# How do you find the domain and range of y = 3(x-2)/x?

Aug 21, 2015

Domain: $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$
Range: $\left(- \infty , 3\right) \cup \left(3 , + \infty\right)$

#### Explanation:

Right from the start, you can say that the domain of the function will not include $x = 0$, since that would make the denominator of the fraction equal to zero.

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

To find if the range of the function has any restrictions, calculate the inverse of $y$ by solving for $x$, then switching $x$ with $y$

$y = \frac{3 x - 6}{x}$

$y \cdot x = 3 x - 6$

$x \left(y - 3\right) = 6 \implies x = \frac{6}{y - 3}$

The inverse function will thus be

$y = \frac{6}{x - 3}$

As you can see, this is not defined for $x = 3$, which means that your original function cannot take the value $y = 3$. The range of the function will thus be $\mathbb{R} - \left\{3\right\}$, or $\left(- \infty , 3\right) \cup \left(3 , + \infty\right)$.

graph{(3(x-2))/x [-10, 10, -5, 5]}