Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 2: Limits and Continuity - Section 2.1 - Rates of Change and Tangents to Curves - Exercises 2.1 - Page 46: 11


a. $12$ b. $ y=12x-16$

Work Step by Step

Given $ y=x^3, P(2,8)$ a. slope=$\frac{(2+h)^3-2^3}{h}=\frac{3\times4h+6h^2+h^3}{h}=12+6h+h^2=12$ (let $h$ be zero) b. tangent line: $ y-8=12(x-2)$, which is $ y=12x-16$
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