Calculus: Early Transcendentals 8th Edition

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

Diagnostic Tests - C Diagnostic Tests: Functions - Page xxix: 2

Answer

$12+6h+h^2$

Work Step by Step

$f(x) = x^3$ We can evaluate the difference quotient: $\frac{f(2+h)- f(2)}{h}$ $= \frac{(2+h)^3- (2)^3}{h}$ $= \frac{(2+h)(2+h)(2+h)- 8}{h}$ $= \frac{(4+4h+h^2)(2+h)- 8}{h}$ $= \frac{8+4h+8h+4h^2+2h^2+h^3- 8}{h}$ $= \frac{12h+6h^2+h^3}{h}$ $= \frac{(h)(12+6h+h^2)}{h}$ $= 12+6h+h^2$
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