Education.com
Try
Brainzy
Try
Plus

# Extreme Value Theorem for AP Calculus

(not rated)
By McGraw-Hill Professional
Updated on Oct 24, 2011

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

If f is a continuous function on a closed interval [a, b], then f has both a maximum and a minimum value on the interval.

### Example 1

If f (x) = x 3 + 3x2 – 1, find the maximum and minimum values of f on [–2, 2]. Since f(x) is a polynomial, it is a continuous function everywhere. Enter y1 = x3 + 3x2 – 1. The graph of y 1 indicates that f has a minimum of – 1 at x = 0 and a maximum value of 19 at x = 2. (See Figure 7.1-7.)

### Example 2

If , find any maximum and minimum values of f on [0, 3]. Since f (x) is a rational function, it is continuous everywhere except at values where the denominator is 0. In this case, at x = 0, f (x) is undefined. Since f (x ) is not continuous on [0, 3], the Extreme Value Theorem may not be applicable. Enter . The graph of y 1 shows that as x → 0+, f (x) increases without bound (i.e., f(x) goes to infinity). Thus f has no maximum value. The minimum value occurs at the endpoint x = 3 and the minimum value is . (See Figure 7.1-8.)

Practice problems for these concepts can be found at:

Graphs of Functions and Derivatives Practice Problems for AP Calculus

### Ask a Question

150 Characters allowed

### Related Questions

#### Q:

See More Questions

### Today on Education.com

Top Worksheet Slideshows