Question

If the domain of the real valued function $f$ given below
$f\left(x\right)=\sqrt{\frac{-x+1}{x+3}}$ is $(-\infty ,-x)\cup (-x,1]$.Find $x$

Answer 1

$f\left(x\right)=\sqrt{\frac{-x+1}{x+3}}$

Domain of $f\left(x\right)=[-\infty ,x)\cup (x,1]$

$f\left(x\right)=\sqrt{-R}$ and ${\sqrt{R}}_0$ is not possible , whre $R$ is the set of real numbers.

So,$x+3$ should not be zero

$x+3=0$

$x=-3$

$\sqrt{-x+1}$should not be$\sqrt{-R}$

So, x should be less or equal to 1 and less than -3 and greater than -3 upto 1.

Domain of $f\left(x\right)=[-\infty ,x)\cup (-x,1]$

So, $x=-3$