1. The problem provides several piecewise functions and evaluates them at specific points. We will verify each. 2. Problem 49: Given $$f(x) = \begin{cases} x^2 + 2 & \text{if } x < 0 \\ x & \text{if } x \geq 0 \end{cases}$$ Evaluate: - $f(-3) = (-3)^2 + 2 = 9 + 2 = 11$ - $f(0) = 0$ - $f(2) = 2$ 3. Problem 50: Given $$f(x) = \begin{cases} 5 & \text{if } x < 2 \\ \frac{1}{2}x - 3 & \text{if } x \geq 2 \end{cases}$$ Evaluate: - $f(-3) = 5$ (since $-3 < 2$) - $f(0) = 5$ (since $0 < 2$) - $f(2) = \frac{1}{2} \times 2 - 3 = 1 - 3 = -2$ 4. Problem 51: Given $$f(x) = \begin{cases} x + 1 & \text{if } x \leq -1 \\ x^2 & \text{if } x > -1 \end{cases}$$ Evaluate: - $f(-3) = -3 + 1 = -2$ - $f(0) = 0^2 = 0$ - $f(2) = 2^2 = 4$ 5. Problem 52: Given $$f(x) = \begin{cases} -1 & \text{if } x \leq 1 \\ 7 - 2x & \text{if } x > 1 \end{cases}$$ Evaluate: - $f(-3) = -1$ - $f(0) = -1$ - $f(2) = 7 - 2 \times 2 = 7 - 4 = 3$ 6. Problem 53: Given the absolute value function $$|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}$$ Define $$f(x) = x + |x| = \begin{cases} x + x = 2x & \text{if } x \geq 0 \\ x + (-x) = 0 & \text{if } x < 0 \end{cases}$$ This means: - For $x \geq 0$, $f(x) = 2x$, a line with slope 2 - For $x < 0$, $f(x) = 0$, the x-axis 7. Graphs: - Problem 49: Parabola $y = x^2 + 2$ for $x < 0$, line $y = x$ for $x \geq 0$. - Problem 50: Horizontal line $y=5$ for $x < 2$, line $y=\frac{1}{2}x -3$ for $x \geq 2$ with open circle at $(2,5)$. - Problem 51: Line $y = x + 1$ for $x \leq -1$, parabola $y = x^2$ for $x > -1$, meeting at $(-1,0)$. - Problem 52: Horizontal $y=-1$ for $x \leq 1$, line $y=7-2x$ for $x > 1$, passing through $(1,5)$. - Problem 53: Line $y=2x$ for $x \geq 0$, line $y=0$ for $x < 0$. Final answers summarized: - 49: $f(-3)=11$, $f(0)=0$, $f(2)=2$ - 50: $f(-3)=5$, $f(0)=5$, $f(2)=-2$ - 51: $f(-3)=-2$, $f(0)=0$, $f(2)=4$ - 52: $f(-3)=-1$, $f(0)=-1$, $f(2)=3$ - 53: $f(x) = \begin{cases} 2x & x \geq 0 \\ 0 & x < 0 \end{cases}$