Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

Published by Pearson
ISBN 10: 0-32193-104-1
ISBN 13: 978-0-32193-104-7

Chapter 13 - A Preview of Calculus: The Limit, Derivative, and Integral of a Function - Chapter Review - Review Exercises - Page 926: 33

Answer

$-24$

Work Step by Step

The general formula for the average rate of change from $x$ to $y$ can be written as: $\dfrac{f(y)-f(x)}{y-x}$. As $y\rightarrow x$, the rate of change becomes instantaneous (derivative). Here, we have: $f(x)=-4x^2+5$ $\lim\limits_{x\to 3}\dfrac{f(x)-f(3)}{x-3}=\lim\limits_{x\to 3}\dfrac{(-4x^2+5)-(-4(3)^2+5)}{x-3} \\=\lim\limits_{x\to 3}\dfrac{-4(x^2-9)}{x-3} \\=\lim\limits_{x\to 3}\dfrac{-4(x-3)(x+3)}{x-3} \\=\lim\limits_{x\to 3} -4 (x+3) \\=-4(3+3) \\=(-4)(6) \\=-24$
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