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 147

Example 2

Find the relative extrema for the functionf(x)=
x^3
3
−x^2 − 3 x.

Step 1: Findf′(x).

f′(x)=x^2 − 2 x− 3

Step 2: Find all critical numbers off(x).
Note that f′(x) is defined for all real numbers.
Setf′(x)=0:x^2 − 2 x− 3 = 0 ⇒(x−3)(x+1)= 0 ⇒x=3orx=−1.

Step 3: Findf′′(x):f′′(x)= 2 x−2.

Step 4: Apply the Second Derivative Test.
f′′(3) =2(3)− 2 = 4 ⇒ f(3) is a relative minimum.
f′′(−1)=2(−1)− 2 =− 4 ⇒ f(−1) is a relative maximum.

f(3)=

33

3

−(3)^2 −3(3)=−9 and f(−1)=

5

3

.

Therefore,−9 is a relative minimum value offand

5

3

is a relative maximum value.

(See Figure 8.2-12.)

[–5, 7] by [–10, 10]

Figure 8.2-12

Example 3

Find the relative extrema for the functionf(x)=(x^2 −1)^2 /^3.

Using the First Derivative Test
Step 1: Findf′(x).

f′(x)=

2

3

(x^2 −1)−^1 /^3 (2x)=

4 x

3(x^2 −1)^1 /^3

Step 2: Find all critical numbers off.
Setf′(x)=0. Thus, 4x=0orx=0.
Setx^2 − 1 =0. Thus,f′(x) is undefined atx=1 andx=−1. Therefore, the critical
numbers are−1, 0, and 1.