Which of the following functions is a surjection?

Options

A. `f:R to R" given by "f(x)=x^(3)+2"for all x"in R`
B. `g:R to R" given by "g(x)=x^(2)+2"for all x"in R`
C. `h:Z to Z" given by "h(x)=3x+2"for all x"in Z`
D. `phi: R to R" given by "f(x)=x^(2)-3x+2"for all x"in R`

Explanation

To determine which of the given functions is a surjection, we need to check if the image of each function is equal to its codomain. A function \( f: A \to B \) is a surjection if for every element \( b \) in \( B \), there exists at least one element \( a \) in \( A \) such that \( f(a) = b \).

Let's analyze each function step by step.

### Step 1: Analyze the first function \( f(x) = x^3 + 2 \)

1. **Set the function equal to \( y \)**:
\[ y = x^3 + 2 \]
Rearranging gives:
\[ x^3 = y - 2 \]
Therefore:
\[ x = (y - 2)^{1/3} \]

2. **Determine the range**:
Since \( x \) can take any real value, \( y \) can take any value in \( \mathbb{R} \) as \( x^3 \) can produce all real numbers. Thus, the image of \( f \) is all of \( \mathbb{R} \).

3. **Conclusion**:
The function \( f(x) = x^3 + 2 \) is a surjection.

### Step 2: Analyze the second function \( g(x) = x^2 + 2 \)

1. **Set the function equal to \( y \)**:
\[ y = x^2 + 2 \]
Rearranging gives:
\[ x^2 = y - 2 \]
Therefore:
\[ x = \sqrt{y - 2} \quad \text{or} \quad x = -\sqrt{y - 2} \]

2. **Determine the range**:
The expression \( y - 2 \) must be non-negative for \( x \) to be real, which means:
\[ y \geq 2 \]
Thus, the image of \( g \) is \( [2, \infty) \).

3. **Conclusion**:
Since not all real numbers can be achieved (e.g., \( y = 1 \) is not in the image), the function \( g(x) = x^2 + 2 \) is not a surjection.

### Step 3: Analyze the third function \( h(x) = 3x + 2 \)

1. **Set the function equal to \( y \)**:
\[ y = 3x + 2 \]
Rearranging gives:
\[ 3x = y - 2 \]
Therefore:
\[ x = \frac{y - 2}{3} \]

2. **Determine the range**:
Since \( x \) must be an integer (as \( h: \mathbb{Z} \to \mathbb{Z} \)), \( y - 2 \) must be a multiple of 3 for \( x \) to be an integer. Therefore, not all integers can be achieved (e.g., \( y = 0 \) gives \( x = -\frac{2}{3} \), which is not an integer).

3. **Conclusion**:
The function \( h(x) = 3x + 2 \) is not a surjection.

### Step 4: Analyze the fourth function \( \phi(x) = x^2 - 3x + 2 \)

1. **Set the function equal to \( y \)**:
\[ y = x^2 - 3x + 2 \]
Rearranging gives:
\[ x^2 - 3x + (2 - y) = 0 \]

2. **Use the quadratic formula**:
The solutions for \( x \) are given by:
\[ x = \frac{3 \pm \sqrt{9 - 4(1)(2 - y)}}{2} \]
Simplifying gives:
\[ x = \frac{3 \pm \sqrt{1 + 4y}}{2} \]

3. **Determine the range**:
For \( x \) to be real, the discriminant must be non-negative:
\[ 1 + 4y \geq 0 \implies y \geq -\frac{1}{4} \]
However, since the function is quadratic and opens upwards, it can achieve a maximum value at its vertex, which limits the range.

4. **Conclusion**:
The function \( \phi(x) = x^2 - 3x + 2 \) is not a surjection.

### Final Conclusion
The only function that is a surjection is:
- **Option 1: \( f(x) = x^3 + 2 \)**