## Calculus 10th Edition

Published by Brooks Cole

# Chapter 2 - Differentiation - 2.1 Exercises - Page 103: 37

#### Answer

$y=-\frac{1}{2}x +\frac{3}{2}$

#### Work Step by Step

Begin by taking the derivative of $f(x)=\frac{1}{\sqrt x}$ $f'(x)= -\frac{1}{2\sqrt x^3}$ Find the equation of the given line $x+2y-6=0$ $y=-\frac{1}{2}x+3$ Set the slope of the line equal to the tangent line slope $-\frac{1}{2\sqrt x^3} = -\frac{1}{2}$ Solve for x $\frac{1}{\sqrt x^3} =1$, $\sqrt[3] 1 = \sqrt[3] (\sqrt x^3)$, $1=\sqrt x$ $1=x$ Plugging x back into the function to find a point $f(1)=1$, $(1,1)$ Use point-slope to solve $y-1=-\frac{1}{2}(x-1)$ $y=-\frac{1}{2}x + \frac{3}{2}$ This yields a line tangent to the graph and parallel to the given line.

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