1. **State the problem:** Find the global maximum of the function $$f(x) = 4 + 6x - 3x^2$$. 2. **Formula and rules:** To find the global maximum of a quadratic function $$f(x) = ax^2 + bx + c$$ where $$a < 0$$, the vertex represents the global maximum point. The vertex $$x$$-coordinate is given by $$x = -\frac{b}{2a}$$. 3. **Identify coefficients:** Here, $$a = -3$$, $$b = 6$$, and $$c = 4$$. 4. **Calculate vertex $$x$$-coordinate:** $$x = -\frac{6}{2 \times (-3)} = -\frac{6}{-6} = 1$$. 5. **Calculate vertex $$y$$-coordinate (maximum value):** $$f(1) = 4 + 6(1) - 3(1)^2 = 4 + 6 - 3 = 7$$. 6. **Conclusion:** The global maximum of the function is $$7$$ at $$x = 1$$.