6. Since f '(x ) exists, f has a tangent at every point. Therefore, f has a point of inflection at x = – 1, 3, and 6.

1. f ' (x )=0 at x = – 4, 2, 4, 8. Thus f has a horizontal tangent at these values.
2. Summarize the information of f ' on a number line:
3. The function f is increasing on (–8, –4], [2, 4], and [8, ∞) and is decreasing on [–4, 2] and [4, 8].
4. Summarize the information of f " on a number line:
5. The function f is concave upward on (–1, 3) and (6, ∞) and concave downward on (–∞, –1) and (3, 6).

### Example 4

A function f is continuous on the interval [–4, 3] with f(–4)=6 and f(3)=2 and the following properties:

1. Find the intervals on which f is increasing or decreasing.
2. Find where f has its absolute extrema.
3. Find where f has the points of inflection.
4. Find the intervals where f is concave upward or downward.
5. Sketch a possible graph of f.

Solution:

1. The graph of f is increasing on [1, 3] and decreasing on [–4, –2] and [–2, 1].
2. At x = – 4, f (x )=6. The function decreases until x =1 and increases back to 2 at x =3. Thus, f has its absolute maximum at x = – 4 and its absolute minimum at x =1.
3. A change of concavity occurs at x = –2, and since f '(–2)=0 which implies a tangent line exists at x = – 2, f has a point of inflection at x = – 2.
4. The graph of f is concave upward on (–4, –2) and concave downward on (–2, 1) and (1, 3).
5. A possible sketch of f is shown in Figure 7.4-5.

### Example 5

If f (x )= |ln(x +1)|, find f '(x). (See Figure 7.4-6.)

The domain of f is (–1,∞).

f (0) = |ln(0+1)| = |ln(1)| = 0

Practice problems for these concepts can be found at:

Graphs of Functions and Derivatives Practice Problems for AP Calculus

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