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

Jul 29, 2017

The domain of $f \left(x\right)$ is ${d}_{f} \left(x\right) = \mathbb{R} - \left\{- 1\right\}$
The range is $f \left(x\right) \in \mathbb{R}$

#### Explanation:

Our function is

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

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

Therefore,

The domain of $f \left(x\right)$ is ${D}_{f} \left(x\right) = \mathbb{R} - \left\{- 1\right\}$

Let,

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

So,

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

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

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

In order for this quadratic equations to have solutions, the discriminant

$\Delta \ge 0$, $\implies$, ${b}^{2} - 4 a c \ge 0$

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

$4 - 4 y + {y}^{2} + 4 y \ge 0$

${y}^{2} + 4 \ge 0$

$\forall y \in \mathbb{R}$, ${y}^{2} + 4 \ge 0$

The range is $y \in \mathbb{R}$

graph{(x^2+2x)/(x+1) [-12.66, 12.65, -6.33, 6.33]}