Answer
a. $-9$
b. $ y=-9x-2$
Work Step by Step
Given $ y=x^3-12x, P(1,-11)$
a. slope=$\frac{(1+h)^3-12(1+h)-1^3+12}{h}=\frac{3h+3h^2+h^3-12h}{h}=-9+3h+h^2=-9$ (let $h$ be zero)
b. tangent line: $ y+11=-9(x-1)$, which gives $ y=-9x-2$
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: 13
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