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: 8

Answer

(a). $-4$ (b). $y=-4x+11$

Work Step by Step

(a). Given $f(x)=y=7-x^2, P(2,3)$, slope=$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to0}\frac{7-(2+h)^2-(7-2^2)}{h}=\lim_{h\to0}\frac{-4h-h^2}{h}==\lim_{h\to0}(-4-h)=-4$ (b). With the kown slope and point P, the tangent line is given by: $y-3=-4(x-2)$ which gives $y=-4x+11$
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