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 - Section 13.2 Algebra Techniques for Finding Limits - 13.2 Assess Your Understanding - Page 903: 52

Answer

$$-2$$

Work Step by Step

The general formula for average rate of change from $a$ to $b$ can be written as: $\dfrac{f(b)-f(a)}{b-a}$ Here, we have: $f(x)=\dfrac{1}{x^2}$ Thus, we find the average rate of change as: $\lim\limits_{x\to 1}\dfrac{f(x)-f(1)}{x-1}=\lim\limits_{x\to 1}\dfrac{\dfrac{1}{x^2}-1}{x-1} \\=\lim\limits_{x\to 1}\dfrac{\dfrac{1-x^2}{x^2}}{x-1} $ In order to simplify the above expression, we will use the following rules. $(a) \lim\limits_{x \to a} \dfrac{p(x)}{q(x)}=\dfrac{\lim\limits_{x \to a} p(x)}{\lim\limits_{x \to a} q(x)} \\ (b) \lim\limits_{x \to a} k(x)=k(a)$ ; where $a$ is a constant. $=\dfrac{\lim\limits_{x\to 1}-(x-1)(x+1)}{\lim\limits_{x\to 1} x^2(x-1)} \\=-\dfrac{\lim\limits_{x\to 1} (x+1)} {\lim\limits_{x\to 1} x^2} \\=-\dfrac{(1+1)}{(1)^2} \\=-2 $
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.