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: 34

Answer

The equation of the tangent line to the curve at $(0,2)$ is $$(l):y=3x+2$$

Work Step by Step

$$y=2e^x+x$$ 1) First, find the derivative of y $$y'=2\frac{d}{dx}(e^x)+\frac{d}{dx}x$$ $$y'=2e^x+1$$ 2) Find the slope of the tangent line to the curve at the point $(0,2)$, or in fact, $y′(0)$: $$y′(0)=2(e^0)+1=2\times1+1=3$$ 3) The tangent line to the curve at the point $(0,2)$ would have the following equation: $$(l):y=y′(0)x+b$$ $$(l):y=3x+b$$ Since the point $(0,2)$ also lies in the tangent line $(l)$, we can apply that point to the equation of $(l)$ to find $b$. In detail, $$3\times0+b=2$$$$b=2$$ Therefore, the equation of the tangent line to the curve at $(0,2)$ is $$(l):y=3x+2$$
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