1. **State the problem:** We want to analyze the function $f(x) = x \sqrt{-2x + 5}$. 2. **Domain:** The expression inside the square root must be non-negative for real values, so: $$-2x + 5 \geq 0$$ $$-2x \geq -5$$ Dividing both sides by $\cancel{-2}$ (remember to flip the inequality sign because we divide by a negative): $$x \leq \frac{5}{2}$$ 3. **Function behavior:** The function is defined for all $x$ such that $x \leq \frac{5}{2}$. 4. **Simplify the function:** Write $f(x)$ explicitly: $$f(x) = x \sqrt{-2x + 5}$$ 5. **Find critical points:** To find maxima or minima, differentiate $f(x)$ using the product and chain rules. Let $g(x) = x$ and $h(x) = \sqrt{-2x + 5} = (-2x + 5)^{1/2}$. Then: $$f'(x) = g'(x)h(x) + g(x)h'(x)$$ Calculate derivatives: $$g'(x) = 1$$ $$h'(x) = \frac{1}{2}(-2x + 5)^{-1/2} \cdot (-2) = -\frac{1}{(-2x + 5)^{1/2}}$$ So: $$f'(x) = 1 \cdot (-2x + 5)^{1/2} + x \cdot \left(-\frac{1}{(-2x + 5)^{1/2}}\right) = \sqrt{-2x + 5} - \frac{x}{\sqrt{-2x + 5}}$$ 6. **Set derivative to zero to find critical points:** $$\sqrt{-2x + 5} - \frac{x}{\sqrt{-2x + 5}} = 0$$ Multiply both sides by $\sqrt{-2x + 5}$: $$(-2x + 5) - x = 0$$ $$-2x + 5 - x = 0$$ $$-3x + 5 = 0$$ $$3x = 5$$ $$x = \frac{5}{3}$$ 7. **Check if $x=\frac{5}{3}$ is in the domain:** Since $\frac{5}{3} \approx 1.666 \leq 2.5$, it is valid. 8. **Evaluate $f(x)$ at critical point:** $$f\left(\frac{5}{3}\right) = \frac{5}{3} \sqrt{-2 \cdot \frac{5}{3} + 5} = \frac{5}{3} \sqrt{-\frac{10}{3} + 5} = \frac{5}{3} \sqrt{\frac{5}{3}} = \frac{5}{3} \cdot \frac{\sqrt{15}}{3} = \frac{5 \sqrt{15}}{9}$$ 9. **Summary:** The function $f(x) = x \sqrt{-2x + 5}$ is defined for $x \leq \frac{5}{2}$, has a critical point at $x = \frac{5}{3}$ where it attains a local maximum of $\frac{5 \sqrt{15}}{9}$. Final answer: The domain is $(-\infty, \frac{5}{2}]$ and the local maximum is at $x=\frac{5}{3}$ with value $f\left(\frac{5}{3}\right) = \frac{5 \sqrt{15}}{9}$.