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

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

142 STEP 4. Review the Knowledge You Need to Score High

no maximum value. The minimum value occurs at the endpointx=3 and the minimum

value is

1

9

. (See Figure 8.1-8.)

[–1, 4] by [–1, 6]

Figure 8.1-8

8.2 Determining the Behavior of Functions

Main Concepts:Test for Increasing and Decreasing Functions, First Derivative Test and

Second Derivative Test for Relative Extrema, Test for Concavity and

Points of Inflection

Test for Increasing and Decreasing Functions

Letf be a continuous function on the closed interval [a,b] and differentiable on the open

interval (a,b).

1. Iff′(x)>0on(a,b), thenf is increasing on [a,b].


2. Iff′(x)<0on(a,b), thenf is decreasing on [a,b].


3. Iff′(x) =0on(a,b), then f is constant on [a,b].

Definition:Let f be a function defined at a numberc. Thencis a critical number off if

eitherf′(c)=0orf′(c) does not exist. (See Figure 8.2-1.)

f(x)

f′ < 0

f decreasing

f′ < 0

f′ > 0 f decreasing^

f increasing

f′ = 0

f′ = 0

constant

y

0 x

f′ = 0

Figure 8.2-1

Example 1

Find the critical numbers off(x)= 4 x^3 + 2 x^2.

To find the critical numbers of f(x), you have to determine where f′(x)=0 and where

f′(x) does not exist. Notef′(x)= 12 x^2 + 4 x, and f′(x) is defined for all real numbers. Let

f′(x)=0 and thus 12x^2 + 4 x=0, which implies 4x(3x+1)= 0 ⇒x=− 1 /3orx=0.

Therefore, the critical numbers off are 0 and− 1 /3. (See Figure 8.2-2.)