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 - Page 180: 39

Answer

$y = 3x - 1$
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Work Step by Step

In order to find the equation of a tangent line, first you need to take the derivative. For this equation, just use the Power Rule. $y=3x^{2}-x^{3}$ Power Rule: $\frac{dy}{dx} = nx^{n-1}$ $\frac{dy}{dx} = 3(2)x-(3)x^{2}$ $\frac{dy}{dx}=6x-3x^{2}$ The derivative is your equation of the slope. To find the slope at the point (1,2), plug in x=1 into the derivative equation: $\frac{dy}{dx} \Bigr|_{\substack{x=1}}=6(1)-3(1)^{2} = 3.$ Now to get the equation of the tangent line, use point-slope form: $y-y1 = m(x-x1)$ $y-(2) = (3)(x-(1))$ $y = 3x-3-2$ $y = 3x - 1$
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