# What is the domain of f(x) = (3x-1) / (x^2-x) ?

Sep 7, 2015

#### Answer:

Domain: $\left(- \infty , 0\right) \cup \left(0 , 1\right) \cup \left(1 , + \infty\right)$

#### Explanation:

Right from the start, you know that the domain of the function must not include any value of $x$ that will make the denominator of the fraction equal to zero.

So basically, any value of $x$ for which

${x}^{2} - x \ne 0$

wil be excluded from the function's domain.

You can rewrite this equation like this

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

$x \cdot \left(x - 1\right) = 0 \implies \left\{\begin{matrix}x = 0 \\ x = 1\end{matrix}\right.$

This means that the domain of the function will be $\mathbb{R} - \left\{0 , 1\right\}$, or, in interval notation, $\left(- \infty , 0\right) \cup \left(0 , 1\right) \cup \left(1 , + \infty\right)$.

graph{(3x - 1)/(x^2 - x) [-46.22, 46.22, -23.1, 23.15]}