Answer
a. $-3$
b. $ y=-3x+4$
Work Step by Step
Given $ y=2-x^3, P(1,1)$
a. slope=$\frac{1^3-(1+h)^3}{h}=\frac{-3h-3h^2-h^3}{h}=-3-3h-h^2=-3$ (let $h$ be zero)
b. tangent line: $ y-1=-3(x-1)$, which is $ y=-3x+4$
Thomas' Calculus 13th Edition
by Thomas Jr., George B.
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 47: 12
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