## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 3: Derivatives - Section 3.1 - Tangents and the Derivative at a Point - Exercises 3.1 - Page 108: 13

#### Answer

$y=−2x+9$

#### Work Step by Step

The question asks us to find the equation of the tangent of the function $f(x)=\frac{x}{x-2}$ at the point (3,3) Given that we are looking for the equation of a tangent, the first reasonable step would be to find the gradient at the points given, which can be found using the derivative function. Using the power rule and quotient rule: $f′(x)=\frac{1}{x-2}=\frac{-x}{(x-2)^2}=\frac{-2}{(x-2)^2}$ $f′(3)=−2$ Thus, the gradient of that function at the value x=3 is -2. Using this information, we can find the intercept value of the tangent line. The equation of the gradient is given by: $\frac{y_{1}−y_{0}}{x_{1}−x_{0}}=m$ We can now use the points (3,3) and (0,c) (where c is the y-intercept). $\frac{3−c}{3-0}=-2$ Solving for c gives us c=9 Thus, the equation of the tangent at (3,3) is $y=−2x+9$

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