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: 19

Answer

$a) \frac{3}{4}$ $b) -3$ $c) \frac{11}{8}$ $d)\frac{13}{9}$ $e)\frac{2a+2h-3}{a+h-4}$ $f)$ Dividing a number by zero is not allowed by the rules of division.

Work Step by Step

$a) f(0)=\frac{2(0)-3}{0-4}=\frac{-3}{-4}=\frac{3}{4}$ $b) f(3)=\frac{2(3)-3}{3-4}=\frac{3}{-1}=-3$ $c) f(-4)=\frac{2(-4)-3}{-4-4}=\frac{-11}{-8}=\frac{11}{8}$ $d) f(-5)=\frac{2(-5)-3}{-5-4}=\frac{-13}{-9}=\frac{13}{9}$ $e)f(a+b)=\frac{2(a+h)-3}{a+h-4}=\frac{2a+2h-3}{a+h-4}$
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