Calculus: Early Transcendentals (2nd Edition)

Published by Pearson
ISBN 10: 0321947347
ISBN 13: 978-0-32194-734-5

Chapter 1 - Functions - 1.1 Review of Functions - 1.1 Exercises: 57



Work Step by Step

$f(x) = x^2$ $\frac{f(x+h)-f(x)}{h}=\frac{(x+h)^2-x^2}{h} =\frac{x^2+2hx+h^2-x^2}{h} =\frac{2hx+h^2}{h} =\frac{h(2x+h)}{h}=\frac{(2x+h)}{1}=2x+h$
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