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: 91

Answer

The soluton is $$\frac{f(x+h)-f(x)}{h}=3\frac{1}{\sqrt{x(x+h)}(\sqrt{x+h}+\sqrt{x})}. $$

Work Step by Step

The expression becomes $$\frac{f(x+h)-f(x)}{h}=\frac{-\frac{3}{\sqrt{x+h}}-\left(-\frac{3}{\sqrt{x}}\right)}{h}=3\frac{\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{x+h}}}{h}=3\frac{\frac{\sqrt{x+h}-\sqrt{x}}{\sqrt{x(x+h)}}}{h}=3\frac{\sqrt{x+h}-\sqrt{x}}{h\sqrt{x(x+h)}}=3\frac{\sqrt{x+h}-\sqrt{x}}{h\sqrt{x(x+h)}}\cdot\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}=3\frac{\sqrt{x+h}^2-\sqrt{x}^2}{h\sqrt{x(x+h)}(\sqrt{x+h}+\sqrt{x})}=3\frac{x+h-x}{h\sqrt{x(x+h)}(\sqrt{x+h}+\sqrt{x})}=3\frac{h}{h\sqrt{x(x+h)}(\sqrt{x+h}+\sqrt{x})}=3\frac{1}{\sqrt{x(x+h)}(\sqrt{x+h}+\sqrt{x})}. $$
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