1. **State the problem:** Find the domains of the functions $$f(x) = \sqrt[4]{3x - 3}$$ and $$g(x) = \sqrt[3]{x + 3}$$ and express them in interval notation. 2. **Recall domain rules for radicals:** - For even roots (like the 4th root), the radicand (expression inside the root) must be \(\geq 0\) because we cannot take an even root of a negative number in the real number system. - For odd roots (like the cube root), the radicand can be any real number because odd roots of negative numbers are defined. 3. **Find the domain of $$f(x)$$:** - Since $$f(x)$$ is a 4th root, set the radicand $$3x - 3 \geq 0$$. - Solve the inequality: $$3x - 3 \geq 0$$ $$3x \geq 3$$ $$x \geq 1$$ - So, the domain of $$f$$ is all real numbers $$x$$ such that $$x \geq 1$$. - In interval notation, this is $$[1, \infty)$$. 4. **Find the domain of $$g(x)$$:** - Since $$g(x)$$ is a cube root, the radicand $$x + 3$$ can be any real number. - Therefore, the domain of $$g$$ is all real numbers. - In interval notation, this is $$(-\infty, \infty)$$. **Final answers:** - Domain of $$f$$: $$[1, \infty)$$ - Domain of $$g$$: $$(-\infty, \infty)$$