Intermediate Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-894-7
ISBN 13: 978-0-13417-894-3

Chapter 8 - Section 8.5 - Polynomial and Rational Inequalities - Concept and Vocabulary Check - Page 648: 5


Therefore, the solution set for $f(x)= \frac{x-1}{x+2}\geq0$ is $(- ∞ , -2) ∪ [1, ∞ )$.

Work Step by Step

We are interested in getting the interval for which $f(x)= \frac{x-1}{x+2}\geq0$. On the given number line, we only take the intervals for which the test value yields a positive number. These are: $(-∞,-2)$ and $(1,∞)$ Since the inequality sign is $\geq$ (greater than or $equal$ $to$), we also need to include the solution of $f (x) = 0$, namely the value that we obtained when we set the numerator of $f$ equal to zero. Thus, we must include $1$ in the solution set. Therefore, the solution set for $f(x)= \frac{x-1}{x+2}\geq0$ is $(- ∞ , -2) ∪ [1, ∞ )$.
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