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: 4

Answer

a. (i) = $-2$; (ii) = $-2$ b. $y = -2x +2$ c. Graph; this graph is where the line and the curve coincide.
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Work Step by Step

Definition 1: $m = \lim\limits_{x \to a} \frac{f(x) - f(a)}{x-a}$ Since we are asked to evaluate the slope at $(1,0)$, so $a = 1$. $m = \lim\limits_{x \to 1} \frac{f(x) - f(1)}{x-1}$ Now change $f(x)$ to $(x-x^{3})$. $m = \lim\limits_{x \to 1} \frac{x-x^{3} - (1-1^{3})}{x-1}$ $m = \lim\limits_{x \to 1} \frac{x-x^{3} - 0}{x-1}$ $m = \lim\limits_{x \to 1} \frac{x-x^{3}}{x-1}$ Simplify: $m = \lim\limits_{x \to 1} \frac{x(1-x^{2})}{x-1}$ Expand: $m = \lim\limits_{x \to 1} \frac{x(1-x)(1+x)}{x-1}$ Now multiply the numerator by $-1$. $m = \lim\limits_{x \to 1} \frac{-x(x-1)(1+x)}{x-1}$ Cancel out: $x-1$. $m = \lim\limits_{x \to 1} -x(1+x)$ Now replace $x$ to $1$. $m = -1(1+1) = -2$ (ii) Using Equation 2 $m = \lim\limits_{h \to 0} \frac{f(a+h)-f(a)}{h}$ $m = \lim\limits_{h \to 0} \frac{((a+h)-(a+h)^{3} - (a-a^{3})}{h}$ Now expand $(a+h)^{3}$ $m = \lim\limits_{h \to 0} \frac{(a+h)-(a^{3} +3a^{2}h + 3ah^{2} + h^{3}) - (a-a^{3})}{h}$ Now replace $a$ with $1$. $m = \lim\limits_{h \to 0} \frac{(1+h)-(1^{3} +3(1)^{2}h + 3(1)h^{2} + h^{3}) - (1-1^{3})}{h}$ $m = \lim\limits_{h \to 0} \frac{-h^{3} - 3h^{2} - 2h}{h}$ Simplify: $m = \lim\limits_{h \to 0} \frac{h(-h^{2} - 3h - 2)}{h}$ Cancel $h$. $m = \lim\limits_{h \to 0} -h^{2} - 3h -2$ Replace $h$ for $0$. $m = \lim\limits_{h \to 0} -0^{2} - 3(0) -2 = -2$ b. Find the tangent line Given: $m = -2$, $x = 1$, $y = 0$ Formula: $y = mx + b$ $0 = -2(1) + b$ $b = 2$ Tangent line: $y = -2x + 2$ c. This graph is the normal graph of the function and the tangent line.
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