Calculus: Early Transcendentals 8th Edition

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

Chapter 4 - Section 4.2 - The Mean Value Theorem - 4.2 Exercises - Page 292: 34

Answer

$2sin^{-1}~x = cos^{-1}(1-2x^2)~~~~$ $x \geq 0$

Work Step by Step

Let $f(x) = 2sin^{-1}~x-cos^{-1}(1-2x^2)~~~~$ $x \geq 0$ We can find $f'(x)$: $f'(x) = \frac{2}{\sqrt{1-x^2}}-\frac{-(-4x)}{\sqrt{1-(1-2x^2)^2}}$ $f'(x) = \frac{2}{\sqrt{1-x^2}}-\frac{4x}{\sqrt{1-(1-4x^2+4x^4)}}$ $f'(x) = \frac{2}{\sqrt{1-x^2}}-\frac{4x}{\sqrt{4x^2-4x^4}}$ $f'(x) = \frac{2}{\sqrt{1-x^2}}-\frac{4x}{2x\sqrt{1-x^2}}$ $f'(x) = \frac{2}{\sqrt{1-x^2}}-\frac{2}{\sqrt{1-x^2}}$ $f'(x) = 0$ Therefore, $f(x) = C$, where $C$ is some constant. Consider $f(0)$: $f(0) = 2sin^{-1}~0-cos^{-1}(1-2(0)^2)$ $f(0) = 2(0)-cos^{-1}(1)$ $f(0) = 0-0$ $f(0) = 0$ Thus $C = 0$ Then: $f(x) = 2sin^{-1}~x-cos^{-1}(1-2x^2) = 0$ $2sin^{-1}~x = cos^{-1}(1-2x^2)~~~~$ $x \geq 0$
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