Question

The domain of the function $f\left(x\right)=\sqrt{2-2x-x^2}$ is

(a) $\left[-\sqrt{3},\sqrt{3}\right]$
(b) $\left[-1-\sqrt{3},-1+\sqrt{3}\right]$
(c) [−2, 2]
(d) $\left[-2-\sqrt{3},-2+\sqrt{3}\right]$

Answer 1

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

$f\left(x\right)=\sqrt{2-2x-x^2}$
$$\begin{gathered} \mathrm{Since},2-2x-x^2\ge 0, \\ {x}^2+2x-2\le 0 \\ \Rightarrow x^2-2x-2+1-1\le 0 \\ \Rightarrow {\left(x-1\right)}^2-{\left(\sqrt{3}\right)}^2\le 0 \\ \Rightarrow \left[x-\left(-1-\sqrt{3}\right)\right]\left[x-\left(-1+\sqrt{3}\right)\right]\le 0 \\ \Rightarrow \left(-1-\sqrt{3}\right)\le x\le \left(-1+\sqrt{3}\right) \\ \mathrm{Thus},\mathrm{dom}\left(f\right)=\left[-1-\sqrt{3},-1+\sqrt{3}\right]. \end{gathered}$$