How do you find the domain and range of f(x) = sqrt{(x - 1)/(x + 4)}?

Jan 26, 2017

The domain is D_f(x)=x in ]-oo,-4[ uu [1,+oo[
The range is R_f(x)=[0, +oo[

Explanation:

As you cannot divide by $0$, $x \ne - 4$

and what's under the sqrt sign is $\ge 0$

$\frac{x - 1}{x + 4} \ge 0$

Let $g \left(x\right) = \frac{x - 1}{x + 4}$

We build a sign chart

$\textcolor{w h i t e}{a a a a}$$x$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a a}$$- 4$$\textcolor{w h i t e}{a a a a a}$$1$$\textcolor{w h i t e}{a a a a a}$$+ \infty$

$\textcolor{w h i t e}{a a a a}$$x + 4$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$+$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$x - 1$$\textcolor{w h i t e}{a a a a a a}$$-$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$g \left(x\right)$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a a}$$| |$$\textcolor{w h i t e}{a a}$$-$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$g \left(x\right) \ge 0$ when x in ]-oo,-4[ uu [1,+oo[

The domain is D_f(x)=x in ]-oo,-4[ uu [1,+oo[

${\lim}_{x \to \pm \infty} g \left(x\right) = {\lim}_{x \to \pm \infty} \frac{x}{x} = {\lim}_{x \to \pm \infty} 1 = 1$

${\lim}_{x \to - {4}^{-}} g \left(x\right) = {\lim}_{x \to - {4}^{-}} - \frac{5}{{0}^{-}} = + \infty$

$g \left(1\right) = \frac{0}{5} = 0$

So,

the range is R_f(x)=[0, +oo[