1. **Problem 1: Evaluate the piecewise function** Given: $$f(x) = \begin{cases} 2x + 1 & \text{for } x \leq 2 \\ -4 & \text{for } x > 2 \end{cases}$$ We need to find $f(5)$, $f(2)$, and $f(-1)$. 2. **Step 1: Evaluate $f(5)$** Since $5 > 2$, use the second case: $$f(5) = -4$$ 3. **Step 2: Evaluate $f(2)$** Since $2 \leq 2$, use the first case: $$f(2) = 2(2) + 1 = 4 + 1 = 5$$ 4. **Step 3: Evaluate $f(-1)$** Since $-1 \leq 2$, use the first case: $$f(-1) = 2(-1) + 1 = -2 + 1 = -1$$ --- 5. **Problem 2: Evaluate the piecewise function** Given: $$g(x) = \begin{cases} 3 - 2x & \text{for } x < -3 \\ x^2 & \text{for } x > -3 \end{cases}$$ We need to find $g(-4)$, $g(0)$, and $g(3)$. 6. **Step 1: Evaluate $g(-4)$** Since $-4 < -3$, use the first case: $$g(-4) = 3 - 2(-4) = 3 + 8 = 11$$ 7. **Step 2: Evaluate $g(0)$** Since $0 > -3$, use the second case: $$g(0) = 0^2 = 0$$ 8. **Step 3: Evaluate $g(3)$** Since $3 > -3$, use the second case: $$g(3) = 3^2 = 9$$ --- **Final answers:** - $f(5) = -4$, $f(2) = 5$, $f(-1) = -1$ - $g(-4) = 11$, $g(0) = 0$, $g(3) = 9$