# How do you find the domain and range of f(x)= sqrt(4-x)/( (x+1)(x^2+1))?

Jun 30, 2018

The domain is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , 4\right]$. The range is $y \in \left(- \infty , + \infty\right)$

#### Explanation:

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

Therefore,

$x + 1 \ne 0$

$\implies$, $x \ne - 1$

Also, what's under the sqrt sign must be $\ge 0$

Therefore,

$4 - x \ge 0$

$\implies$, $x \le 4$

Therefore,

The domain is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , 4\right]$

To find the range,

$f \left(4\right) = 0$

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

The limits are as follows :

${\lim}_{x \to - \infty} y = {O}^{-}$

${\lim}_{x \to - {1}^{-}} y = - \infty$

${\lim}_{x \to - {1}^{+}} y = + \infty$

The range is $y \in \left(- \infty , + \infty\right)$

graph{sqrt(4-x)/((x+1)(x^2+1)) [-10, 10.01, -5, 5]}