# How do you find the domain and range of f(x) = 2 / (1 - x²)?

Jun 24, 2015

Domain is defined by inequalities:
$x \ne \pm 1$ or a union of three intervals
$\left(- \infty , - 1\right)$, $\left(- 1 , 1\right)$, $\left(1 , + \infty\right)$

Range is a combination of two intervals
$- \infty < f \left(x\right) < 0$ and $2 \le f \left(x\right) < + \infty$

#### Explanation:

Domain of this function is determined by the fact that denominator might be equal to zero and the function would be undefined in this case. So, we are talking about the following restriction on domain:
$1 - {x}^{2} \ne 0$ or
${x}^{2} \ne 1$ or
$x \ne \pm 1$
In other words, the domain is a union of three intervals:
$- \infty < x < - 1$;
$- 1 < x < 1$;
$1 < x < + \infty$
or, in another notation,
$\left(- \infty , - 1\right)$, $\left(- 1 , 1\right)$, $\left(1 , + \infty\right)$

To determine the range, it's most useful to graph this function.
The easiest way is to start by drawing a graph of a function $y = 1 - {x}^{2}$, invert it into $y = \frac{1}{1 - {x}^{2}}$ and multiply the latter by a factor 2 to get $y = \frac{2}{1 - {x}^{2}}$.

$y = 1 - {x}^{2}$

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

$y = \frac{1}{1 - {x}^{2}}$

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

$y = \frac{2}{1 - {x}^{2}}$

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

As you see, on the left and on the right graph is located below the X-axis and goes to negative infinity, while in the middle it has a minimum of $2$ at $x = 0$ and from that goes to positive infinity.

That's why the range is represented as two intervals:

$- \infty < f \left(x\right) < 0$ and $2 \le f \left(x\right) < + \infty$

or, in another notation,

$\left(- \infty , 0\right)$ and $\left[2 , + \infty\right)$

Notice, that value $y = 2$ is included into the second interval.