## Intermediate Algebra for College Students (7th Edition)

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