College Algebra 7th Edition

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

Chapter 2, Functions - Section 2.1 - Functions - 2.1 Exercises - Page 193: 66


Domain: $[0, \frac{1}{2})\cup (\frac{1}{2},\ \infty)$

Work Step by Step

We are given: $g(x)= \frac{\sqrt{x}}{2x^{2}+x-1}$ The domain of a function consists of all values that $x$ is allowed to have. In this case, we can not take the square root of a negative number, so: $x\geq0$ We also can not have division by $0$: $2x^{2}+x-1\neq 0$ $(2x-1)(x+1)\neq 0$ $ 2x-1\neq 0$ or $x+1 \neq 0$ $ x\neq\frac{1}{2}$ or $x\neq-1$ We can ignore the $x\neq -1$ solution, since we already have $x\geq 0$. Therefore, the domain is: $[0, \frac{1}{2})\cup (\frac{1}{2},\ \infty)$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.