1. **State the problem:** We have a piecewise function $$f(x) = \begin{cases} \frac{x \sqrt{x}}{x-2} & x < 2 \\ \frac{(x^2 - 2)^2}{(x^2 - 2)^2} & x \geq 2 \end{cases}$$ We need to find the values of $f(-8)$, $f(-1)$, $f(0)$, $f(2)$, and $f(3)$. 2. **Analyze the function parts:** - For $x < 2$, $f(x) = \frac{x \sqrt{x}}{x-2}$. - For $x \geq 2$, $f(x) = \frac{(x^2 - 2)^2}{(x^2 - 2)^2}$. 3. **Important note:** The expression $\sqrt{x}$ is only defined for $x \geq 0$ in the real numbers. So for $x < 0$, $f(x)$ is not defined because $\sqrt{x}$ is not real. 4. **Evaluate $f(-8)$:** Since $-8 < 2$ but $-8 < 0$, $\sqrt{-8}$ is not real. So $f(-8)$ is **not defined** in real numbers. 5. **Evaluate $f(-1)$:** Similarly, $-1 < 2$ but $-1 < 0$, so $\sqrt{-1}$ is not real. $f(-1)$ is **not defined**. 6. **Evaluate $f(0)$:** $0 < 2$ and $0 \geq 0$, so we can use the first part: $$f(0) = \frac{0 \cdot \sqrt{0}}{0 - 2} = \frac{0 \cdot 0}{-2} = 0$$ 7. **Evaluate $f(2)$:** Since $2 \geq 2$, use the second part: $$f(2) = \frac{(2^2 - 2)^2}{(2^2 - 2)^2} = \frac{(4 - 2)^2}{(4 - 2)^2} = \frac{2^2}{2^2} = \frac{4}{4} = 1$$ 8. **Evaluate $f(3)$:** Since $3 \geq 2$, use the second part: $$f(3) = \frac{(3^2 - 2)^2}{(3^2 - 2)^2} = \frac{(9 - 2)^2}{(9 - 2)^2} = \frac{7^2}{7^2} = \frac{49}{49} = 1$$ **Final answers:** - $f(-8)$ is not defined (no real value) - $f(-1)$ is not defined (no real value) - $f(0) = 0$ - $f(2) = 1$ - $f(3) = 1$