Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.2 - The Product and Quotient Rules - 3.2 Exercises - Page 189: 53


There are two tangent lines. The tangent lines touch the curve at these points: $(-2-\sqrt{3},\frac{1+\sqrt{3}}{2})$ $(-2+\sqrt{3},\frac{1-\sqrt{3}}{2})$

Work Step by Step

$y = \frac{x}{x+1}$ $\frac{dy}{dx} = \frac{(x+1)-x}{(x+1)^2} = \frac{1}{(x+1)^2}$ Note that $\frac{dy}{dx}$ is the slope of the graph at each point $x$ If a line through the point $(1,2)$ is tangent to the curve at $x$, then the slope of the line is $~~m = \frac{1}{(x+1)^2}$ The equation of the line is: $y-2 = \frac{1}{(x+1)^2}(x-1)$ $y = \frac{1}{(x+1)^2}(x-1)+2$ We can find the points of intersection: $\frac{x}{x+1} = \frac{1}{(x+1)^2}(x-1)+2$ $\frac{x}{x+1}-2 = \frac{x-1}{(x+1)^2}$ $\frac{x}{x+1}-\frac{2x+2}{x+1} = \frac{x-1}{(x+1)^2}$ $\frac{-x-2}{x+1} = \frac{x-1}{(x+1)^2}$ $(-x-2)(x+1) = x-1$ $-x^2-3x-2 = x-1$ $x^2+4x+1 = 0$ Using the quadratic formula: $x = \frac{-4\pm\sqrt{4^2-4(1)(1)}}{2(1)}$ $x = \frac{-4\pm\sqrt{12}}{2}$ $x = -2\pm\sqrt{3}$ There are two solutions. We can find $y$ when $x = -2-\sqrt{3}$: $y = \frac{x}{x+1}$ $y = \frac{-2-\sqrt{3}}{-2-\sqrt{3}+1}$ $y = \frac{-2-\sqrt{3}}{-1-\sqrt{3}}\cdot \frac{-1+\sqrt{3}}{-1+\sqrt{3}}$ $y = \frac{-1-\sqrt{3}}{-2}$ $y = \frac{1+\sqrt{3}}{2}$ We can find $y$ when $x = -2+\sqrt{3}$: $y = \frac{x}{x+1}$ $y = \frac{-2+\sqrt{3}}{-2+\sqrt{3}+1}$ $y = \frac{-2+\sqrt{3}}{-1+\sqrt{3}}\cdot \frac{-1-\sqrt{3}}{-1-\sqrt{3}}$ $y = \frac{-1+\sqrt{3}}{-2}$ $y = \frac{1-\sqrt{3}}{2}$ The tangent lines touch the curve at these points: $(-2-\sqrt{3},\frac{1+\sqrt{3}}{2})$ $(-2+\sqrt{3},\frac{1-\sqrt{3}}{2})$
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