Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1 - Page 108: 27


$y=-x-1$ and $y=-x+3$

Work Step by Step

Given the function $f(x)=\frac{1}{x-1}$, we are asked find the equation of all the straight lines with a slope of $-1$. Since we are asked to find lines with a specific slope, the best start would be to find the derivative function to go about finding which x values have a slope of $-1$. $f(x)={(x-1)}^{-1}$ $f'(x)=(-1)(x-1)^{-2}=\frac{-1}{(x-1)^2}$ We equate $f'(x)$ to $-1$ $\frac{-1}{(x-1)^2}=-1$ $(x-1)^2=1$ $x^2-2x+1=1$ $x(x-2)=0$ $x=0$ or $x=2$ It turns out that the function has two points at which the tangent line would have a slope of $-1$. One of the coordinates with a slope of $-1$ is $(0,-1)$ (the y-value was found using $f(0)$). As such, the coordinate also readily gives up information of the y intercep. Therefore, the equation of the first tangent is $y=-x-1$ The second coordinate can be found using $f(2)=1$ which corresponds to a point $(2,1)$ The y intercept for this tangent is found as per the following: $\frac{1-c}{2-0}=-1$ $c=3$ In the end, the equation of the second tangent is $y=-x+3$
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