1. **State the problem:** Determine which of the given functions are one-to-one functions. 2. **Recall the definition:** A function $f$ is one-to-one (injective) if for every $a$ and $b$ in the domain, $f(a) = f(b)$ implies $a = b$. 3. **Analyze each function:** - $f(x) = x^3 - 15$ - The cubic function $x^3$ is strictly increasing and one-to-one. - Subtracting 15 shifts the graph but does not affect injectivity. - Therefore, $f(x) = x^3 - 15$ is one-to-one. - $f(x) = (5x + 1)^3 + 6$ - The inner function $5x + 1$ is linear and one-to-one. - Cubing preserves strict monotonicity. - Adding 6 shifts the graph but does not affect injectivity. - Therefore, $f(x) = (5x + 1)^3 + 6$ is one-to-one. - $f(x) = \sqrt[3]{x + 9}$ - Cube root function is strictly increasing and one-to-one. - Adding 9 inside shifts the graph but does not affect injectivity. - Therefore, $f(x) = \sqrt[3]{x + 9}$ is one-to-one. - $f(x) = \sqrt[3]{2x^2 + 7}$ - The expression inside the cube root, $2x^2 + 7$, is not one-to-one because $x^2$ is even: $f(-a) = f(a)$. - Cube root is one-to-one, but composition with a non-injective function is not one-to-one. - Therefore, $f(x) = \sqrt[3]{2x^2 + 7}$ is not one-to-one. 4. **Final answer:** The one-to-one functions are: $$ \boxed{f(x) = x^3 - 15, \quad f(x) = (5x + 1)^3 + 6, \quad f(x) = \sqrt[3]{x + 9}} $$ and the function $$ f(x) = \sqrt[3]{2x^2 + 7} $$ is not one-to-one.