Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 13 - Section 13.3 - Tangent Lines and Derivatives - 13.3 Exercises - Page 922: 17

Answer

$y=\dfrac{1}{4}x+\dfrac{7}{4}$ The graph of the function and the tangent line is shown below:

Work Step by Step

$y=\sqrt{x+3},$ at $(1,2)$ The slope of the tangent line at the point $P(a,f(a))$ is given by $m=\lim_{h\to0}\dfrac{f(a+h)-f(a)}{h}$ In this case, $a=1$ Find $f(a+h)$ by substituting $x$ by $1+h$ in $f(x)$ and simplifying: $f(1+h)=\sqrt{1+h+3}=\sqrt{4+h}$ Since $f(a)$ is the $y$-coordinate of the point given, $f(a)=2$ Substitute the known values into the formula that gives the slope of the tangent line and evaluate: $m=\lim_{h\to0}\dfrac{\sqrt{4+h}-2}{h}\cdot\dfrac{\sqrt{4+h}+2}{\sqrt{4+h}+2}=...$ $...=\lim_{h\to0}\dfrac{(\sqrt{4+h})^{2}-2^{2}}{h(\sqrt{4+h}+2)}=\lim_{h\to0}\dfrac{4+h-4}{h(\sqrt{4+h}+2)}=...$ $...=\lim_{h\to0}\dfrac{h}{h(\sqrt{4+h}+2)}=\lim_{h\to0}\dfrac{1}{\sqrt{4+h}+2}=...$ $...=\dfrac{1}{\sqrt{4+0}+2}=\dfrac{1}{2+2}=\dfrac{1}{4}$ The point-slope form of the equation of a line is $y-y_{0}=m(x-x_{0})$. Both the slope of the line and a point through which it passes are known. Substitute them into the formula to obtain the equation of the tangent line: $y-2=\dfrac{1}{4}(x-1)$ $y-2=\dfrac{1}{4}x-\dfrac{1}{4}$ $y=\dfrac{1}{4}x-\dfrac{1}{4}+2$ $y=\dfrac{1}{4}x+\dfrac{7}{4}$
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