Find the derivative at each critical point and determine the y = x23(x2 - 16); x 20 Select one: A. Critical Pt derivative Extremum|Value x=0 Undefined local max 4 x=2 minimum -19.049 B. Critical Pt derivative Extremum|Value Undefined local max x=0 x= 2 minimum -19.049 C Critical Pt derivative Extremum|Value x = 0 x = 2 maximum 0 minimum |-19.049 D. Critical Pt derivative Extremum|Value Unde fine d local max 0 x = 0 x = 2 minimum |31748

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)
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Chapter1: Functions
Section1.2: Functions Given By Tables
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Find the derivative at each critical point and determine the local extreme values.
y = x23(x2 - 16); x 20
Select one:
A. Critical Pt derivative Extremum|Value
|Unde fine d local max | 4
minimum |-19.049
x =0
x= 2
B. Critical Pt derivative Extremum|Value
x = 0
x = 2
Unde fined local max
minimum -19.049
C. Critical Pt. derivative Extremum|Value
maximum 0
|minimum -19.049
x = 0
x = 2
|0
D. Critical Pt derivative Extremum Value
|Undefined local max 0
minimum 31748
x = 0
x = 2
Transcribed Image Text:Find the derivative at each critical point and determine the local extreme values. y = x23(x2 - 16); x 20 Select one: A. Critical Pt derivative Extremum|Value |Unde fine d local max | 4 minimum |-19.049 x =0 x= 2 B. Critical Pt derivative Extremum|Value x = 0 x = 2 Unde fined local max minimum -19.049 C. Critical Pt. derivative Extremum|Value maximum 0 |minimum -19.049 x = 0 x = 2 |0 D. Critical Pt derivative Extremum Value |Undefined local max 0 minimum 31748 x = 0 x = 2
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