a. Answer The local maximum at ( 0 , 0 ) The local minimum at ( − 2 , 6 − 2 3 ) Explanation Given information : The given function is g ( x ) = x 1 3 ( x + 8 ) Concept used : Extreme values occur only at critical points and end points. A point in the interior of the domain of a function f at which f ′ = 0 or f ′ does not exist is called a critical point of f . Calculation : The domain of the function is { real numbers } Take derivative of the function g ′ ( x ) = x + 8 3 x 2 3 + x 1 3 Equate the derivative to 0. x + 8 3 x 2 3 + x 1 3 = 0 x + 8 3 x 2 3 = − x 1 3 x + 8 = − x 1 3 ⋅ 3 x 2 3 x + 8 = − 3 x 4 x = − 8 x = − 8 4 x = − 2 Value at x = − 2 g ( − 2 ) = ( − 2 ) 1 3 ( ( − 2 ) + 8 ) g ( − 2 ) = 6 − 2 3 x = 0 Is also a critical point and value at x = 0 g ( 0 ) = 0 1 3 ( 0 + 8 ) g ( 0 ) = 0 Therefore, The local maximum at ( 0 , 0 ) The local minimum at ( − 2 , 6 − 2 3 ) b. To determine To find the intervals on which the function is increasing. Answer The function increases on [ − 2 , ∞ ) . Explanation Given information : The given function is g ( x ) = x 1 3 ( x + 8 ) Concept used : If f ′ > 0 at each point of ( a , b ) , then f increases on [ a , b ] It is assumed that f be continuous on [ a , b ] and differentiable on ( a , b ) . Calculation : For increasing function f ′ ( x ) must be greater than 0. x + 8 3 x 2 3 + x 1 3 > 0 x + 8 3 x 2 3 > − x 1 3 x + 8 > − x 1 3 ⋅ 3 x 2 3 x + 8 > − 3 x 4 x > − 8 x > − 8 4 x > − 2 Therefore, The function increases on [ − 2 , ∞ ) . c. To determine To find the intervals on which the function is decreasing. Answer The function decreases on ( − ∞ , − 2 ] Explanation Given information: The given function is g ( x ) = x 1 3 ( x + 8 ) Concept used : If f ′ < 0 at each point of ( a , b ) , then f decreases on [ a , b ] It is assumed that f be continuous on [ a , b ] and differentiable on ( a , b ) . Calculation : For decreasing function f ′ ( x ) must be less than 0. x + 8 3 x 2 3 + x 1 3 < 0 x + 8 3 x 2 3 < − x 1 3 x + 8 < − x 1 3 ⋅ 3 x 2 3 x + 8 < − 3 x 4 x < − 8 x < − 8 4 x < − 2 Therefore, The function decreases on ( − ∞ , − 2 ]

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

4th Edition

ISBN: 9780133178579

Author: Ross L. Finney

Publisher: PEARSON

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Question

Chapter 5.2, Problem 24E

To determine

To find the local extrema.

Expert Solution

Answer to Problem 24E

The local maximum at (0,0)

The local minimum at (−2,6−23)

Explanation of Solution

Given information:

The given function is g(x)=x13(x+8)

Concept used:

Extreme values occur only at critical points and end points.

A point in the interior of the domain of a function f at which f′=0 or f′ does not exist is called a critical point of f .

Calculation :

The domain of the function is {real numbers}

Take derivative of the function

g′(x)=x+83x23+x13

Equate the derivative to 0.

x+83x23+x13=0x+83x23=−x13x+8=−x13⋅3x23x+8=−3x4x=−8x=−84x=−2

Value at x=−2

g(−2)=(−2)13((−2)+8)g(−2)=6−23

x=0 Is also a critical point and value at x=0

g(0)=013(0+8)g(0)=0

Therefore,

The local maximum at (0,0)

The local minimum at (−2,6−23)

To determine

To find the intervals on which the function is increasing.

Expert Solution

Answer to Problem 24E

The function increases on [−2,∞) .

Explanation of Solution

Given information:

The given function is g(x)=x13(x+8)

Concept used:

If f′>0 at each point of (a,b) , then f increases on [a,b]

It is assumed that f be continuous on [a,b] and differentiable on (a,b) .

Calculation :

For increasing function f′(x) must be greater than 0.

x+83x23+x13>0x+83x23>−x13x+8>−x13⋅3x23x+8>−3x4x>−8x>−84x>−2

Therefore,

The function increases on [−2,∞) .

To determine

To find the intervals on which the function is decreasing.

Expert Solution

Answer to Problem 24E

The function decreases on (−∞,−2]

Explanation of Solution

Given information:

The given function is g(x)=x13(x+8)

Concept used:

If f′<0 at each point of (a,b) , then f decreases on [a,b]

It is assumed that f be continuous on [a,b] and differentiable on (a,b) .

Calculation :

For decreasing function f′(x) must be less than 0.

x+83x23+x13<0x+83x23<−x13x+8<−x13⋅3x23x+8<−3x4x<−8x<−84x<−2

Therefore,

The function decreases on (−∞,−2]

Chapter 5.2, Problem 23E

Chapter 5.2, Problem 25E

Chapter 5 Solutions

Calculus 2012 Student Edition (by Finney/Demana/Waits/Kennedy)

Ch. 5.1 - Prob. 1QR Ch. 5.1 - Prob. 2QR Ch. 5.1 - Prob. 3QR Ch. 5.1 - Prob. 4QR Ch. 5.1 - Prob. 5QR Ch. 5.1 - Prob. 6QR Ch. 5.1 - Prob. 7QR Ch. 5.1 - Prob. 8QR Ch. 5.1 - Prob. 9QR Ch. 5.1 - Prob. 10QR

Ch. 5.1 - Prob. 11QR Ch. 5.1 - Prob. 12QR Ch. 5.1 - Prob. 1E Ch. 5.1 - Prob. 2E Ch. 5.1 - Prob. 3E Ch. 5.1 - Prob. 4E Ch. 5.1 - Prob. 5E Ch. 5.1 - Prob. 6E Ch. 5.1 - Prob. 7E Ch. 5.1 - Prob. 8E Ch. 5.1 - Prob. 9E Ch. 5.1 - Prob. 10E Ch. 5.1 - Prob. 11E Ch. 5.1 - Prob. 12E Ch. 5.1 - Prob. 13E Ch. 5.1 - Prob. 14E Ch. 5.1 - Prob. 15E Ch. 5.1 - Prob. 16E Ch. 5.1 - Prob. 17E Ch. 5.1 - Prob. 18E Ch. 5.1 - Prob. 19E Ch. 5.1 - Prob. 20E Ch. 5.1 - Prob. 21E Ch. 5.1 - Prob. 22E Ch. 5.1 - Prob. 23E Ch. 5.1 - Prob. 24E Ch. 5.1 - Prob. 25E Ch. 5.1 - Prob. 26E Ch. 5.1 - Prob. 27E Ch. 5.1 - Prob. 28E Ch. 5.1 - Prob. 29E Ch. 5.1 - Prob. 30E Ch. 5.1 - Prob. 31E Ch. 5.1 - Prob. 32E Ch. 5.1 - Prob. 33E Ch. 5.1 - Prob. 34E Ch. 5.1 - Prob. 35E Ch. 5.1 - Prob. 36E Ch. 5.1 - Prob. 37E Ch. 5.1 - Prob. 38E Ch. 5.1 - Prob. 39E Ch. 5.1 - Prob. 40E Ch. 5.1 - Prob. 41E Ch. 5.1 - Prob. 42E Ch. 5.1 - Prob. 43E Ch. 5.1 - Prob. 44E Ch. 5.1 - Prob. 45E Ch. 5.1 - Prob. 46E Ch. 5.1 - Prob. 47E Ch. 5.1 - Prob. 48E Ch. 5.1 - 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