Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 2 - Differentiation - 2.2 Exercises - Page 116: 95

Answer

Average rate of change for interval [1,2]: $\frac{1}{2}$ Instantaneous rate of change for endpoint 1: $1$ Instantaneous rate of change for endpoint 2: $\frac{1}{4}$

Work Step by Step

Average rate of change is given by the formula: $$\frac{f(x_2) - f(x_1)}{x_2 - x_1}$$ for points $(x_1, f(x_1))$ and $(x_2, f(x_2))$. In this case, for endpoints $[1,2]$: $$f(x) = -\frac{1}{x}$$ $$f(1) =-\frac{1}{(1)} = -1$$ $$f(2) = -\frac{1}{(2)} = -\frac{1}{2}$$ giving an average rate of change: $$AverageRate = \frac{-0.5 - (-1)}{2 - 1} = \frac{1}{2}$$ The instantaneous rate of change is given by the first derivative of the function: $$f(x) = -\frac{1}{x} = x^{-1}$$ $$f'(x) = x^{-1-1} =x^{-2} = \frac{1}{x^{2}}$$ which, for the endpoints of the interval [1,2] are as follows: $$f'(2) = \frac{1}{(2)^2} = \frac{1}{4}$$ $$f'(1) = \frac{1}{(1)^{2}} = 1$$
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