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

Aug 16, 2015

Domain: $\left(- \infty , 2\right) \cup \left(2 , + \infty\right)$
Range: $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$

#### Explanation:

Your function is defined for any value of $\in \mathbb{R}$ except the one that can make the denominator equal to zero.

$x - 2 = 0 \implies x = 2$

This means that $x = 2$ will be excluded from the domain of the function, which will thus be $\mathbb{R} - \left\{2\right\}$, or $\left(- \infty , 2\right) \cup \left(2 , + \infty\right)$.

The range of the function will be affected by the fact that the only way a fraction can be equal to zero is if the numerator is equal to zero.

In your case, the numerator is constant, euqal to $1$ regardless of the value of $x$, which implies that the function can never be equal to zero

$f \left(x\right) \ne 0 \text{, } \left(\forall\right) x \in \mathbb{R} - \left\{2\right\}$

The range of the function will thus be $\mathbb{R} - \left\{0\right\}$, or $\left(- \infty , 0\right) \cup \left(0 , + \infty\right)$.

graph{1/(x-2) [-10, 10, -5, 5]}