Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 4 - Section 4.1 - Maximum and Minimum Values - 4.1 Exercises - Page 284: 67

Answer

(a) The maximum value is approximately $0.30$ The minimum value is $0$ (b) The maximum value is $0.32$ The minimum value is $0$

Work Step by Step

(a) $f(x) = x~\sqrt{x-x^2}$ We can use the zoom function on a graphing calculator to estimate the maximum and minimum values of the function in the interval $[0,1]$ where the function is defined. The maximum value is approximately $0.30$ The minimum value is $0$ (b) $f(x) = x~\sqrt{x-x^2}$ $f'(x) = \sqrt{x-x^2}+(x)~(\frac{1-2x}{2\sqrt{x-x^2}})$ $f'(x) = \frac{2x-2x^2}{2\sqrt{x-x^2}}+\frac{x-2x^2}{2\sqrt{x-x^2}}$ $f'(x) = \frac{3x-4x^2}{2\sqrt{x-x^2}} = 0$ $3x-4x^2 = 0$ $x(3-4x) = 0$ $x = 0~~$ or $~~x = \frac{3}{4}$ When $x = 0$: $f(0) = (0)~\sqrt{(0)-(0)^2} = 0$ The minimum value is $0$ When $x = \frac{3}{4}$: $f(x) = (\frac{3}{4})~\sqrt{(\frac{3}{4})-(\frac{3}{4})^2} = 0.32$ The maximum value is $0.32$
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