Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 2 - Nonlinear Functions - 2.1 Properties of Functions - 2.1 Exercises - Page 55: 56

Answer

$$ f(x) =-\frac{1}{x^{2}} $$ (a) $$ \begin{aligned} f(x+h) =-\frac{1}{x^{2}+2hx+h^{2}} \end{aligned} $$ (b) $$ \begin{aligned} f(x+h)-f(x) = \frac{2 x h+h^{2}}{x^{2}\left(x^{2}+2 x h+h^{2}\right)} \end{aligned} $$ (c) $$ \begin{aligned} \frac{f(x+h)-f(x)}{h} =\frac{2 x+h}{x^{2}\left(x^{2}+2 x h+h^{2}\right)} \end{aligned} $$

Work Step by Step

$$ f(x) =-\frac{1}{x^{2}} $$ (a) $$ \begin{aligned} f(x+h) &=-\frac{1}{(x+h)^{2}} \\ &=-\frac{1}{x^{2}+2hx+h^{2}} \end{aligned} $$ (b) $$ \begin{aligned} f(x+h)-f(x) & =-\frac{1}{x^{2}+2 x h+h^{2}}-\left(-\frac{1}{x^{2}}\right) \\=&-\frac{1}{x^{2}+2 x h+h^{2}}+\frac{1}{x^{2}} \\=&-\frac{x^{2}}{x^{2}\left(x^{2}+2 x h+h^{2}\right)} \\ &+\frac{\left(x^{2}+2 x h+h^{2}\right)}{x^{2}\left(x^{2}+2 x h+h^{2}\right)} \\=& \frac{-x^{2}+x^{2}+2 x h+h^{2}}{x^{2}\left(x^{2}+2 x h+h^{2}\right)} \\=& \frac{2 x h+h^{2}}{x^{2}\left(x^{2}+2 x h+h^{2}\right)} \end{aligned} $$ (c) $$ \begin{aligned} \frac{f(x+h)-f(x)}{h} &=\frac{\frac{h(2 x+h)}{x^{2}\left(x^{2}+2 x h+h^{2}\right)}}{h} \\ &=\frac{2 x+h}{x^{2}\left(x^{2}+2 x h+h^{2}\right)} \end{aligned} $$
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.