Graphs of Functions and Derivatives 113
Example 2
Find the relative extrema for the functionf(x)=
x^3
3
−x^2 − 3 x.
Step 1: Find f′(x).
f′(x)=x^2 − 2 x− 3
Step 2: Find all critical numbers of f(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: Find f′′(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 7.2-12.)
Figure 7.2-12
Example 3
Find the relative extrema for the functionf(x)=(x^2 −1)^2 /^3.
Using the First Derivative Test
Step 1: Find f′(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 of f.
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.