Explanation Given information: f is a continuous function has exactly one critical point. j ' ( 3 ) i s u n d e f i n e d , j ' ( x ) = 2 f o r x < 3 a n d j ' ( x ) = − 2 f o r x > 3 o n 1 ≤ x ≤ 5 Facts used: The critical points for a function f are numbers in the domain of a function where the derivative f ' is either 0 or it fails to exist. 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: j ' ( 3 ) i s u n d e f i n e d , j ' ( x ) = 2 f o r

6th Edition

ISBN: 9781119399353

Author: DA

Publisher: JOHN WILEY+SONS,INC.-CONSIGNMENT

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Differential Equations

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Chapter 4.3, Problem 31P

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The global maximum and the global minimum occur in the interval 1≤x≤5.

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Chapter 4.3, Problem 30P

Chapter 4.3, Problem 32P

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The critical points of a function f(x) occur at x = 2 & x = 6 and its 8 second order derivative is given by f "(x) = (4 – x)3 - If the Local Minimum of the function occurs at x = b , find b. Note: If say b = 4 , just enter 4. Enter an integer or a fully reduced fraction such as 4, -16 ,3/7 , 1/9 etc. No Spaces please.

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The global maximum and the global minimum occur in the interval 1 ≤ x ≤ 5 .

Solution Summary: The author explains that f is a continuous function with exactly one critical point.

Concept explainers

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To find:

The global maximum and the global minimum occur in the interval 1≤x≤5.

Want to see the full answer?

Chapter 4 Solutions

The global maximum and the global minimum occur in the interval 1 ≤ x ≤ 5 .

Solution Summary: The author explains that f is a continuous function with exactly one critical point.

Concept explainers

Videos

To find: The global maximum and the global minimum occur in the interval 1≤x≤5.

Want to see the full answer?

Chapter 4 Solutions

To find:

The global maximum and the global minimum occur in the interval 1≤x≤5.