Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.1 Tangent Lines And Rates Of Change - Exercises Set 2.1 - Page 120: 15

Answer

$ (a) \quad m_{tan} = 2x_0. \\ (b) \quad m_{tan} = -2. $

Work Step by Step

Let us find the slope of the tangent line to the graph of $ f(x) = x^2 - 1 $ at a general point $x = x_0$. $$ \begin{align} m_{tan} &= \lim\limits_{h \to 0} \frac{f(x_0+h) - f(x_0)}{h} \\ &= \lim\limits_{h \to 0} \frac{(x_0+h)^2 - 1 - (x_0^2 - 1)}{h} \\ &= \lim\limits_{h \to 0} \frac{x_0^2 +2x_0h + h^2 - x_0^2}{h} \\ &= \lim\limits_{h \to 0} \frac{2x_0h + h^2}{h} \\ &= \lim\limits_{h \to 0} 2x_0 + h \\ &= 2x_0 \end{align}$$ Now, let us find the the slope of the tangent line to the graph of $f$ at the point $x_0 = -1$. $$ m_{tan} = 2x_0 = 2 \cdot (-1) = -2 $$
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