A function `f :Rto R` is defined as `f (x) =3x ^(2) +1.` then `f ^(-1)(x)` is :

Options

A. `(sqrt(x-1))/(3)`
B. `(1/2 sqrtx-1`
C. `f ^(-1)` does not exist
D. `sqrt((x-1)/(3))`

Answer

To find the inverse of the function \( f(x) = 3x^2 + 1 \), we need to determine if the function is one-to-one (injective) and onto (surjective).

### Step-by-Step Solution:

1. **Determine if the function is one-to-one**:
- A function is one-to-one if \( f(x_1) = f(x_2) \) implies \( x_1 = x_2 \).
- Let \( f(x_1) = f(x_2) \):
\[ 3x_1^2 + 1 = 3x_2^2 + 1 \]
- Subtracting 1 from both sides:
\[ 3x_1^2 = 3x_2^2 \]
- Dividing both sides by 3:
\[ x_1^2 = x_2^2 \]
- This implies:
\[ x_1 = x_2 \quad \text{or} \quad x_1 = -x_2 \]
- Since \( x_1 \) can be equal to \( -x_2 \), the function is not one-to-one.

2. **Determine if the function is onto**:
- A function is onto if for every \( y \) in the codomain, there exists an \( x \) in the domain such that \( f(x) = y \).
- The range of \( f(x) = 3x^2 + 1 \) is \( [1, \infty) \) since \( 3x^2 \) is always non-negative and the minimum value occurs at \( x = 0 \).
- Therefore, the function does not cover all real numbers (it only covers values starting from 1), which means it is not onto.

3. **Conclusion**:
- Since the function \( f(x) = 3x^2 + 1 \) is neither one-to-one nor onto, it does not have an inverse.

### Final Answer:
The inverse function \( f^{-1}(x) \) does not exist.