To determine which of the given functions from \( \mathbb{Z} \) (the set of integers) to \( \mathbb{Z} \) are bijective, we need to check each function for two properties: injectivity (one-to-one) and surjectivity (onto). A function is bijective if it is both injective and surjective.
### Step-by-Step Solution:
1. **Understanding Bijectivity**:
- A function \( f: A \to B \) is **injective** (one-to-one) if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \) for all \( x_1, x_2 \in A \).
- A function is **surjective** (onto) if for every \( y \in B \), there exists at least one \( x \in A \) such that \( f(x) = y \).
2. **Analyzing Each Function**:
**Option 1: \( f(x) = x^3 \)**
- **Injectivity**: Assume \( f(x_1) = f(x_2) \) implies \( x_1^3 = x_2^3 \). Taking cube roots gives \( x_1 = x_2 \). Thus, the function is injective.
- **Surjectivity**: For any integer \( y \), we need \( x^3 = y \). The cube root of an integer is not necessarily an integer (e.g., \( 3^{1/3} \)). Hence, not all integers can be expressed as \( x^3 \). Thus, the function is not surjective.
- **Conclusion**: Not bijective.
**Option 2: \( f(x) = x + 2 \)**
- **Injectivity**: Assume \( f(x_1) = f(x_2) \) implies \( x_1 + 2 = x_2 + 2 \). Cancelling \( 2 \) gives \( x_1 = x_2 \). Thus, the function is injective.
- **Surjectivity**: For any integer \( y \), we can find \( x \) such that \( f(x) = y \). Setting \( x + 2 = y \) gives \( x = y - 2 \), which is an integer. Thus, the function is surjective.
- **Conclusion**: Bijective.
**Option 3: \( f(x) = 2x + 1 \)**
- **Injectivity**: Assume \( f(x_1) = f(x_2) \) implies \( 2x_1 + 1 = 2x_2 + 1 \). Cancelling \( 1 \) gives \( 2x_1 = 2x_2 \), leading to \( x_1 = x_2 \). Thus, the function is injective.
- **Surjectivity**: For any integer \( y \), we need \( 2x + 1 = y \). Rearranging gives \( x = \frac{y - 1}{2} \). If \( y \) is even, \( x \) is an integer; if \( y \) is odd, \( x \) is not an integer. Hence, not all integers can be expressed this way. Thus, the function is not surjective.
- **Conclusion**: Not bijective.
**Option 4: \( f(x) = x^2 + 1 \)**
- **Injectivity**: Assume \( f(x_1) = f(x_2) \) implies \( x_1^2 + 1 = x_2^2 + 1 \). Cancelling \( 1 \) gives \( x_1^2 = x_2^2 \), which implies \( x_1 = x_2 \) or \( x_1 = -x_2 \). Thus, the function is not injective (e.g., \( f(1) = f(-1) = 2 \)).
- **Conclusion**: Not bijective.
3. **Final Conclusion**:
- The only bijective function among the options is **Option 2: \( f(x) = x + 2 \)**.
