What is the domain and range of y= 1 / (x-3) ?

Mar 20, 2017

Domain: $\mathbb{R} - \left\{3\right\}$, or $\left(- \infty , 3\right) \cup \left(3 , \infty\right)$
Range: $\mathbb{R} - \left\{0\right\}$, or $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$

Explanation:

You can't divide by zero, meaning the denominator of the fraction cannot be zero, so
$x - 3 \ne 0$
$x \ne 3$
Thus, the domain of the equation is $\mathbb{R} - \left\{3\right\}$, or $\left(- \infty , 3\right) \cup \left(3 , \infty\right)$
Alternately, to find the domain and range, look at a graph:
graph{1/(x-3) [-10, 10, -5, 5]}
As you can see, the x never equals 3, there is a gap at that point, so the domain doesn't include 3 - and there is a vertical gap in the range of the graph at y=0, so the range doesn't include 0.
So, again, the domain is $\mathbb{R} - \left\{3\right\}$, or $\left(- \infty , 3\right) \cup \left(3 , \infty\right)$
And the range is $\mathbb{R} - \left\{0\right\}$, or $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$.

NOTE: Another way to find y that may or may not be allowed (solving for x):
Multiply both sides by x:
$y \left(x - 3\right) = 1$
Divide by y:
$x - 3 = \frac{1}{y}$
$x = \frac{1}{y} + 3$
Since you can't divide by zero, $y \ne 0$, and the range of y is $\mathbb{R} - \left\{0\right\}$ or $\left(- \infty , 0\right) \cup \left(0 , \infty\right)$.