1. **Stating the problem:** We have a piecewise function defined as: $$f(x) = \begin{cases} x^2 & x \leq 0 \\ \sqrt{x} & 0 < x < 4 \\ \frac{x}{2} & x \geq 4 \end{cases}$$ We need to identify the correct graph among options A to E that matches this function. 2. **Understanding domain and continuity at boundaries:** - For $x \leq 0$, $f(x) = x^2$ is a parabola with a **closed circle at $x=0$** because the domain includes $0$. - For $0 < x < 4$, $f(x) = \sqrt{x}$ is defined only for strictly greater than 0 and less than 4, so there are **open circles at $x=0$ and $x=4$**. - For $x \geq 4$, $f(x) = \frac{x}{2}$ is a line starting at $x=4$ with a **closed circle at $x=4$**. 3. **Evaluating function values at boundary points:** - At $x=0$, from the parabola side: $f(0) = 0^2 = 0$ (closed circle). - At $x=0$, from the square root side: $f(0) = \sqrt{0} = 0$ but domain excludes 0, so open circle. - At $x=4$, from the square root side: $f(4) = \sqrt{4} = 2$ (open circle). - At $x=4$, from the line side: $f(4) = \frac{4}{2} = 2$ (closed circle). 4. **Checking continuity:** - At $x=0$, the parabola and square root meet at 0, but since the square root excludes 0, the graph has a closed circle from parabola and open circle from square root. - At $x=4$, the square root and line meet at 2, with an open circle from square root and closed circle from line. 5. **Matching with graphs:** - Graph B shows: - Parabola continuous with closed circle at $x=0$. - Square root curve continuous between 0 and 4 with open circles at 0 and 4. - Line starting at $x=4$ with closed circle at 2. 6. **Conclusion:** Graph B correctly represents the piecewise function $f(x)$ with the correct domain inclusions and boundary points. **Final answer:** The correct graph is **Graph B**.