Calculus: Early Transcendentals (2nd Edition)

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

Chapter 3 - Derivatives - 3.1 Introducing the Derivative - 3.1 Execises: 58

Answer

This expression represents the slope of the tangent of the function $f(x)=\sqrt{x}$ at $a=2$ and its value is $m_{tan}=\frac{1}{2\sqrt{2}}$.

Work Step by Step

The definition of the slope of the tangent at $a$ is given by $$m_{tan}=\lim_{h\to0}\frac{f(a+h)-f(a)}{h}.$$ We see that if we take $f(x)=\sqrt{x}$ and $a=2$ (and also we calculate $f(a)=f(2)=\sqrt{2}$) the upper expression becomes the expression for the slope of $f(x)=\sqrt{x}$ at $a=2$. Let us calculate this slope: $$m_{tan}=\lim_{h\to0}\frac{\sqrt{2+h}-\sqrt{2}}{h}=\lim_{h\to0}\frac{\sqrt{2+h}-\sqrt{2}}{h}\cdot\frac{\sqrt{2+h}+\sqrt{2}}{\sqrt{2+h}+\sqrt{2}}=\lim_{h\to0}\frac{\sqrt{2+h}^2-\sqrt{2}^2}{h(\sqrt{2+h}+\sqrt{2})}=\lim_{h\to0}\frac{2+h-2}{h(\sqrt{2+h}+\sqrt{2})}=\lim_{h\to0}\frac{h}{h(\sqrt{2+h}+\sqrt{2})}=\lim_{h\to0}\frac{1}{\sqrt{2+h}+\sqrt{2}}=\frac{1}{\sqrt{2+0}+\sqrt{2}}=\frac{1}{2\sqrt{2}}.$$
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