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: 65

Answer

(a) The maximum value is approximately $2.20$ The minimum value is approximately $1.80$ (b) The maximum value is $2.21$ The minimum value is approximately $1.81$
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Work Step by Step

(a) $f(x) = x^5-x^3+2$ We can use the zoom function on a graphing calculator to estimate the maximum and minimum values of the function in the interval $[-1,1]$ The maximum value is approximately $2.20$ The minimum value is approximately $1.80$ (b) $f(x) = x^5-x^3+2$ $f'(x) = 5x^4-3x^2 = 0$ $x^2(5x^2-3) = 0$ $x=0$ or $5x^2 = 3$ $x=0$ or $x = \pm \sqrt{\frac{3}{5}}$ When $x = -\sqrt{\frac{3}{5}}$: $(-\sqrt{\frac{3}{5}})^5-(-\sqrt{\frac{3}{5}})^3+2 = 2.19$ The maximum value is $2.21$ When $x = \sqrt{\frac{3}{5}}$: $(\sqrt{\frac{3}{5}})^5-(\sqrt{\frac{3}{5}})^3+2 = 1.81$ The minimum value is approximately $1.81$
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