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 - Page 10: 32



Work Step by Step

We know that $f(x)=x^2−4$, so: $\frac{f(2+h)-f(2)}{h}=\frac{[(2+h)^2-4]-[2^2-4]}{h}=\frac{4+4h+h^2-4-0}{h}=4+h$
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