Explanation Given information: f ( x ) = x + 1 x , x > 0 . Fact used: If f ' changes from negative to positive at p, then f has a local minimum at p. If f ' changes from positive to negative at p, then f has a local maximum at p. Calculation: f ( x ) = x + 1 x , x > 0 (Given) Differentiating with respect to ‘x’, we get f ' ( x ) = 1 − 1 x 2 Equating f ' ( x ) = 0 ⇒ 1 − 1 x 2

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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Chapter 4.3, Problem 40P

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The exact global maximum and minimum values of the function f(x)=x+1x,x>0.

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Chapter 4.3, Problem 39P

Chapter 4.3, Problem 41P

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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 ( x ) = x + 1 x , x &gt; 0 .

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To find: The exact global maximum and minimum values of the function f(x)=x+1x,x>0.

Want to see the full answer?

Chapter 4 Solutions

The exact global maximum and minimum values of the function f ( x ) = x + 1 x , x > 0 .

To find:

The exact global maximum and minimum values of the function f(x)=x+1x,x>0.