# What is the domain and range of f(x) =(3x^2-2x-8)/(2x^3+x^2-3x)?

Apr 27, 2017

Domain: $\left(- \setminus \infty , - \frac{3}{2}\right) \setminus \cup \left(- \frac{3}{2} , 0\right) \setminus \cup \left(0 , 1\right) \setminus \cup \left(1 , \setminus \infty\right)$
Range: $\left(- \setminus \infty , \setminus \infty\right)$

#### Explanation:

To find the domain, we have to look for any cases where division by zero can occur. In this case, we have to make sure $2 {x}^{3} + {x}^{2} - 3 x \setminus \ne 0$ To solve this we can simplify by factoring out an $x$.
$x \left(2 {x}^{2} + x - 3\right) \setminus \ne 0$

Solving we have two options
$x \setminus \ne 0$ and $2 {x}^{2} + x - 3 \setminus \ne 0$

We have to solve the second equation to get
$\setminus \frac{- \left(1\right) \setminus \pm \setminus \sqrt{{\left(1\right)}^{2} - 4 \left(2\right) \left(- 3\right)}}{2 \left(2\right)}$
$\setminus \frac{- 1 \setminus \pm \setminus \sqrt{1 + 24}}{4}$
$\setminus \frac{- 1 \setminus \pm 5}{4}$
$\setminus \frac{- 1 + 5}{4} = \frac{4}{4} = 1$
$\setminus \frac{- 1 - 5}{4} = - \frac{6}{4} = - \frac{3}{2}$

So the function is undefined at $x = - \frac{3}{2} , 0 , 1$
This means our domain is
$\left(- \setminus \infty , - \frac{3}{2}\right) \setminus \cup \left(- \frac{3}{2} , 0\right) \setminus \cup \left(0 , 1\right) \setminus \cup \left(1 , \setminus \infty\right)$

As you get closer to any of those x-values we found, the denominator gets closer to 0. As the denominator gets closer to 0, the resulting value goes to positive or negative infinity so the range is $\left(- \setminus \infty , \setminus \infty\right)$.