1. **State the problem:** Find the range of the function $$f(x) = 2 - \sqrt{-x^2 + 2x + 15}$$. 2. **Understand the domain restriction:** The expression inside the square root must be non-negative: $$-x^2 + 2x + 15 \geq 0$$. 3. **Rewrite the inequality:** Multiply both sides by -1 (remember to reverse inequality): $$x^2 - 2x - 15 \leq 0$$. 4. **Factor the quadratic:** $$x^2 - 2x - 15 = (x - 5)(x + 3)$$. 5. **Determine the domain:** The quadratic is less than or equal to zero between its roots: $$-3 \leq x \leq 5$$. 6. **Find the range:** Since $$f(x) = 2 - \sqrt{-x^2 + 2x + 15}$$, the square root term is always non-negative, so: $$f(x) \leq 2$$. 7. **Find minimum value of $$f(x)$$:** The maximum of the square root term occurs at the vertex of the quadratic inside the root. 8. **Vertex of $$-x^2 + 2x + 15$$:** $$x = -\frac{b}{2a} = -\frac{2}{2 \times (-1)} = 1$$. 9. **Evaluate inside root at $$x=1$$:** $$-1^2 + 2(1) + 15 = -1 + 2 + 15 = 16$$. 10. **Maximum square root value:** $$\sqrt{16} = 4$$. 11. **Minimum $$f(x)$$ value:** $$2 - 4 = -2$$. 12. **Range conclusion:** $$-2 \leq f(x) \leq 2$$. **Final answer:** The range of $$f(x)$$ is $$[-2, 2]$$.