Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.1 Review of Functions - 1.1 Exercises: 90

Answer

The solution is $$\frac{f(x+h)-f(x)}{h}=\frac{-2}{(\sqrt{1-2x-2h}+\sqrt{1-2x})}$$

Work Step by Step

The expression becomes $$\frac{f(x+h)-f(x)}{h}=\frac{\sqrt{1-2(x+h)}-\sqrt{1-2x}}{h}=\frac{\sqrt{1-2x-2h}-\sqrt{1-2x}}{h}=\frac{\sqrt{1-2x-2h}-\sqrt{1-2x}}{h}\cdot\frac{\sqrt{1-2x-2h}+\sqrt{1-2x}}{\sqrt{1-2x-2h}+\sqrt{1-2x}}=\frac{\sqrt{1-2x-2h}^2-\sqrt{1-2x}^2}{h(\sqrt{1-2x-2h}+\sqrt{1-2x})}=\frac{1-2x-2h-1+2x}{h(\sqrt{1-2x-2h}+\sqrt{1-2x})}=\frac{-2h}{h(\sqrt{1-2x-2h}+\sqrt{1-2x})}=\frac{-2}{\sqrt{1-2x-2h}+\sqrt{1-2x}}.$$
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