The domain of the function f(x)

(b) $[-1-\sqrt{3},-1+\sqrt{3}]$

$f(x)=\sqrt{2-2 x-x^{2}}$

Since, $2-2 x-x^{2} \geq 0$

$x^{2}+2 x-2 \leq 0$

$\Rightarrow x^{2}-2 x-2+1-1 \leq 0$

$\Rightarrow(x-1)^{2}-(\sqrt{3})^{2} \leq 0$

$\Rightarrow[x-(-1-\sqrt{3})][x-(-1+\sqrt{3})] \leq 0$

$\Rightarrow(-1-\sqrt{3}) \leq x \leq(-1+\sqrt{3})$

Thus, $\operatorname{dom}(f)=[-1-\sqrt{3},-1+\sqrt{3}]$.