Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.1 - Derivatives of Polynomials and Exponential Functions - 3.1 Exercises: 58

Answer

The equation of the tangent line that needs to be found is $$(l_0):y=32x-47$$

Work Step by Step

$$y=x^4+1$$ 1) First, find the derivative $y$, or $y'$ $$y'=\frac{d}{dx}(x^4)+\frac{d}{dx}(1)$$ $$y'=4x^3$$ 2) The slope of the tangent line $(l)$ to the curve $y$ at point $A(a,b)$ would be $y'(a)$. Consider the mentioned line $(d)$: $$(d): 32x-y=15$$ $$(d):y=32x-15$$ The slope of line $(d)$ equals $32$. If we want the tangent line $(l)$ to be parallel with line $(d)$, both of their slopes must be equal. In other words, $$y'(a)=32$$ $$4a^3=32$$ $$a^3=8$$ $$a=2$$ So, $b=2^4+1=17$ Therefore, the tangent line $(l_0)$ at point $A(2,17)$ is parallel with line $(d)$. The equation of tangent line $(l_0)$ is $$(l_0):(y-b)=y'(a)(x-a)$$ $$(l_0):y-17=32(x-2)$$ $$(l_0):y-17=32x-64$$ $$(l_0):y=32x-47$$
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