Calculus: Early Transcendentals 8th Edition

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

Chapter 2 - Section 2.8 - The Derivative as a Function - 2.8 Exercises - Page 162: 19

Answer

(a) $f'(0) = 0$ $f'(\frac{1}{2}) = 1$ $f'(1) = 2$ $f'(2) = 4$ (b) $f'(-\frac{1}{2}) = -1$ $f'(-1) = -2$ $f'(-2) = -4$ (c) $f'(x) =2x$ (d) $f'(x) = 2x$
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Work Step by Step

(a) $f(x) = x^2$ Using the graph, we can estimate the slope at the following points: $f'(0) = 0$ $f'(\frac{1}{2}) = 1$ $f'(1) = 2$ $f'(2) = 4$ (b) We can see that the graph is even. Then $f(-x) = f(x)$ and $f'(-x) = -f'(x)$ Using the graph and symmetry, we can estimate the slope at the following points: $f'(-\frac{1}{2}) = -1$ $f'(-1) = -2$ $f'(-2) = -4$ (c) We could guess that $f'(x) = 2x$ (d) We can find an expression for $f'(x)$: $f'(x) = \lim\limits_{h \to 0}\frac{(x+h)^2-x^2}{h}$ $f'(x) = \lim\limits_{h \to 0}\frac{(x^2+2xh+h^2)-x^2}{h}$ $f'(x) = \lim\limits_{h \to 0}\frac{2xh+h^2}{h}$ $f'(x) = \lim\limits_{h \to 0}(2x+h)$ $f'(x) = 2x$
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