Question:

<p><strong>Step 1:</strong> Solve the inequality to find the natural domain where the function is defined.</p><p>From $-1 \leq \log_2\left(\frac{x^2}{2}\right) \leq 1$:</p><p>$2^{-1} \leq \frac{x^2}{2} \leq 2^1$</p><p>$\frac{1}{2} \leq \frac{x^2}{2} \leq 2$</p><p>$1 \leq x^2 \leq 4$</p><p>$x \in [-2,-1] \cup [1,2]$ ✓ (matches given domain D)</p><p><strong>Step 2:</strong> Analyze $f(x) = \log_2\left(\frac{x^2}{2}\right)$ on domain D.</p><p>Let $f(x) = \log_2(x^2) - 1 = 2\log_2|x| - 1$</p><p><strong>Step 3:</strong> Find monotonicity. Since $f'(x) = \frac{2}{x\ln 2}$:</p><p>- On $[-2,-1]$: $x < 0$, so $f'(x) < 0$ (decreasing)</p><p>- On $[1,2]$: $x > 0$, so $f'(x) > 0$ (increasing)</p><p><strong>Step 4:</strong> Evaluate at critical points and endpoints:</p><p>- $f(-2) = 2\log_2(2) - 1 = 2(1) - 1 = 1$</p><p>- $f(-1) = 2\log_2(1) - 1 = 0 - 1 = -1$</p><p>- $f(1) = 2\log_2(1) - 1 = 0 - 1 = -1$</p><p>- $f(2) = 2\log_2(2) - 1 = 2(1) - 1 = 1$</p><p><strong>Step 5:</strong> On $[-2,-1]$ (decreasing): range is $[-1, 1]$</p><p>On $[1,2]$ (increasing): range is $[-1, 1]$</p><p>∴ Range = $[-1, 1]$</p>