## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# 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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