1. **State the problem:** Find the domain and range of the function $$f(x) = \sqrt{7x^2 + 25} + 9$$. 2. **Domain:** The domain of a function involving a square root requires the expression inside the root to be non-negative: $$7x^2 + 25 \geq 0$$ Since $$7x^2 \geq 0$$ for all real $$x$$ and $$25 > 0$$, the sum $$7x^2 + 25$$ is always positive. 3. Therefore, the domain is all real numbers: $$\text{Domain: } x \in \mathbb{R}$$ 4. **Range:** The square root function outputs values $$\geq 0$$. The minimum value inside the root is when $$x=0$$: $$7(0)^2 + 25 = 25$$ So the minimum value of the square root term is: $$\sqrt{25} = 5$$ 5. Adding 9 to this minimum gives the minimum value of $$f(x)$$: $$5 + 9 = 14$$ 6. Since $$\sqrt{7x^2 + 25}$$ increases without bound as $$|x|$$ increases, the range is all values $$y$$ such that: $$y \geq 14$$ **Final answers:** Domain: $$x \in \mathbb{R}$$ Range: $$y \geq 14$$