Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 2 - Section 2.1 - Functions - Exercises - Page 157: 55

Answer

$\{x\in R: x \neq 3\} =(-\infty , 3) \cup (3, \infty).$

Work Step by Step

The domain of $f(x)=\frac{1}{x-3}$ is the set of all real numbers $x$ for which $\frac{1}{3-x}$ is defined as a real number. We know a rational expression is not defined where the denominator is zero, so we must find all values of $x$ for which $$3-x =0.$$ We add $x$ to both sides to get $$3=x.$$ Thus we know $\frac{1}{3-x}$ is not defined when $x=3$. This means the domain of $f(x)=\frac{1}{3-x}$ is the set af all real numbers $x$ for which $x\neq 3$ In set notation, the domain is $\{x\in R: x \neq 3\}.$ In interval notation, the domain is $(-\infty , 3) \cup (3, \infty).$
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