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

Sep 20, 2015

The domain of $f \left(x\right)$ is $\left(- \infty , 0\right) \cup \left(0 , 5\right) \cup \left(5 , \infty\right)$ and

the range of $f \left(x\right)$ is $\left(- \infty , - \frac{1}{5}\right) \cup \left(- \frac{1}{5} , 0\right) \cup \left(0 , \infty\right)$.

#### Explanation:

$f \left(x\right) = \frac{x}{{x}^{2} - 5 x} = \frac{x}{x \left(x - 5\right)} = \frac{1}{x - 5}$ with exclusion $x \ne 0$

The denominator of $f \left(x\right)$ is zero when $x = 0$ or $x = 5$.

Let $y = f \left(x\right) = \frac{1}{x - 5}$. Then $x = \frac{1}{y} + 5$.

Therefore $y = 0$ is an excluded value. Also $y = - \frac{1}{5}$ is an excluded value, since it would result in $x = 0$, which is an excluded value.

So the domain of $f \left(x\right)$ is $\left(- \infty , 0\right) \cup \left(0 , 5\right) \cup \left(5 , \infty\right)$ and

the range of $f \left(x\right)$ is $\left(- \infty , - \frac{1}{5}\right) \cup \left(- \frac{1}{5} , 0\right) \cup \left(0 , \infty\right)$.