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: 94

Answer

Average rate of change for interval [3, 3.1]: $6.1$ Instantaneous rate of change for endpoint 3: $6$ Instantaneous rate of change for endpoint 3.1: $6.2$

Work Step by Step

Average rate of change is given by the formula $$\frac{f(t_2) - f(t_1)}{t_2 - t_1} $$ for $(t_1, f(t_1))$ and $(t_2, f(t_2))$. In this case, for endpoints [3, 3.1]: $$f(t) = t^{2} - 7$$ $$f(3) = 3^{2} - 7 = 2$$ $$f(3.1) = (3.1)^{2} - 7 = 2.61$$ Therefore, the average rate of change for the interval [3, 3.1] is given by: $$AverageRate = \frac{2.61 - 2}{3.1-3} = \frac{0.61}{0.1} = 6.1$$ The instantaneous rate of change is given by the first derivative of the function: $$f(t) = t^{2} - 7$$ $$f'(t) = 2t^{2-1} - 0$$ $$f'(t) = 2t$$ which for the endpoints of the interval [3, 3.1] yields: $$f'(3) = 2(3) = 6$$ $$f'(3.1) = 2(3.1) = 6.2$$
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