Is the function \( f: \mathbb{R} \rightarrow \mathbb{R}, f(x) = x^3 \) surjective?
Answer
Step 4: Since \( \sqrt[3]{y} \) is defined for all \( y \) in \( \mathbb{R} \), we can conclude that for every \( y \) in \( \mathbb{R} \), there exists an \( x \) in \( \mathbb{R} \) such that \( f(x) = y \).
Step 1 ：Step 1: To determine if a function is surjective, we need to show that for every element \( y \) in the codomain \( \mathbb{R} \), there exists an element \( x \) in the domain \( \mathbb{R} \) such that \( f(x) = y \).
Step 2 ：Step 2: Let's take an arbitrary \( y \) in \( \mathbb{R} \). We need to find an \( x \) such that \( f(x) = y \). So we solve the equation \( x^3 = y \) for \( x \).
Step 3 ：Step 3: By taking the cube root on both sides, we find that \( x = \sqrt[3]{y} \).
Step 4 ：Step 4: Since \( \sqrt[3]{y} \) is defined for all \( y \) in \( \mathbb{R} \), we can conclude that for every \( y \) in \( \mathbb{R} \), there exists an \( x \) in \( \mathbb{R} \) such that \( f(x) = y \).
