Explanation Given: The function is f ( x ) = x 3 sin ( 1 x ) for x ≠ 0 and f ( 0 ) = 0 . Calculation: The given function is f ( x ) = x 3 sin ( 1 x ) for x ≠ 0 and also given that f ( 0 ) = 0 . If f ′ is differentiable at x = 0 then f ′ is continuous at x = 0 . Label the given function as follows. f ( x ) = x 3 sin ( 1 x ) ..... (1) Differentiate the function in equation (1) as shown below. f ′ ( x ) = d ( x 3 sin ( 1 x ) ) d x = 3 x 2 sin ( 1 x ) + x 3 ⋅ cos ( 1 x ) ⋅ ( − 1 x 2 ) = 3 x 2 sin ( 1 x ) − x cos ( 1 x ) = x ( 3 x sin ( 1 x ) − cos ( 1 x ) ) Take limit both sides of the above expression as shown below. lim x → 0 f ′ ( x ) = lim x → 0 ( x ( 3 x sin ( 1 x ) − cos ( 1 x ) ) ) = 0 The left hand limit of f ′ ( x ) = x ( 3 x sin ( 1 x ) − cos ( 1 x ) ) can be calculated as follows To determine (b) To draw: The graphs of function f and f ′ , also explain whether first derivative test is applied or not. To determine (c) To show: The function f ( 0 ) is neither a local min nor a local max.

4th Edition

ISBN: 9781319379421

Author: Rogawski

Publisher: MAC HIGHER

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Question

Chapter 4.3, Problem 78E

To determine

(a)

To show:

The derivative of f(x)=x3sin(1x) is continuous at x=0 and that x=0 is a critical point of f.

To determine

(b)

To draw:

The graphs of function f and f′, also explain whether first derivative test is applied or not.

To determine

(c)

To show:

The function f(0) is neither a local min nor a local max.

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Chapter 4.3, Problem 77E

Chapter 4.3, Problem 79E

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Chapter 4 Solutions

CALCULUS 4E (HC) W/ ACHIEVE ACCESS

Ch. 4.1 - Prob. 1PQ Ch. 4.1 - Prob. 2PQ Ch. 4.1 - Prob. 3PQ Ch. 4.1 - Prob. 4PQ Ch. 4.1 - Prob. 1E Ch. 4.1 - Prob. 2E Ch. 4.1 - Prob. 3E Ch. 4.1 - Prob. 4E Ch. 4.1 - Prob. 5E Ch. 4.1 - Prob. 6E

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(a) To show: The derivative of f ( x ) = x 3 sin ( 1 x ) is continuous at x = 0 and that x = 0 is a critical point of f .

Solution Summary: The author explains that the derivative of f(x)=x 3mathrmsin

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(a)

To show:

The derivative of f(x)=x3sin(1x) is continuous at x=0 and that x=0 is a critical point of f.

(b)

To draw:

The graphs of function f and f′, also explain whether first derivative test is applied or not.

(c)

To show:

The function f(0) is neither a local min nor a local max.

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Chapter 4 Solutions

(a) To show: The derivative of f ( x ) = x 3 sin ( 1 x ) is continuous at x = 0 and that x = 0 is a critical point of f .

Solution Summary: The author explains that the derivative of f(x)=x 3mathrmsin

Concept explainers

Videos

(a) To show: The derivative of f(x)=x3sin(1x) is continuous at x=0 and that x=0 is a critical point of f.

(b) To draw: The graphs of function f and f′, also explain whether first derivative test is applied or not.

(c) To show: The function f(0) is neither a local min nor a local max.

Want to see the full answer?

Chapter 4 Solutions

(a)

To show:

The derivative of f(x)=x3sin(1x) is continuous at x=0 and that x=0 is a critical point of f.

(b)

To draw:

The graphs of function f and f′, also explain whether first derivative test is applied or not.

(c)

To show:

The function f(0) is neither a local min nor a local max.