If F:Q `rarr` Q is given by f(x) =`x^(2)` then

Options

A. `f^(-1)(9)={-3,3}`
B. `f^(-1)(-5)=phi`
C. `f^(-1)(0)={0}`
D. `f^(-1)(16)={-4,4}`

Answer

To solve the problem, we need to analyze the function \( f: \mathbb{Q} \rightarrow \mathbb{Q} \) defined by \( f(x) = x^2 \) and find its inverse.

### Step-by-Step Solution:

1. **Understanding the Function**:
The function \( f(x) = x^2 \) takes a rational number \( x \) and maps it to its square. Since the domain and codomain are both the set of rational numbers \( \mathbb{Q} \), we need to find the inverse function \( f^{-1}(y) \).

**Hint**: Identify the relationship between \( x \) and \( y \) in the function.

2. **Setting Up the Equation**:
To find the inverse, we set \( y = f(x) = x^2 \). We need to express \( x \) in terms of \( y \).

**Hint**: Rearrange the equation to isolate \( x \).

3. **Solving for \( x \)**:
From the equation \( y = x^2 \), we can take the square root of both sides:
\[ x = \pm \sqrt{y} \]
This means that the inverse function can be expressed as:
\[ f^{-1}(y) = \pm \sqrt{y} \]

**Hint**: Remember that the square root function has both positive and negative roots.

4. **Evaluating Specific Values**:
Now we can evaluate the inverse function for specific values:
- For \( f^{-1}(9) \):
\[ f^{-1}(9) = \pm \sqrt{9} = \pm 3 \]
- For \( f^{-1}(5) \):
\[ f^{-1}(5) = \pm \sqrt{5} \]
Since \( \sqrt{5} \) is not a rational number, we note that \( f^{-1}(5) \) does not exist in \( \mathbb{Q} \).
- For \( f^{-1}(0) \):
\[ f^{-1}(0) = \pm \sqrt{0} = 0 \]
- For \( f^{-1}(16) \):
\[ f^{-1}(16) = \pm \sqrt{16} = \pm 4 \]

**Hint**: Substitute the values into the inverse function to find the results.

5. **Conclusion**:
From our calculations:
- \( f^{-1}(9) = \pm 3 \) (valid)
- \( f^{-1}(5) \) does not exist in \( \mathbb{Q} \) (not valid)
- \( f^{-1}(0) = 0 \) (valid)
- \( f^{-1}(16) = \pm 4 \) (valid)

Thus, the valid outputs for the inverse function are confirmed, and we conclude that the options regarding \( f^{-1} \) are evaluated correctly.