1. **State the problem:** Find the range of the function $$f(x) = 3 - 2x^2$$ for $$x \leq 2$$. 2. **Analyze the function:** This is a quadratic function with a negative leading coefficient (-2), so the parabola opens downward. 3. **Find the vertex:** The vertex of $$f(x) = 3 - 2x^2$$ is at $$x=0$$ because the function is in the form $$f(x) = a(x-h)^2 + k$$ with $$h=0$$ and $$k=3$$. 4. **Calculate the vertex value:** $$f(0) = 3 - 2(0)^2 = 3$$. This is the maximum value of the function since the parabola opens downward. 5. **Evaluate the function at the boundary $$x=2$$:** $$f(2) = 3 - 2(2)^2 = 3 - 8 = -5$$. 6. **Determine the range:** Since the domain is $$x \leq 2$$, the function values range from the minimum at $$x=2$$ to the maximum at the vertex $$x=0$$. Therefore, the range is $$[-5, 3]$$. **Final answer:** The range of $$f(x)$$ for $$x \leq 2$$ is $$[-5, 3]$$.