How to find the domain and range of $f(x) = \sqrt {\frac{x+1}{x+2}}$

Question

Find the domain and range of $$f(x) = \sqrt {\frac{x+1}{x+2}}$$

I got the domain $[-1, \infty)$ but the answer contains $(-\infty, -2)$ along with it. And how to calculate range?

Answer 1

We must have $$\frac{x+1}{x+2}\geq 0.$$ We consider the following cases:

Case 1. Suppose that $x+1\geq 0$ and $x+2>0$. Then $x\geq -1$ and $x>-2$. Thus, $$SS_1=[-1,\infty).$$

Case 2. Suppose that $x+1\leq 0$ and $x+2<0$. Then $x\leq -1$ and $x<-2$. Thus, $$SS_2=[-\infty,-2).$$

Hence, domain$=SS_1\cup SS_2$

Let $y=f(x)=\sqrt{\frac{x+1}{x+2}}$. Note that $y\geq 0$. Now, $$y^2=\frac{x+1}{x+2}=1-\frac{1}{x+2}$$ and we get $$x=\frac{1}{1-y^2}-2\qquad;y\neq\pm 1$$ Because $y\geq 0$, it follows that the range is $\Bbb R\smallsetminus\{1\}$. For completeness the graph is given below:

Answer 2

The sign of $\dfrac{x+1}{x+2}$ is the same as the sign of $(x+1)(x+2)$. The latter is a parabola with positive concavity and roots at $-2$ and $-1$ therefore is positive for $x < -2$ or $x>-1$.

In your case $x=-1$ allowed while $x=-2$ is not. So the domain you are looking for is $(-\infty,-2) \cup [-1,\infty)$.