1. **State the problem:** Find the domains of the functions $$f(x) = \sqrt[4]{x - 9}$$ and $$g(x) = \sqrt[3]{3x - 9}$$ and express them in interval notation. 2. **Recall domain rules for roots:** - For an even root (like the 4th root), the radicand (expression inside the root) must be \(\geq 0\) because even roots of negative numbers are not real. - For an odd root (like the cube root), the radicand can be any real number because odd roots of negative numbers are defined. 3. **Find domain of $$f(x)$$:** - Since $$f(x) = \sqrt[4]{x - 9}$$ is a 4th root, require: $$x - 9 \geq 0$$ - Solve inequality: $$x \geq 9$$ - So the domain of $$f$$ is all real numbers from 9 to infinity, including 9. - In interval notation: $$[9, \infty)$$ 4. **Find domain of $$g(x)$$:** - Since $$g(x) = \sqrt[3]{3x - 9}$$ is a cube root, the radicand can be any real number. - So no restrictions on $$x$$. - Domain of $$g$$ is all real numbers. - In interval notation: $$(-\infty, \infty)$$ **Final answers:** - Domain of $$f$$: $$[9, \infty)$$ - Domain of $$g$$: $$(-\infty, \infty)$$