Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 3 - The Derivative - 3.4 Definition of the Derivative - 3.4 Exercises - Page 176: 33

Answer

$$ f(x)=\sqrt {x} $$ Now use the rules for limits to get $$ \begin{split} f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\ &= \lim\limits_{h \to 0}\frac{\sqrt {x+h} -\sqrt {x}}{h}\\ &=\lim\limits_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ &=\lim\limits_{h \to 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} \\ &=\lim\limits_{h \to 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x}) } \\ &=\lim\limits_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}\\ &=\frac{1}{2\sqrt {x}} \end{split} $$ Use this result to find $$ f^{\prime}(2)=\frac{1}{2\sqrt {2}} $$ $$ f^{\prime}(16)=\frac{1}{2\sqrt {16}} =\frac{1}{8} $$ and $$ f^{\prime}(-3)=\frac{1}{2\sqrt {-3}} \\ \quad\quad\quad\quad[\text { is not a real number , hence }] $$ $f^{\prime}(-3)$ does not exist.
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Work Step by Step

$$ f(x)=\sqrt {x} $$ Go through the four steps to find $f^{\prime}(x)$ Step 1 Find $f(x+h)$ by replacing $x$ with $x +h$ $$ f(x +h)=\sqrt {x+h} $$ Step 2 $$ f(x +h) -f(x)=\sqrt {x+h} -\sqrt {x} $$ Step 3 $$ \frac{f(x +h)- f(x)}{h}= \frac{\sqrt {x+h} -\sqrt {x}}{h} $$ Step 4 Now use the rules for limits to get $$ \begin{split} f^{\prime}(x) &=\lim\limits_{h \to 0} \frac{f(x +h)- f(x)}{h}\\ &= \lim\limits_{h \to 0}\frac{\sqrt {x+h} -\sqrt {x}}{h}\\ &=\lim\limits_{h \to 0}\frac{\sqrt{x+h}-\sqrt{x}}{h} \cdot \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \\ &=\lim\limits_{h \to 0} \frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})} \\ &=\lim\limits_{h \to 0} \frac{h}{h(\sqrt{x+h}+\sqrt{x}) } \\ &=\lim\limits_{h \to 0} \frac{1}{\sqrt{x+h}+\sqrt{x}}\\ &=\frac{1}{2\sqrt {x}} \end{split} $$ Use this result to find $$ f^{\prime}(2)=\frac{1}{2\sqrt {2}} $$ $$ f^{\prime}(16)=\frac{1}{2\sqrt {16}} =\frac{1}{8} $$ and $$ f^{\prime}(-3)=\frac{1}{2\sqrt {-3}} \\ \quad\quad\quad\quad[\text { is not a real number , hence }] $$ $f^{\prime}(-3)$ does not exist.
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