To determine which of the functions from \( \mathbb{Z} \) to itself are bijections, we need to check if each function is both one-to-one (injective) and onto (surjective). A function is bijective if it satisfies both conditions.
### Step-by-Step Solution:
1. **Understanding the Functions**:
We have four functions to analyze. Each function is defined from \( \mathbb{Z} \) (the set of all integers) to \( \mathbb{Z} \).
2. **Analyzing the First Function: \( f(x) = x^3 \)**:
- **Injective**: We check if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
\[
x_1^3 = x_2^3 \implies x_1 = x_2 \quad \text{(true for integers)}
\]
- **Surjective**: We need to check if every integer \( y \) has a pre-image \( x \) such that \( f(x) = y \).
- For example, \( f(0) = 0 \), \( f(1) = 1 \), \( f(2) = 8 \), but there are integers (like 2, 3, 4, etc.) that cannot be expressed as \( x^3 \) for any integer \( x \).
- **Conclusion**: This function is not onto. Therefore, it is not bijective.
3. **Analyzing the Second Function: \( f(x) = x + 2 \)**:
- **Injective**: Check if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
\[
x_1 + 2 = x_2 + 2 \implies x_1 = x_2 \quad \text{(true)}
\]
- **Surjective**: For any integer \( y \), we can find \( x = y - 2 \) such that \( f(x) = y \).
- **Conclusion**: This function is both one-to-one and onto. Therefore, it is bijective.
4. **Analyzing the Third Function: \( f(x) = 2x + 1 \)**:
- **Injective**: Check if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
\[
2x_1 + 1 = 2x_2 + 1 \implies 2x_1 = 2x_2 \implies x_1 = x_2 \quad \text{(true)}
\]
- **Surjective**: For any integer \( y \), we can solve \( y = 2x + 1 \) to find \( x = \frac{y - 1}{2} \). However, \( x \) must be an integer, which is not guaranteed for all integers \( y \) (e.g., \( y = 2 \) gives \( x = \frac{1}{2} \)).
- **Conclusion**: This function is not onto. Therefore, it is not bijective.
5. **Analyzing the Fourth Function: \( f(x) = x^2 + 1 \)**:
- **Injective**: Check if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
\[
x_1^2 + 1 = x_2^2 + 1 \implies x_1^2 = x_2^2 \implies x_1 = x_2 \text{ or } x_1 = -x_2 \quad \text{(not injective)}
\]
- **Surjective**: The function can only produce values \( y \geq 1 \) (since \( x^2 \geq 0 \)). Thus, it cannot reach all integers (e.g., \( y = 0 \)).
- **Conclusion**: This function is neither injective nor surjective. Therefore, it is not bijective.
### Final Conclusion:
The only function that is a bijection from \( \mathbb{Z} \) to \( \mathbb{Z} \) is the second function, \( f(x) = x + 2 \).
