# What is the domain and range of f(x)=(x^2+1)/(x+1)?

Sep 4, 2017

The domain is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right)$
The range is $y \in \left(- \infty , - 2 - \sqrt{8}\right] \cup \left[- 2 + \sqrt{8} , + \infty\right)$

#### Explanation:

As we cannot divide by $0$, $x \ne - 1$

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

Let $y = \frac{{x}^{2} + 1}{x + 1}$

So,

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

${x}^{2} + y x + 1 - y = 0$

In order for this equation to have solutions, the discriminant is

$\Delta \le 0$

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

$y = \frac{- 4 \pm \left(16 - 4 \cdot \left(- 4\right)\right)}{2}$

$y = \frac{- 4 \pm \sqrt{32}}{2} = \left(- 2 \pm \sqrt{8}\right)$

${y}_{1} = - 2 - \sqrt{8}$

${y}_{2} = - 2 + \sqrt{8}$

Therefore the range is

$y \in \left(- \infty , - 2 - \sqrt{8}\right] \cup \left[- 2 + \sqrt{8} , + \infty\right)$

graph{(x^2+1)/(x+1) [-25.65, 25.66, -12.83, 12.84]}