1. **Problem:** Find the range of the function $$f(x) = \sqrt{x^2 + 2x + 5}$$. 2. **Formula and rules:** The function inside the square root must be non-negative for real values, but since it is a quadratic plus a constant, it is always positive. The range of $$f(x)$$ is all values $$y$$ such that $$y = \sqrt{g(x)}$$ where $$g(x) = x^2 + 2x + 5$$. 3. **Intermediate work:** Complete the square for $$g(x)$$: $$ g(x) = x^2 + 2x + 5 = (x+1)^2 + 4 $$ Since $$(x+1)^2 \geq 0$$ for all $$x$$, the minimum value of $$g(x)$$ is 4. 4. **Range of $$f(x)$$:** Since $$f(x) = \sqrt{g(x)}$$, the minimum value of $$f(x)$$ is $$\sqrt{4} = 2$$. 5. **Maximum value:** As $$x \to \pm \infty$$, $$(x+1)^2 \to \infty$$, so $$f(x) \to \infty$$. 6. **Final answer:** The range of $$f(x)$$ is $$[2, \infty)$$.