College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Chapter P Review - Exercises: 96

Answer

$\{x|x\geq 3\}$

Work Step by Step

We know that $\displaystyle \frac{\sqrt{x-3}}{x^{2}-4x+4}$ is undefined whenever the denominator is 0, so: $x^2-4x+4=(x-2)^2=0$ $x=2$ We also know that the equation is undefined for negative roots: $x-3<0$ $x<3$ So $x$ must be greater than or equal to 3 (which also ensures that it's not 2). Thus the domain is: $\{x|x\geq 3\}$
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