Question

Find the domain & the range for $f\left(x\right)=\frac{1}{1-x^2}$

Answer 1

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

As the function $f\left(x\right)$ has to be defined at all values of $x$.

Hence, $xϵR^{+}-\{1,-1\}$

Range ,

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

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

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

Hence,

$\frac{y-1}{y}>0$

$y(y-1)>0$

$y<0$ and $y\ge 1$ is the required range