1. **State the problem:** Determine the domain of the function $$f(x) = \frac{\log(-2x^2 + 8)}{2x - 1}$$. 2. **Recall domain rules:** - The argument of the logarithm must be positive: $$-2x^2 + 8 > 0$$. - The denominator cannot be zero: $$2x - 1 \neq 0$$. 3. **Solve the inequality for the logarithm argument:** $$-2x^2 + 8 > 0$$ $$-2x^2 > -8$$ Divide both sides by -2 (remember to reverse inequality): $$\cancel{-2}x^2 < \cancel{-8}4$$ So, $$x^2 < 4$$ 4. **Find the interval for $x$:** $$-2 < x < 2$$ 5. **Exclude values where denominator is zero:** $$2x - 1 = 0 \Rightarrow x = \frac{1}{2} = 0.5$$ 6. **Domain of $f(x)$ is:** $$(-2, 0.5) \cup (0.5, 2)$$ **Final answer:** The function is defined for $$(-2, 0.5) \cup (0.5, 2)$$, which corresponds to option (a).