1. **Problem Statement:** Given two functions $f(x) = x^3$ and $g(x) = x + 1$, find the nature of the composite function $(g \circ f)(x) = g(f(x))$. 2. **Formula and Explanation:** The composite function $(g \circ f)(x)$ means applying $f$ first, then $g$ to the result. So, $$ (g \circ f)(x) = g(f(x)) = g(x^3) = x^3 + 1. $$ 3. **Check if the function is odd:** A function $h(x)$ is odd if $h(-x) = -h(x)$. Calculate: $$ (g \circ f)(-x) = (-x)^3 + 1 = -x^3 + 1. $$ Compare with $-(g \circ f)(x) = -(x^3 + 1) = -x^3 - 1$. Since $-x^3 + 1 \neq -x^3 - 1$, the function is **not odd**. 4. **Check if the function is even:** A function $h(x)$ is even if $h(-x) = h(x)$. We have: $$ (g \circ f)(-x) = -x^3 + 1, \quad (g \circ f)(x) = x^3 + 1. $$ Since $-x^3 + 1 \neq x^3 + 1$ for all $x$, the function is **not even**. 5. **Check if the function is one-to-one:** A function is one-to-one if each $y$ value corresponds to exactly one $x$ value. The function is: $$ y = x^3 + 1. $$ Since $x^3$ is strictly increasing and adding 1 shifts the graph vertically, the function remains strictly increasing and thus one-to-one. 6. **Check if the function is not one-to-one:** From above, it is one-to-one, so this option is false. **Final answer:** The function $(g \circ f)(x) = x^3 + 1$ is **one-to-one**.