1. **Problem Statement:** Find the domain and range of the function $$f(x) = \sqrt{7x^2 + 25} + 9$$. 2. **Understanding the function:** The function involves a square root, so the expression inside the root must be non-negative for real values of $$f(x)$$. 3. **Domain:** Set the radicand $$7x^2 + 25 \geq 0$$. Since $$7x^2 \geq 0$$ for all real $$x$$ and $$25 > 0$$, the sum $$7x^2 + 25$$ is always positive. Therefore, the domain is all real numbers: $$\text{Domain} = \mathbb{R}$$. 4. **Range:** The square root function outputs values $$\geq 0$$. Minimum value inside the root is when $$x=0$$: $$7(0)^2 + 25 = 25$$. So, minimum of $$\sqrt{7x^2 + 25}$$ is $$\sqrt{25} = 5$$. Adding 9 shifts the range up by 9, so minimum value of $$f(x)$$ is $$5 + 9 = 14$$. As $$x$$ grows large, $$7x^2$$ dominates, so $$\sqrt{7x^2 + 25} \to \infty$$. Hence, $$f(x) \to \infty$$ as $$x \to \pm \infty$$. Therefore, the range is $$f(x) \geq 14$$. 5. **Final answers:** $$\text{Domain} = \mathbb{R}$$ $$\text{Range} = f(x) \geq 14$$