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

Nov 29, 2017

The domain is $\mathbb{R} - \left\{- 2 , 2\right\}$.
The range is $y \in \left(- \infty , - 0.93\right] \cup \left[- 0.67 , + \infty\right)$

#### Explanation:

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

${x}^{2} - 4 = \left(x + 2\right) \left(x - 2\right) \ne 0$

Therefore,

The domain is $\mathbb{R} - \left\{- 2 , 2\right\}$

To determine the range, proceed as follows

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

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

$y {x}^{2} - x - 4 \left(y + 4\right) = 0$

This is a quadratic equation in ${x}^{2}$ and in order for this equation to have solutions, the discriminant $\Delta \ge 0$

$\Delta = {b}^{2} - 4 a c = {\left(- 1\right)}^{2} - 4 \left(y\right) \left(- 4 \left(y + 4\right)\right) \ge 0$

$1 + 16 {y}^{2} + 16 y \ge 0$

$16 {y}^{2} + 16 y + 1 \ge 0$

$y = \frac{- 16 \pm \sqrt{{16}^{2} - 4 \cdot 16}}{2 \cdot 16} = \frac{- 16 \pm \sqrt{192}}{32}$

${y}_{1} = \frac{- 16 + \sqrt{192}}{32} = - 0.67$

${y}_{2} = \frac{- 16 - \sqrt{192}}{32} = - 0.93$

Therefore,

The range is $y \in \left(- \infty , - 0.93\right] \cup \left[- 0.67 , + \infty\right)$

graph{(x+4)/(x^2-4) [-8.89, 8.89, -4.444, 4.445]}