Explanation Given information: f ( x ) = x 3 ( 1 − x ) 4 Definition used: For any function f , a point p in the domain of f where f ' ( p ) = 0 or f ' ( p ) is undefined is called a critical point of the function. Fact used: Second derivative test: If f ' ' is positive, f has a local minimum at the critical point. f ' ' is negative, f has a local maximum at the critical point. Calculation: f ( x ) = x 3 ( 1 − x ) 4 (Given) Differentiating with respect to ‘x’, we get Equating f ' ( x ) = 0 , we have Therefore, the critical points are x = 0 , 1 , 3 7 . Classification: The second derivative is given by

6th Edition

ISBN: 9781119399353

Author: DA

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 4.1, Problem 31P

To determine

To find and classify:

The critical points of f(x)=x3(1−x)4 as local maxima and minima.

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Chapter 4.1, Problem 30P

Chapter 4.1, Problem 32P

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To find and classify: The critical points of f ( x ) = x 3 ( 1 − x ) 4 as local maxima and minima .

Solution Summary: The author explains how to find and classify the critical points of f(x)=x3(1-x), which are local maxima and minima.

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To find and classify:

The critical points of f(x)=x3(1−x)4 as local maxima and minima.

Want to see the full answer?

Chapter 4 Solutions

To find and classify: The critical points of f ( x ) = x 3 ( 1 − x ) 4 as local maxima and minima .

Solution Summary: The author explains how to find and classify the critical points of f(x)=x3(1-x), which are local maxima and minima.

Concept explainers

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To find and classify: The critical points of f(x)=x3(1−x)4 as local maxima and minima.

Want to see the full answer?

Chapter 4 Solutions

To find and classify:

The critical points of f(x)=x3(1−x)4 as local maxima and minima.