Intermediate Algebra for College Students (7th Edition)

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

Chapter 2 - Section 2.1 - Introduction to Functions - Exercise Set - Page 112: 20

Answer

$a.\displaystyle \quad \frac{1}{5}$ $b.\quad -4$ $c.\displaystyle \quad \frac{5}{4}$ $d.\displaystyle \quad \frac{29}{5}$ $e.\displaystyle \quad \frac{3a+3h-1}{a+h-5}$

Work Step by Step

$a.$ $f(0)=\displaystyle \frac{3(0)-1}{0-5}=\frac{-1}{-5}=\frac{1}{5}$ $b.$ $f(3)=\displaystyle \frac{3(3)-1}{3-5}=\frac{9-1}{3-5}=\frac{8}{-2}=-4$ $c.$ $f(-3)=\displaystyle \frac{3(-3)-1}{-3-5}=\frac{-9-1}{-3-5}=\frac{-10}{-8}=\frac{5}{4}$ $d.$ $f(10)=\displaystyle \frac{3(10)-1}{10-5}=\frac{30-1}{10-5}=\frac{29}{5}$ $e.$ $f(a+h)=\displaystyle \frac{3(a+h)-1}{a+h-5}=\frac{3a+3h-1}{a+h-5}$
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