Answered: By the extreme value theorem, f(x) =…

By the extreme value theorem, f(x) = x/4 – x² has extreme values on [0, 3/2]. What are the minimum and maximum of f(x) on the interval [0, 3/2]. Explain.

College Algebra (MindTap Course List)

12th Edition

ISBN:9781305652231

Author:R. David Gustafson, Jeff Hughes

Publisher:R. David Gustafson, Jeff Hughes

Chapter3: Functions

Section3.3: More On Functions; Piecewise-defined Functions

Problem 99E: Determine if the statemment is true or false. If the statement is false, then correct it and make it...

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Question

By the extreme value theorem, f(x) = x/4 – x² has extreme values on
[0, 3/2]. What are the minimum and maximum of f(x) on the interval [0, 3/2].
Explain.

Expert Solution

Step 1

Maxima and minima:

$f (x) = x \sqrt{4 - x^{2}} f' (x) = x . \frac{- 2 x}{2 \sqrt{4 - x^{2}}} + \sqrt{4 - x^{2}} f' (x) = \frac{- x^{2}}{\sqrt{4 - x^{2}}} + \sqrt{4 - x^{2}} f' (x) = 0 \frac{- x^{2}}{\sqrt{4 - x^{2}}} = - \sqrt{4 - x^{2}} x^{2} = 4 - x^{2} 2 x^{2} = 4 x^{2} = \pm \sqrt{2}$

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