A function `f: R to R` is defined as`f(x)=x^(2)+1`. Evaluate each of the following:
`(i) f^(-1)(-2) " "(ii) f^(-1)(17)`
`(iii) f^(-1) (26,50)" "(iv) f^(-1)(0)`

Answer

To solve the given problem, we need to evaluate the inverse function \( f^{-1}(y) \) for the function \( f(x) = x^2 + 1 \).

### Step 1: Find the inverse function \( f^{-1}(y) \)

1. Start with the equation \( y = f(x) = x^2 + 1 \).
2. Rearranging gives us \( x^2 = y - 1 \).
3. Taking the square root of both sides, we find:
\[ x = \pm \sqrt{y - 1} \]
4. Thus, the inverse function is:
\[ f^{-1}(y) = \pm \sqrt{y - 1} \]

### Step 2: Evaluate \( f^{-1}(-2) \)

1. Substitute \( y = -2 \) into the inverse function:
\[ f^{-1}(-2) = \pm \sqrt{-2 - 1} = \pm \sqrt{-3} \]
2. Since the square root of a negative number is not defined in the real numbers, we conclude:
\[ f^{-1}(-2) \text{ is undefined (null)} \]

### Step 3: Evaluate \( f^{-1}(17) \)

1. Substitute \( y = 17 \) into the inverse function:
\[ f^{-1}(17) = \pm \sqrt{17 - 1} = \pm \sqrt{16} \]
2. This simplifies to:
\[ f^{-1}(17) = \pm 4 \]

### Step 4: Evaluate \( f^{-1}(26) \) and \( f^{-1}(50) \)

1. For \( f^{-1}(26) \):
\[ f^{-1}(26) = \pm \sqrt{26 - 1} = \pm \sqrt{25} = \pm 5 \]

2. For \( f^{-1}(50) \):
\[ f^{-1}(50) = \pm \sqrt{50 - 1} = \pm \sqrt{49} = \pm 7 \]

### Step 5: Evaluate \( f^{-1}(0) \)

1. Substitute \( y = 0 \) into the inverse function:
\[ f^{-1}(0) = \pm \sqrt{0 - 1} = \pm \sqrt{-1} \]
2. Since the square root of a negative number is not defined in the real numbers, we conclude:
\[ f^{-1}(0) \text{ is undefined (null)} \]

### Summary of Results:
- \( f^{-1}(-2) \) is undefined.
- \( f^{-1}(17) = \pm 4 \).
- \( f^{-1}(26) = \pm 5 \).
- \( f^{-1}(50) = \pm 7 \).
- \( f^{-1}(0) \) is undefined.