Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 6 - Section 6.3 - Complex Rational Expressions - Exercise Set - Page 436: 49

Answer

$ -\frac{3}{a(a+h)}$.

Work Step by Step

The given function is $\Rightarrow f(x)=\frac{3}{x}$ Replace $x$ with $a+h$. $\Rightarrow f(a+h)=\frac{3}{a+h}$ Replace $x$ with $a$. $\Rightarrow f(a)=\frac{3}{a}$ Substitute all the values into $\Rightarrow \frac{f(a+h)-f(a)}{h}$ $\Rightarrow \frac{\frac{3}{a+h}-\frac{3}{a}}{h}$ Solve the numerator. $=\frac{3}{a+h}-\frac{3}{a}$ The LCD of the denominators is $a(a+h)$. $=\frac{3a}{a(a+h)}-\frac{3(a+h)}{a(a+h)}$ $=\frac{3a-3(a+h)}{a(a+h)}$ $=\frac{3a-3a-3h}{a(a+h)}$ Simplify. $=\frac{-3h}{a(a+h)}$ Back substitute all values into the original fraction. $\Rightarrow \frac{-3h}{a(a+h)h}$ Simplify. $\Rightarrow -\frac{3}{a(a+h)}$.
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.