Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 2 - Section 2.7 - Derivatives and Rates of Change - 2.7 Exercises: 3

Answer

(i) Using the first definition $m = \lim\limits_{x \to 1} = \frac{f(x) - f(1)}{x-1} = 2$ (ii) Using the second definition $m = \lim\limits_{h \to 0} = \frac{f(1+h) - f(1)}{h} = 2$ (b) Tangent line $y = 2x+1$

Work Step by Step

(i) Using the first definition $m = \lim\limits_{x \to 1} \frac{f(x) - f(1)}{x-1}$ $m = \lim\limits_{x \to 1} \frac{(4x-x^{2}) - (4(1)-1^{2})}{x-1}$ $m = \lim\limits_{x \to 1} \frac{-x^{2} +4x - 3}{x-1}$ $m = \lim\limits_{x \to 1} \frac{(x-1)(3-x)}{x-1}$ Cancel out $(x-1)$ $ m = \lim\limits_{x \to 1} (3 -x)$ $ m = (3 -1)$ $ m = 2$ (ii) Using the second definition $m = \lim\limits_{h \to 0} \frac{f(1+h) - f(1)}{h}$ $m = \lim\limits_{h \to 0} \frac{[4(1+h) - (1+h)^{2}] - [4(1) - 1^{2}]}{h}$ $m = \lim\limits_{h \to 0} \frac{2h - h^{2}}{h}$ Simplify: $m = \lim\limits_{h \to 0} \frac{h(2- h)}{h}$ $m = \lim\limits_{h \to 0} 2 - h$ $m = \lim\limits_{h \to 0} 2 - 0$ $m = \lim\limits_{h \to 0} 2= 2$ (b) Tangent Line $ m = \frac{y-3}{x-1} $ $m = 2$ $2 = \frac{y-3}{x-1} $ $2(x-1) = \frac{y-3}{x-1} (x-1)$ $2x - 2 = y-3$ $ y = 2x -2+3$ $ y = 2x + 1$
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