1. The problem asks to determine the nature of the composite function $ (g \circ f)(x) $ where $ f(x) = x^3 $ and $ g(x) = x + 1 $. 2. The composite function $ (g \circ f)(x) $ means $ g(f(x)) $. Substitute $ f(x) $ into $ g $: $$ (g \circ f)(x) = g(f(x)) = g(x^3) = x^3 + 1 $$ 3. Now analyze the properties: - Odd function: $ f(-x) = -f(x) $ - Even function: $ f(-x) = f(x) $ - One-to-one function: each input maps to a unique output. 4. Check if $ (g \circ f)(x) = x^3 + 1 $ is odd: $$ (g \circ f)(-x) = (-x)^3 + 1 = -x^3 + 1 $$ This is not equal to $ -(x^3 + 1) = -x^3 - 1 $, so it is not odd. 5. Check if it is even: $$ (g \circ f)(-x) = -x^3 + 1 \neq x^3 + 1 = (g \circ f)(x) $$ So it is not even. 6. Check if it is one-to-one: Since $ x^3 $ is one-to-one and adding 1 is a one-to-one operation, the composite function is one-to-one. 7. Therefore, the correct answer is (c) one-to-one. Final answer: The function $ (g \circ f)(x) = x^3 + 1 $ is one-to-one.