Question

Find the domain of definition of the following function:
$y=\frac{\log (3-2x-x^2)}{\sqrt{x}}$

Answer 1

$y=\frac{log(3-2x-x^2)}{\sqrt{x}}$

For $y$ to be real and defined.

$x$ must be $>0$ $\Rightarrow x\in (0,\infty )$

$3-2x-x^2>0\Rightarrow x^2+2x-3<0\Rightarrow (x+3)(x-1)<0$

Therefore, $x\in (-3,1)$

Domain of $y$ is $x\in (0,1).$