Explanation Given information: f ( x ) = a x e b x has a critical point at x = 3. 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. Calculation: f ( x ) = a x e b x (Given) Differentiating with respect to ‘x’, we get f ' ( x ) = a e b x + a b x e b x = a e b x ( 1 + b x ) f ( x ) has critical point ⇒ f ' ( x ) = 0 ⇒ a e b x ( 1 + b x ) = 0 ⇒ ( 1 + b x ) = 0

6th Edition

ISBN: 9781119399353

Author: DA

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Chapter 4.1, Problem 54P

To determine

To find:

The values of aand b so that f(x)=axebx has f(13)=1 and f has a local maximum at x=13.

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Chapter 4.1, Problem 53P

Chapter 4.1, Problem 55P

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Chapter 4 Solutions

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The values of aand b so that f ( x ) = a x e b x has f ( 1 3 ) = 1 and f has a local maximum at x = 1 3 .

Solution Summary: The author explains that the value of aand b is f(x)=axebx.

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To find:

The values of aand b so that f(x)=axebx has f(13)=1 and f has a local maximum at x=13.

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Chapter 4 Solutions

The values of aand b so that f ( x ) = a x e b x has f ( 1 3 ) = 1 and f has a local maximum at x = 1 3 .

Solution Summary: The author explains that the value of aand b is f(x)=axebx.

Concept explainers

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To find: The values of aand b so that f(x)=axebx has f(13)=1 and f has a local maximum at x=13.

Want to see the full answer?

Chapter 4 Solutions

To find:

The values of aand b so that f(x)=axebx has f(13)=1 and f has a local maximum at x=13.