# What is the domain of  f(x)= 1/(x^2-4x)?

Oct 15, 2017

All real numbers except $x = 0$ and $x = 4$

#### Explanation:

The domain of a function is simply the set of all $x$-values that will output real $y$-values. In this equation, not all $x$-values will work as we cannot divide by $0$. Thus, we need to find when the denominator will be $0$.

${x}^{2} - 4 x = 0$

$x \cdot \left(x - 4\right) = 0$

Using the Zero Property of Multiplication, if $x = 0$ or $x - 4 = 0$, then ${x}^{2} - 4 x = 0$ will be $0$.

Thus, $x = 0$ and $x = 4$ should not be part of the domain as they would result in a non-existent $y$-value.

This means the domain is all real numbers except $x = 0$ and $x = 4$.

In set notation, this can be written as $x \in \mathbb{R} \text{ such that } x \ne 0 \mathmr{and} x \ne 4$