If a function `f:RtoR` is defined by `f(x)=x^(2)+1`, then pre-images of 17 and -3 respectively, are

Answer

To find the pre-images of the values 17 and -3 for the function \( f(x) = x^2 + 1 \), we will follow these steps:

### Step 1: Understand the concept of pre-image
The pre-image of a value \( y \) under a function \( f \) is the value(s) of \( x \) such that \( f(x) = y \). In mathematical terms, if \( y = f(x) \), then the pre-image is \( x = f^{-1}(y) \).

### Step 2: Set up the equation for pre-images
Given the function \( f(x) = x^2 + 1 \), we need to find \( x \) such that:
1. \( f(x) = 17 \)
2. \( f(x) = -3 \)

### Step 3: Solve for the first pre-image (for \( y = 17 \))
Set the equation:
\[ x^2 + 1 = 17 \]
Subtract 1 from both sides:
\[ x^2 = 17 - 1 \]
\[ x^2 = 16 \]
Now, take the square root of both sides:
\[ x = \pm \sqrt{16} \]
Thus, we find:
\[ x = \pm 4 \]
So, the pre-images of 17 are \( 4 \) and \( -4 \).

### Step 4: Solve for the second pre-image (for \( y = -3 \))
Set the equation:
\[ x^2 + 1 = -3 \]
Subtract 1 from both sides:
\[ x^2 = -3 - 1 \]
\[ x^2 = -4 \]
Since the square of a real number cannot be negative, this implies:
\[ x = \pm \sqrt{-4} \]
This results in imaginary numbers:
\[ x = \pm 2i \]
Thus, there are no real pre-images for \( -3 \), and we conclude that the pre-image is the empty set.

### Final Answer:
The pre-images of 17 are \( 4 \) and \( -4 \), while the pre-image of -3 is the empty set.

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