# What is the domain of f(x)=x/(x^2-5x)?

Sep 19, 2015

$D = - \infty < x < \infty | x \ne 0 , x \ne 5 \mathmr{and} x \in \mathbb{R}$

#### Explanation:

The domain is every value that $x$ can take without having a math error (division by zero, logarithm of a null or negative number, even root of a negative number, etc.)

So the only caveat we have here is that the denominator mustn't be 0. Or
${x}^{2} - 5 x \ne 0$

We can solve this using the quadratic formula, sum and product, or, just do the easy thing and factor it out.
${x}^{2} - 5 x \ne 0$
$x \left(x - 5\right) \ne 0$

Since the product can't be zero, neither can, that is
$x \ne 0$
$x - 5 \ne 0 \rightarrow x \ne 5$

So the domain D, is $D = - \infty < x < \infty , x \ne 0 , x \ne 5 | x \in \mathbb{R}$
Or
$D = - \infty < x < 0 \mathmr{and} 0 < x < 5 \mathmr{and} 5 < x | x \in \mathbb{R}$
Or that same thing in set notation.