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: 92

Answer

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

Work Step by Step

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