Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - Exercises - Page 157: 44

Answer

$f(a) = 3a^2+2 $ $f(a +h)=3a^2+6ah+3h^2 +2 $ $\frac{f(a+h)-f(a)}{h}= 6a+3h$

Work Step by Step

$f(a) =3a^2+2 $ $f(a +h)=3(a+h)^2+2$ ... simplify $=3(a^2+2ah+h^2)+2$ $=3a^2+6ah+3h^2 +2 $ $\frac{f(a+h)-f(a)}{h}=\frac{(3a^2+6ah+3h^2+2)-(3a^2+2)}{h}$ ... simplify $=\frac{3a^2+6ah+3h^2+2-3a^2-2}{h}$ $=\frac{6ah+3h^2}{h}$ $=6a+3h$
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