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: 35

Answer

The equation of tangent line to the curve at $(2,3)$ is $$(l):y=\frac{1}{2}x+2$$

Work Step by Step

$$y=x+\frac{2}{x}$$$$y=x+2x^{-1}$$ 1) Find the derivative of $y$ $$y'=\frac{d}{dx}(x)+2\frac{d}{dx}(x^{-1})$$ $$y'=1+2\times(-1)x^{-2}$$ $$y'=1-2x^{-2}$$ $$y'=1-\frac{2}{x^2}$$ 2) Find the slope of the tangent line to the curve at $(2,3)$, or in fact, $y'(2)$ $$y'(2)=1-\frac{2}{2^2}$$$$y'(2)=1-\frac{1}{2}=\frac{1}{2}$$ 3) The tangent line to the curve at $(2,3)$ would have the following equation: $$(l):y=y'(2)x+b$$$$y=\frac{1}{2}x+b$$ Since $(2,3)$ lies in the tangent line $(l)$, we can use that point to find $b$. In detail, $$\frac{1}{2}\times2+b=3$$$$1+b=3$$$$b=2$$ In conclusion, the equation of tangent line to the curve at $(2,3)$ is $$(l):y=\frac{1}{2}x+2$$
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