Precalculus (6th Edition) Blitzer

Published by Pearson
ISBN 10: 0-13446-914-3
ISBN 13: 978-0-13446-914-0

Chapter 1 - Review Exercises - Page 303: 95

Answer

The domain of the function $g\left( x \right)=\frac{\sqrt{x-2}}{x-5}$ is $\left[ 2,5 \right)\cup \left( 5,\infty \right)$.

Work Step by Step

Now, consider the function $g\left( x \right)=\frac{\sqrt{x-2}}{x-5}$. We can see that this function contains division and division by 0 is not defined. Exclude those values of $x$ from the domain that cause the denominator to be zero. Thus, set the denominator equal to 0, that is, $x-5=0$. This implies that $x=5$. So, exclude 5 from the domain. Next, note that this function also contains the square root in the numerator and the square root of a negative function is not defined; therefore, the term written inside the square root must be non-negative. That is, $x-2\ge 0$ or $x\ge 2$. Therefore, the domain of the function $g\left( x \right)=\frac{\sqrt{x-2}}{x-5}$ is $\left[ 2,5 \right)\cup \left( 5,\infty \right)$.
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