## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises: 54

#### Answer

$y = \frac{1}{2}x + \frac{7}{2}$ $y = \frac{1}{2}x - \frac{1}{2}$

#### Work Step by Step

Solve for $y$: $x - 2y = 2$ $-2y = 2 -x$ $y = \frac{1}{2}x -1$ The slope of the line is $m = \frac{1}{2}$ Then we find the derivative to the curve: $y = \frac{x-1}{x+1}$ $y' = \frac{\frac{d(x-1)}{dx}(x+1) - (x-1)\frac{d(x+1)}{dx}}{(x+1)^{2}}$ $y' = \frac{(1)(x+1) - (x-1)(1)}{(x+1)^{2}}$ $y' = \frac{(x+1) - (x-1)}{(x+1)^{2}}$ $y' = \frac{x+1 - x+1}{(x+1)^{2}}$ $y' = \frac{2}{(x+1)^{2}}$ Now we use $y'$ which is the slope of a tangent line to find where $y' = \frac{1}{2}$ $\frac{1}{2} = \frac{2}{(x+1)^{2}}$ Now multiply: $4 = (x + 1)^{2}$ Now rationalize: $\sqrt 4 = \sqrt (x + 1)^{2}$ $\pm{2} = x+1$ $x = 2 - 1 = 1$ $x = -2 -1 = -3$ Now we calculate the coordinates using the two points we found: $y = \frac{x-1}{x+1}$ $x = -3$ $y = \frac{-3-1}{-3+1}$ $y = \frac{-4}{-2}$ $y = 2$; this becomes $(-3,2)$ $x = 1$ $y = \frac{1-1}{1+1}$ $y = \frac{0}{2}$ $y = 0$; this becomes $(1,0)$ Now we use the line equation to find the tangent line: $y = m(x - x_{1})+y_{1}$ $(-3,2)$: $y = \frac{1}{2}(x -(-3))+2$ $y = \frac{1}{2}x + \frac{3}{2}+2$ $y = \frac{1}{2}x + \frac{7}{2}$ $(1,0)$: $y = \frac{1}{2}(x -1)+0$ $y = \frac{1}{2}x - \frac{1}{2}+0$ $y = \frac{1}{2}x - \frac{1}{2}$

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