Given: f'(x)=x*(x² - 4) Note: This is fX) .... not f (x) (a) What are the x-coordinates of any critical numbers of f(x)?

Calculus: Early Transcendentals
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Author:James Stewart
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Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Given: f'(x)=x*(x² -4)*
5.
Note: This is f (x) .... not f (x)
(a)
What are the x-coordinates of any critical numbers of f(x)?
(b)
What are the x-coordinates of any maximum/minimum numbers f(x)?
(c)
Give the intervals for which f(x) is Increasing.
(d)
Give the intervals for which f(x) is decreasing.
Transcribed Image Text:Given: f'(x)=x*(x² -4)* 5. Note: This is f (x) .... not f (x) (a) What are the x-coordinates of any critical numbers of f(x)? (b) What are the x-coordinates of any maximum/minimum numbers f(x)? (c) Give the intervals for which f(x) is Increasing. (d) Give the intervals for which f(x) is decreasing.
Expert Solution
Step 1

To find out critical values , we need to set f'(X)=0  and solve for x

f'(x)=x2(x2-4)3

set the derivative =0 

x2(x2-3)3=0x2=0, x=0(x2-4)3=0x2-4=0x2=4x=±2x=2, -2

x coordinates of critical numbers are -2,0,2

 

 

Step 2

Now make a sign chart using the x values

Make a number line using x values and make 4 intervals

Pick a number from each interval and plug it in first derivative

f'(x)=x2(x2-4)3(-,-2)x=-3f'(-3)=(-3)2((-3)2-4)3=1125x=-1f'(-1)=(-1)2((-1)2-4)3=-27x=1f'(-3)=(1)2((1)2-4)3=-27x=3f'(3)=(3)2((3)2-4)3=1125

Calculus homework question answer, step 2, image 1

 

 

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