Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.1 Exercises - Page 103: 38



Work Step by Step

The line has an equation: $x+2y+7=0$ $2y=-x-7$ $y=-\frac{1}{2}x-\frac{7}{2}$ The slope of the line is $-\frac{1}{2}$. For a line to be parallel that is tangent to $f$, the tangent should also have a slope of $-\frac{1}{2}$. $f'(x)=-\frac{1}{2}$ $f(x)=\frac{1}{\sqrt{x-1}}$ $f(x)=(x-1)^{-\frac{1}{2}}$ $f'(x)=(-\frac{1}{2})(x-1)^{(-\frac{1}{2}-1)}(1)$ $f'(x)=-\frac{1}{2}(x-1)^{-\frac{3}{2}}$ $f'(x)=-\frac{1}{2(\sqrt{x-1})^3}$ $f'(x)=-\frac{1}{2}$ $-\frac{1}{2}=-\frac{1}{2(\sqrt{x-1})^3}$ $(\sqrt{x-1})^3=1$ $x=2$ $f(2)=\frac{1}{\sqrt{(2)-1}}$ $f(2)=1$ The line that is tangent to $f$ and is parallel to $x+2y+7=0$ has a slope of $-\frac{1}{2}$ and passes through the point $(2,1)$. $y=mx+c$ $(1)=(-\frac{1}{2})(2)+c$ $c=2$ Therefore, the line has the equation, $y=-\frac{1}{2}x+2$ $2y=-x+4$ $x+2y-4=0$
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