5 Steps to a 5 AP Calculus AB 2019 - William Ma

MA 3972-MA-Book April 11, 2018 17:21

Graphs of Functions and Derivatives 141

Set

2

3

(x−1)^1 /^3 = 0 ⇒x=1. Note thatfis not differentiable (a+x=1). Therefore,cdoes

not exist. (See Figure 8.1-6.)

[–8, 8] by [–4, 4]

Figure 8.1-6

TIP • The formula for finding the area of an equilateral triangle isarea=s

2 √ 3

4

wheresis the length of a side. You might need this to find the volume of a solid whose

cross sections are equilateral triangles.

Extreme Value Theorem

Iff is a continuous function on a closed interval [a,b], thenfhas both a maximum and

a minimum value on the interval.

Example 1

Iff(x)=x^3 + 3 x^2 −1, find the maximum and minimum values off on [−2, 2]. Sincef(x)

is a polynomial, it is a continuous function everywhere. Entery 1 =x^3 + 3 x^2 −1. The graph

ofy1 indicates thatfhas a minimum of−1atx=0 and a maximum value of 19 atx=2.

(See Figure 8.1-7.)

[–3, 3] by [–4, 20]

Figure 8.1-7

Example 2

If f(x)=

1

x^2

, find any maximum and minimum values of f on [0, 3]. Since f(x)isa

rational function, it is continuous everywhere except at values where the denominator is

0. In this case, atx=0, f(x) is undefined. Since f(x) is not continuous on [0, 3], the
   Extreme Value Theorem may not be applicable. Entery 1 =

1

x^2

. The graph ofy1 shows
that asx → 0 +, f(x) increases without bound (i.e., f(x) goes to infinity). Thus, f has