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

Jun 20, 2018

Domain: $\left\{x | x \in \mathbb{R} , x \ne 1 , x \ne - 1\right\}$

Range: $\left\{f \left(x\right) | f \left(x\right) \in \mathbb{R} , x < 0 \mathmr{and} x \ge 2\right\}$

#### Explanation:

$f \left(x\right) = \frac{2}{1 - {x}^{2}}$

To find the Domain find where the function is undefined, i.e. $\frac{a}{0}$

To do this we set the denominator equal to zero and solve, notice it is a difference of squares:

$1 - {x}^{2} = 0$

$\left(1 - x\right) \left(1 + x\right) = 0$

$x = 1$ and $x = - 1$ so the domain is all real numbers except those two:

$\left\{x | x \in \mathbb{R} , x \ne 1 , x \ne - 1\right\}$

Now the range, as $x$ gets really big positive or negative the $1$ becomes insignificant so we must only consider $\frac{1}{-} x$, sense we know as

$\frac{2}{-} x \to \pm \infty , f \left(x\right) \to 0$ then:

$\frac{2}{1 - {x}^{2}} \to \pm \infty , f \left(x\right) \to 0$

so we know for numbers above and below the asymptotes the range is $- \infty \text{ to } 0$, what about between the asymptotes? In that case the numbers input are $- 1 \le x \le 1$ so at 0 the range is 2 and the closer we get to -1 or 1 the range is $\frac{2}{\text{some tiny number}} \to \infty$, so that range is $2 \text{ to } \infty$

Putting it all together:

$\left\{f \left(x\right) | f \left(x\right) \in \mathbb{R} , x < 0 \mathmr{and} x \ge 2\right\}$

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