The given function is $f(x) = \frac{1}{2}x^2$ with the domain $\{-2, 0, 4\}$. To find the range, we need to evaluate the function at each value in the domain:

1. For $x = -2$: $f(-2) = \frac{1}{2}(-2)^2 = \frac{1}{2} \times 4 = 2$

2. For $x = 0$: $f(0) = \frac{1}{2}(0)^2 = 0$

3. For $x = 4$: $f(4) = \frac{1}{2}(4)^2 = \frac{1}{2} \times 16 = 8$

Thus, the range of the function is $\{0, 2, 8\}$.

The correct answer is: $\{0, 2, 8\}$.

Would you like more details or have any questions? Here are five related questions for further practice:

1. How do you determine the range of a quadratic function in general?
2. What is the significance of the domain in defining a function’s range?
3. How does the factor $\frac{1}{2}$ affect the shape of the parabola?
4. Can the range of a quadratic function ever include negative numbers?
5. How would the range change if the domain was extended to include more values?

Tip: Always evaluate the function for each value in the domain to find the corresponding range values.