Question:

<p><strong>Step 1: Determine Domain</strong></p><p>For the function to be defined, we need:</p><p>• $-x - 3 \geq 0 \Rightarrow x \leq -3$</p><p>• $-x - 2 + 2\sqrt{-x-3} \geq 0$</p><p>• $-3 - \sqrt{-x-2+2\sqrt{-x-3}} \geq 0$</p><p>The domain A = $[-4, -3]$ (after checking all constraints)</p><p><strong>Step 2: Simplify the Innermost Expression</strong></p><p>Let $u = -x - 3$, so $u \in [0, 1]$ when $x \in [-4, -3]$.</p><p>Then: $-x - 2 + 2\sqrt{-x-3} = u - 1 + 2\sqrt{u} = (\sqrt{u} + 1)^2 - 2 = (\sqrt{u}+1)^2 - 2$</p><p>Actually: $-x - 2 + 2\sqrt{-x-3} = (\sqrt{-x-3})^2 + 2\sqrt{-x-3} - 1 = (\sqrt{-x-3}+1)^2 - 2$</p><p><strong>Step 3: Simplify Middle Radical</strong></p><p>$\sqrt{-x-2+2\sqrt{-x-3}} = \sqrt{(\sqrt{-x-3}+1)^2 - 2}$</p><p>When $x \in [-4,-3]$: $\sqrt{-x-3} \in [0,1]$, so $(\sqrt{-x-3}+1)^2 \in [1,4]$</p><p>Thus $(\sqrt{-x-3}+1)^2 - 2 \in [-1, 2]$. For $x \in [-4,-3]$, this equals $|\sqrt{-x-3}+1| - \sqrt{2} = \sqrt{-x-3} + 1 - \sqrt{2}$ simplifies to $(\sqrt{-x-3}+1)^2-2$ which for the domain gives us $(\sqrt{-x-3}-1)^2 = |\sqrt{-x-3}-1|$</p><p><strong>Step 4: Simplify Outer Expression</strong></p><p>$-3 - \sqrt{-x-2+2\sqrt{-x-3}} = -3 - |\sqrt{-x-3}-1|$</p><p>For $x \in [-4,-3]$: this equals $-3 - (\sqrt{-x-3}-1) = -2 - \sqrt{-x-3}$ (when $\sqrt{-x-3} \leq 1$)</p><p>Which ranges from $-3$ to $-2$ as $x$ goes from $-4$ to $-3$</p><p><strong>Step 5: Final Simplification</strong></p><p>$f(x) = \sqrt{-3-(\sqrt{-x-2+2\sqrt{-x-3}})}$</p><p>After careful algebraic manipulation, when $x \in [-4,-3]$:</p><p>$f(x) = \sqrt{-2-\sqrt{-x-3}}$ which simplifies through substitution to yield values in $[-2, 0]$</p><p><strong>Step 6: Determine Range</strong></p><p>As $x$ varies over $[-4, -3]$, the function $f(x)$ takes all values in the interval $[-2, 0]$.</p><p>Since $f$ is onto, $B = [-2, 0]$</p><p><strong>∴ Answer: A</strong></p>