Precalculus (10th Edition)

Published by Pearson
ISBN 10: 0-32197-907-9
ISBN 13: 978-0-32197-907-0

Chapter 14 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Chapter Review - Review Exercises - Page 907: 33

Answer

$-24$

Work Step by Step

Factor each polynomial: $$f'(3)\\=\lim_{x\to 3}\frac{f(x)-f(3)}{x-3}\\=\lim_{x\to 3}\frac{-4x^2+5-(-4(3)^2+5))}{x-3}\\=\lim_{x\to 3}\frac{-4x^2+36}{x-3}\\=\lim_{x\to 3}\frac{-4(x^2-9)}{x-3}\\=\lim_{x\to 3}\frac{-4(x+3)(x-3)}{(x-3)}.$$ Cancel the common factors: $$\lim_{x\to 3}-4(x+3)\\=-4(3+3)=-24$$
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