Explanation Given information: f ( t ) = t 1 + t 2 . The domain is all real numbers Fact used: If f ' changes from negative to positive at p, then f has a local minimum at p. f ' changes from positive to negative at p, then f has a local maximum at p. Calculation: f ( t ) = t 1 + t 2 (Given) Differentiating with respect to ‘x’, we get f ' ( t ) = ( 1 + t 2 ) ( 1 ) − t ( 2 t ) ( 1 + t 2 ) 2 = ( 1 + t 2 ) − 2 t 2 ( 1 + t 2 ) 2 = 1 − t 2 ( 1

Applied Calculus, 6e WileyPLUS + Loose-leaf

6th Edition

ISBN: 9781119408901

Author: Deborah Hughes-Hallett, Andrew M. Gleason, Patti Frazer Lock, Daniel E. Flath

Publisher: Wiley (WileyPLUS Products)

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Question

Chapter 4.3, Problem 45P

To determine

To find:

The exact global maximum and minimum values of the function f(t)=t1+t2.

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Chapter 4.3, Problem 44P

Chapter 4.3, Problem 46P

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Chapter 4 Solutions

Applied Calculus, 6e WileyPLUS + Loose-leaf

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The exact global maximum and minimum values of the function f ( t ) = t 1 + t 2 .

Solution Summary: The author calculates the global maximum and minimum values of the function f(t)=t1.

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To find:

The exact global maximum and minimum values of the function f(t)=t1+t2.

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Chapter 4 Solutions

The exact global maximum and minimum values of the function f ( t ) = t 1 + t 2 .

Solution Summary: The author calculates the global maximum and minimum values of the function f(t)=t1.

Concept explainers

Videos

To find: The exact global maximum and minimum values of the function f(t)=t1+t2.

Want to see the full answer?

Chapter 4 Solutions

To find:

The exact global maximum and minimum values of the function f(t)=t1+t2.