Problem

Each of the graphs of the functions in Exercise has one relative maximum and one relative...

Each of the graphs of the functions in Exercise has one relative maximum and one relative minimum point. Find these points using the first-derivative test. Use a variation chart as in Example.

Example

Applying the First-Derivative Test Find the local maximum and minimum points of f(x) = − 2x2 + 3x + 1.

SOLUTION

First, we find the critical values and critical points of f:

f′(x) =− 2(2)x + 3 = x2 − 4x + 3
= (x − 1)(x − 3).

The first derivative f′(x) = 0 if x − 1 = 0 or x − 3 = 0. Thus, the critical values are

x = 1 and x = 3.

Substitute the critical values into the expression of f:

f(1) = − 2(1)2 + 3(1) + 1 =
f(3) = − 2(3)2 + 3(3) + 1 = 1.

Thus, the critical points are and (3, 1). To determine whether we have a relative maximum, minimum, or neither at a critical point, we shall apply the first derivative test. This requires a careful study of the sign of f′(x), which can be facilitated with the help of a chart. Here is how you can set up the chart:

• Divide the real line into intervals with the critical values as endpoints.

• Since the sign of f′ depends on the signs of its two factors x − 1 and x − 3, determine the signs of the factors of f′ over each interval. Usually, we do this by testing the sign of a factor at points selected from each interval.

• In each interval, use a plus sign if the factor is positive or a minus sign if it is negative. Then determine the sign of f′ over each interval by multiplying the signs of the factors and using

(+) ・ (+) = +; (+) ・ (−) = −; (−) ・ (+) = −; (−) ・ (−) = +.

A plus sign of f′ corresponds to an increasing portion of the graph of f and a minus sign to a decreasing portion. Denote an increasing portion with an upward arrow and a decreasing portion with a downward arrow. The sequence of arrows should convey the general shape of the graph and, in particular, tell you whether or not your critical values correspond to extreme points. Here is the chart:

You can see from the chart that the sign of f′ varies from positive to negative at x = 1. Thus, according to the first-derivative test, f has a local maximum at x = 1. Also, the sign of f′ varies from negative to positive at x = 3; so f has a local minimum at x = 3. You can confirm these assertions by following the direction of the arrows in the last row of the chart. In conclusion, f has a local maximum at and a local minimum at (3, 1). (See Figure.)

Figure 1

The previous example illustrates cases (a) and (b) in Figure which correspond to local extreme points on the graph at each critical value. The following example illustrates cases (c) and (d) in Figure.

Figure 2