1. The problem asks for the domain and range of the function $$f(x) = \sqrt{x + 7} - 2$$. 2. The domain of a square root function $$\sqrt{g(x)}$$ requires the expression inside the root to be non-negative: $$g(x) \geq 0$$. 3. For $$f(x)$$, set the inside of the root $$x + 7 \geq 0$$. 4. Solve for $$x$$: $$x + 7 \geq 0$$ $$x \geq -7$$ 5. So, the domain is all $$x$$ values greater than or equal to $$-7$$, or $$[-7, \infty)$$. 6. To find the range, consider the output values of $$f(x)$$. 7. The square root function $$\sqrt{x + 7}$$ outputs values $$\geq 0$$. 8. Since $$f(x) = \sqrt{x + 7} - 2$$, the smallest value occurs when $$\sqrt{x + 7} = 0$$, which is at $$x = -7$$. 9. Evaluate $$f(-7)$$: $$f(-7) = 0 - 2 = -2$$ 10. As $$x$$ increases, $$\sqrt{x + 7}$$ increases without bound, so $$f(x)$$ increases without bound. 11. Therefore, the range is $$[-2, \infty)$$. 12. The correct answer is option B: domain = $$[-7, \infty)$$ and range = $$[-2, \infty)$$.