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: 80


Equation of tangent: $$32x-y-47=0$$

Work Step by Step

Given curve, $y=x^4+1$. The tangent line to this curve is parallel to $32x-y=15$, which means that the slope of the tangent line is $32$ (the slope of the line it is parallel to). Differentiating $y$ with respect to $x$, $$\frac{dy}{dx} = 4x^3$$This slope must be equal to $32$, that is, the slope of the required tangent. $$4x^3=32\implies x^3=8\implies x=2.$$Thus, the tangent touches the curve $y$ at $(2, y(2))\equiv(2,17)$. Thus, using point-slope form, the equation of the tangent is, $$y-17=32(x-2)=32x-64$$$$32x-y-47=0$$
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