Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 3 - Derivatives - 3.4 The Product and Quotient Rules - 3.4 Exercises: 33


The equation of the tangent line at the given point is $y=-\dfrac{3}{2}x+\dfrac{17}{2}$

Work Step by Step

$y=\dfrac{x+5}{x-1}$ $;$ $a=3$ First, evaluate the derivative of the given expression using the quotient rule: $y'=\dfrac{(x-1)(x+5)'-(x+5)(x-1)'}{(x-1)^{2}}=...$ $...=\dfrac{(x-1)(1)-(x+5)(1)}{(x-1)^{2}}=\dfrac{x-1-x-5}{(x-1)^{2}}=-\dfrac{6}{(x-1)^{2}}$ Substitute $x$ by $a=3$ in the derivative found to obtain the slope of the tangent line at the given point: $m_{\text{tan}}=-\dfrac{6}{(3-1)^{2}}=-\dfrac{6}{2^{2}}=-\dfrac{6}{4}=-\dfrac{3}{2}$ Substitute $x$ by $a=3$ in the original expression to obtain the $y$-coordinate of the point given: $y=\dfrac{3+5}{3-1}=\dfrac{8}{2}=4$ The point is $(3,4)$ The slope of the tangent line and a point through which it passes are known. Use the point-slope form of the equation of a line, which is $y-y_{1}=m(x-x_{1})$ to obtain the equation of the tangent line at the given point: $y-4=-\dfrac{3}{2}(x-3)$ $y-4=-\dfrac{3}{2}x+\dfrac{9}{2}$ $y=-\dfrac{3}{2}x+\dfrac{9}{2}+4$ $y=-\dfrac{3}{2}x+\dfrac{17}{2}$ The graph of both the function and the tangent line are shown the answer section.
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