If `f: R to R` is defined by `f (x)= x^2 + 1`, then values of `f^(-1) (17) and f^(-1)(-3)` respectively are

Answer

To solve the problem, we need to find the values of \( f^{-1}(17) \) and \( f^{-1}(-3) \) for the function \( f(x) = x^2 + 1 \).

### Step-by-Step Solution:

1. **Define the function**: We start with the function given in the problem:
\[ f(x) = x^2 + 1 \]

2. **Set \( f(x) \) equal to \( y \)**: To find the inverse, we set:
\[ y = f(x) = x^2 + 1 \]

3. **Solve for \( x \) in terms of \( y \)**: Rearranging the equation gives:
\[ y - 1 = x^2 \]
Taking the square root of both sides, we have:
\[ x = \sqrt{y - 1} \quad \text{(since we consider the principal square root)} \]

4. **Define the inverse function**: Thus, the inverse function is:
\[ f^{-1}(y) = \sqrt{y - 1} \]

5. **Calculate \( f^{-1}(17) \)**:
\[ f^{-1}(17) = \sqrt{17 - 1} = \sqrt{16} = 4 \]

6. **Calculate \( f^{-1}(-3) \)**:
\[ f^{-1}(-3) = \sqrt{-3 - 1} = \sqrt{-4} = \sqrt{-1 \cdot 4} = 2i \]
Since \( 2i \) is not a real number, we conclude that \( f^{-1}(-3) \) does not exist in the real number system.

### Final Answers:
- \( f^{-1}(17) = 4 \)
- \( f^{-1}(-3) \) does not exist.