1. The problem involves analyzing and understanding the given functions and their properties. 2. Let's examine each function one by one. 3. For the polynomial function $$f(x) = 7x^6 + 3x^4 + x^2 - 4$$, this is a sum of powers of $x$ with coefficients. It is continuous and differentiable everywhere. 4. For the polynomial $$f(x) = x^0 + 7x^3 + 2$$, note that $x^0 = 1$ for all $x \neq 0$, so the function simplifies to $$f(x) = 1 + 7x^3 + 2 = 7x^3 + 3$$. 5. For the square root function $$f(x) = \sqrt{x + 3}$$, the domain is $$x + 3 \geq 0 \Rightarrow x \geq -3$$. 6. For the fourth root function $$f(x) = \sqrt[4]{4 - x^3}$$, the radicand must be nonnegative: $$4 - x^3 \geq 0 \Rightarrow x^3 \leq 4 \Rightarrow x \leq \sqrt[3]{4}$$. 7. For the cube root function $$f(x) = \sqrt[3]{x + 1}$$, the domain is all real numbers since cube roots are defined everywhere. 8. For the fifth root function $$f(x) = \sqrt[5]{x^2 + 2x - 1}$$, the radicand can be any real number because odd roots are defined for all real numbers. 9. For the rational function $$f(x) = \frac{4x^2 - 1}{\sqrt[3]{\sqrt{x} - 1}}$$, the denominator is $$\sqrt[3]{\sqrt{x} - 1}$$. 10. The cube root is defined for all real numbers, but $$\sqrt{x}$$ requires $$x \geq 0$$. 11. The denominator cannot be zero, so $$\sqrt{x} - 1 \neq 0 \Rightarrow \sqrt{x} \neq 1 \Rightarrow x \neq 1$$. 12. Therefore, the domain is $$x \geq 0$$ and $$x \neq 1$$. 13. The function can be graphed to observe intercepts and extrema. Final answers: - Polynomial functions are defined for all real $x$. - Square root and fourth root functions have domain restrictions based on radicand nonnegativity. - Cube and fifth root functions are defined for all real $x$. - Rational function domain excludes points making denominator zero and where square root is undefined.