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

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Explanation

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Explanation:

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Jun 6, 2018

The domain is $x \in \left(- \infty , - 1\right) \cup \left(- 1 , + \infty\right)$. The range is $y \in \mathbb{R}$

Explanation:

The denominator must be $\ne 0$.

Therefore,

$x + 1 \ne 0$

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

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

To calculate the range, proceed as follows :

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

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

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

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

In order for this quadratic equation in $x$ to have solutions, the discriminant $\Delta \ge 0$

The discriminant is

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

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

$= {y}^{2} + 4$

Therefore,

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

So,

$\forall y \in \mathbb{R} , \Delta \ge 0$

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

graph{(x^2+2x)/(x+1) [-16.02, 16.01, -8.01, 8.01]}

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