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: 37

Answer

The equation of the tangent line: $$(l):y=2x+2$$ The equation of the normal line: $$(m):y=-\frac{1}{2}x+2$$

Work Step by Step

$$y=f(x)=x^4+2e^x$$ 1) Find the derivative of $y$$$y'=\frac{d}{dx}(x^4)+2\frac{d}{dx}(e^x)$$ $$y'=4x^3+2e^x$$ 2) Find $y'(0)$ $$y'(0)=4\times0^3+2e^0$$$$y'(0)=0+2\times1=2$$ 3) $y'(0)$ is the slope of the tangent line $(l)$ to the curve $f(x)$ at point $(0,2)$ Knowing the slope of the tangent line $(l)$ $y'(0)$ and 1 point of (l) $(0,2)$, the equation of the tangent line $(l)$ would be: $$(l):(y-2)=y'(0)(x-0)$$ $$(l):y-2=2x$$ $$(l):y=2x+2$$ 4) The normal line $(m)$ at $(0,2)$ is the line perpendicular to the tangent line $(l)$ at $(0,2)$. So, the slope $a$ of $(m)$ is: $$a=\frac{-1}{y'(0)}=\frac{-1}{2}$$ (the product of the slopes of 2 perpendicular lines equals $-1$) Knowing the slope of $(m)$ and 1 point of $(m)$ $(0,2)$, the equation of the normal line $(m)$ would be $$(m):(y-2)=a(x-0)$$ $$(m):y-2=-\frac{1}{2}x$$ $$(m):y=-\frac{1}{2}x+2$$
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