Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 2 - Derivatives - 2.3 Differentiation Formulas - 2.3 Exercises - Page 142: 81


The equations of tangent are:$$3x-y-3=0$$ $$3x-y-7=0$$

Work Step by Step

Given curve, $y=x^3-3x^2+3x-3$. The tangent line to this curve is parallel to $3x−y=15$, which means that the slope of the tangent line is $3$ (the slope of the line it is parallel to). Differentiating y with respect to x, $$\frac{dy}{dx}=3x^2-6x+3$$This slope must be equal to $3$, that is, the slope of the required tangent. $$3x^2-6x+3=3\implies3x^2-6x=0$$$$x=0,2$$Thus, the tangent touches the curve y at $(0,y(0))≡(0,-3)$ & $(2,y(2))≡(2,-1)$. Thus, there will be two tangents. For first point,$$y+3=3(x-0)\implies3x-y-3=0.$$For second point, $$y+1=3(x-2)=3x-6$$$$3x-y-7=0.$$
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