Let `f:RrarrR` be defined by `f(x)=x^(2)+1`. Then pre-images of 17 and - 3 respectively, are:

Options

A. `phi,{4,-4}`
B. `{3,-3},phi`
C. `{4,-4},phi`
D. `{4,-4},{2,-2}`

Answer

To find the pre-images of the values 17 and -3 for the function \( f(x) = x^2 + 1 \), we will solve the equations \( f(x) = 17 \) and \( f(x) = -3 \) step by step.

### Step 1: Finding the pre-image of 17
We start with the equation:
\[ f(x) = 17 \]
Substituting the function:
\[ x^2 + 1 = 17 \]

### Step 2: Simplifying the equation
Subtract 1 from both sides:
\[ x^2 = 17 - 1 \]
\[ x^2 = 16 \]

### Step 3: Solving for x
Taking the square root of both sides gives:
\[ x = \pm 4 \]
Thus, the pre-images of 17 are \( x = 4 \) and \( x = -4 \).

### Step 4: Finding the pre-image of -3
Now we solve for the pre-image of -3:
\[ f(x) = -3 \]
Substituting the function:
\[ x^2 + 1 = -3 \]

### Step 5: Simplifying the equation
Subtract 1 from both sides:
\[ x^2 = -3 - 1 \]
\[ x^2 = -4 \]

### Step 6: Solving for x
Taking the square root of both sides gives:
\[ x = \pm \sqrt{-4} \]
This simplifies to:
\[ x = \pm 2i \]
These are imaginary numbers, indicating that there are no real pre-images for -3.

### Final Answer
The pre-images of 17 are \( 4 \) and \( -4 \), while the pre-images of -3 are imaginary: \( 2i \) and \( -2i \).