Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 5 - Cumulative Review Exercises - Page 709: 12


$ 2a+h+3$

Work Step by Step

Step 1. Given $f(x)=x^2+3x-1$, we have $f(a)=a^2+3a-1$ Step 2. We have $f(a+h)=(a+h)^2+3(a+h)-1=a^2+2ah+h^2+3a+3h-1$ Step 3. We have $f(a+h)-f(a)=2ah+h^2+3h$ Step 4. Thus, we have $\frac{f(a+h)-f(a)}{h}=2a+h+3$
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