Question

Find the domain of definition of the following function:
$y=\sqrt{1-x}+\log (x+1).$

Answer 1

$y=\sqrt{1-x}+log(x+1)$

For $y$ to be real and defined.

$1-x$ must be $\ge 0\Rightarrow x\le 1$

$x+1$ must be $>0\Rightarrow x>-1$

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

Domain of $y$ is $x\in (-1,1].$