How do you find the domain and range of g(x)= (3+x^2)/(4-x^2)?

Oct 7, 2017

The domain is $x \in \left(- \infty , - 2\right) \cup \left(- 2 , 2\right) \cup \left(2 , + \infty\right)$
The range is $g \left(x\right) \in \left(- \infty , - 1\right) \cup \left[\frac{3}{4} , + \infty\right)$

Explanation:

As you cannot divide by $0$, the denominator is $\ne 0$

$4 - {x}^{2} \ne 0$

$\left(2 + x\right) \left(2 - x\right) \ne 0$

$\implies$, $x \ne - 2$ and $x \ne 2$

Therefore,

The domain is $x \in \left(- \infty , - 2\right) \cup \left(- 2 , 2\right) \cup \left(2 , + \infty\right)$

Let

$y = \frac{3 + {x}^{2}}{4 - {x}^{2}}$

$y \left(4 - {x}^{2}\right) = 3 + {x}^{2}$

$4 y - y {x}^{2} = 3 + {x}^{2}$

${x}^{2} \left(1 + y\right) = 4 y - 3$

${x}^{2} = \frac{4 y - 3}{1 + y}$

$x = \sqrt{\frac{4 y - 3}{1 + y}}$

Therefore,

$\frac{4 y - 3}{1 + y} \ge 0$

$\implies$, $y \ne - 1$

Let $g \left(y\right) = \frac{4 y - 3}{1 + y}$

We build a sign chart

$\textcolor{w h i t e}{a a a a}$$y$$\textcolor{w h i t e}{a a a a}$$- \infty$$\textcolor{w h i t e}{a a a a a a}$$- 1$$\textcolor{w h i t e}{a a a a a a a}$$\frac{3}{4}$$\textcolor{w h i t e}{a a a a a a a}$$+ \infty$

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

$\textcolor{w h i t e}{a a a a}$$4 y - 3$$\textcolor{w h i t e}{a a a a a}$$-$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$

$\textcolor{w h i t e}{a a a a}$$g \left(y\right)$$\textcolor{w h i t e}{a a a a a a a}$$+$$\textcolor{w h i t e}{a a a}$$| |$$\textcolor{w h i t e}{a a a a}$$-$$\textcolor{w h i t e}{a a}$$0$$\textcolor{w h i t e}{a a a a}$$+$

Therefore,

$g \left(y\right) \ge 0$ when $y \in \left(- \infty , - 1\right) \cup \left[\frac{3}{4} , + \infty\right)$

graph{(3+x^2)/(4-x^2) [-12.66, 12.65, -6.33, 6.33]}