MATH 12002 - CALCULUS I 3.1: Maximum and Minimum Values - Examples - - PowerPoint PPT Presentation

MATH 12002 - CALCULUS I §3.1: Maximum and Minimum Values - Examples

Professor Donald L. White

Department of Mathematical Sciences Kent State University

D.L. White (Kent State University) 1 / 7

Definitions & Theorems

Extreme Value Theorem

If y = f (x) is a continuous function on a closed interval [a, b], then f has both an absolute maximum and an absolute minimum on [a, b].

Theorem

If y = f (x) is continuous on a closed interval [a, b], then its absolute maximum and minimum on [a, b] must occur at one of the endpoints

• r at a point c, a < c < b, where f has a local maximum or minimum.

Theorem

If y = f (x) has a local maximum or minimum at x = c, then either f ′(c) = 0 or f ′(c) is undefined. A number c in the domain of f where f ′(c) = 0 or f ′(c) is undefined is called a critical number for f .

D.L. White (Kent State University) 2 / 7

Definitions & Theorems

Putting these theorems together, we have the following procedure for finding absolute maximum and minimum values

• f a continuous function on a closed interval:

Let y = f (x) be continuous on the closed interval [a, b].

1 Find all critical numbers c of f in (a, b);

that is, all c with a < c < b such that f ′(c) = 0 or f ′(c) is undefined.

2 Compute the y values f (a) and f (b) of f at the endpoints a and b

and the y value f (c) for every critical number c found in Step 1.

3 Compare the y values found in Step 2.

The largest is the absolute maximum value of f on [a, b] and the smallest is the absolute minimum value of f on [a, b].

D.L. White (Kent State University) 3 / 7

Examples

Example 1

Find the absolute maximum and absolute minimum values of f (x) = x4 − 2x2 + 3

• n the interval [−2, 3].

Solution

We first need to find the critical numbers. We have f ′(x) = 4x3 − 4x = 4x(x2 − 1) = 4x(x − 1)(x + 1). Hence f ′(x) is defined everywhere and is 0 when one of the factors

• f f ′(x) is 0, so at x = 0, x = 1, and x = −1.

Therefore, the critical numbers for f are 0, 1, and −1. [Continued →]

D.L. White (Kent State University) 4 / 7

Examples

Example 1 Solution [continued]

The maximum and minimum must occur at one of the critical numbers 0, 1, −1, or at an endpoint −2 or 3. For f (x) = x4 − 2x2 + 3, we compute f (0) = 04 − 2(02) + 3 = 3 f (1) = 14 − 2(12) + 3 = 1 − 2 + 3 = 2 f (−1) = (−1)4 − 2((−1)2) + 3 = 1 − 2 + 3 = 2 f (−2) = (−2)4 − 2((−2)2) + 3 = 16 − 8 + 3 = 11 f (3) = 34 − 2(32) + 3 = 81 − 18 + 3 = 66. Therefore, the absolute maximum value of f (x) is y = 66 at x = 3, and the absolute minimum value of f (x) is y = 2 at x = 1 and x = −1.

D.L. White (Kent State University) 5 / 7

Examples

Example 2

Find the absolute maximum and absolute minimum values of f (x) = x(1 − x)2/5 on the interval 1

2, 2

• .

Solution

We first need to find the critical numbers. We have f ′(x) = 1 · (1 − x)2/5 + x · 2

5(1 − x)−3/5(−1)

= (1 − x)2/5 − 2x 5(1 − x)3/5 = 5(1 − x)3/5(1 − x)2/5 − 2x 5(1 − x)3/5 = 5(1 − x) − 2x 5(1 − x)3/5 = 5 − 7x 5(1 − x)3/5 . [Continued →]

D.L. White (Kent State University) 6 / 7

Examples

Example 2 Solution [continued]

We have f ′(x) =

5−7x 5(1−x)3/5 ,

and so f ′(x) = 0 when 5 − 7x = 0, so at x = 5/7, and f ′(x) is undefined when 5(1 − x)3/5 = 0, so at x = 1. Thus the critical numbers for f are 5/7 and 1. The maximum and minimum must occur at one of the critical numbers 5/7 or 1, or at an endpoint 1/2 or 2. For f (x) = x(1 − x)2/5, we compute f (5

7)

=

5 7(1 − 5 7)2/5 = 5 7(2 7)2/5 ≈ 0.43

f (1) = 1 · (1 − 1)2/5 = 1 · 02/5 = 0 f (1

2)

=

1 2(1 − 1 2)2/5 = 1 2(1 2)2/5 = (1 2)7/5 ≈ 0.38

f (2) = 2 · (1 − 2)2/5 = 2 · (−1)2/5 = 2 · 1 = 2. Therefore, the absolute maximum value of f (x) is y = 2 at x = 2, and the absolute minimum value of f (x) is y = 0 at x = 1.

D.L. White (Kent State University) 7 / 7