1. **State the problem:** Find the domains of the functions $$f(x) = \sqrt[4]{x} - 9$$ and $$g(x) = \sqrt[3]{3x} - 9$$ and write the answers in interval notation. 2. **Recall domain rules for radicals:** - For an even root function like the fourth root $$\sqrt[4]{x}$$, the radicand (expression inside the root) must be \(\geq 0\) because even roots of negative numbers are not real. - For an odd root function like the cube root $$\sqrt[3]{3x}$$, the radicand can be any real number (no restriction). 3. **Find domain of $$f(x)$$:** - Since $$f(x) = \sqrt[4]{x} - 9$$, the radicand is $$x$$. - Set $$x \geq 0$$. - So, domain of $$f$$ is all real numbers $$x$$ such that $$x \geq 0$$. - In interval notation: $$[0, \infty)$$. 4. **Find domain of $$g(x)$$:** - Since $$g(x) = \sqrt[3]{3x} - 9$$, the radicand is $$3x$$. - Cube root allows all real numbers, so no restriction on $$3x$$. - Therefore, domain of $$g$$ is all real numbers $$(-\infty, \infty)$$. **Final answers:** - Domain of $$f$$: $$[0, \infty)$$ - Domain of $$g$$: $$(-\infty, \infty)$$