# What is the domain and range of F(x) = 5/(x-2)?

Mar 11, 2018

$\textrm{D o m a \in} : x \ne 2$
$\textrm{R a n \ge} : f \left(x\right) \ne 0$

#### Explanation:

The domain is the range of $x$ values which give $f \left(x\right)$ a value that is unique, such there is only one $y$ value per $x$ value.

Here, since the $x$ is on the bottom of the fraction, it cannot have any value such that the whole denominator equals zero, i.e. $d \left(x\right) \ne 0$ $d \left(x\right) = \textrm{\mathrm{de} n o \min a \to r o f t h e \frac{t}{i} o n t \hat{i} s a f u n c t i o n o f}$ $x$.

$x - 2 \ne 0$
$x \ne 2$

Now, the range is the set of $y$ values given for when $f \left(x\right)$ is defined. To find any $y$ values that cannot be reached, i.e. holes, asymptotes, etc. We rearrange to make $x$ the subject.

$y = \frac{5}{x - 2}$

$x = \frac{5}{y} + 2$, $y \ne 0$ since this would be undefined, and so there are no values of $x$ where $f \left(x\right) = 0$. Therefore the range is $f \left(x\right) \ne 0$.