1. Understand the problem:
We are given the function \( f(x) = x^2 + 1 \) and need to find the pre-images of 17 and -3, i.e., the set of all \( x \) such that \( f(x) = 17 \) and \( f(x) = -3 \).
2. Find the pre-image of 17:
Solve \( x^2 + 1 = 17 \):
\[ x^2 = 16 \implies x = \pm 4 \]
Thus, the pre-image is \( \{4, -4\} \).
3. Find the pre-image of -3:
Solve \( x^2 + 1 = -3 \):
\[ x^2 = -4 \]
This has no real solutions, so the pre-image is the empty set \( \phi \).
Correct Answer: (C) \(\{4, -4\}, \phi\)
For $f(x) = 17$: \[ x^2 + 1 = 17 \implies x^2 = 16 \implies x = \pm 4 \implies \{4, -4\}. \] For $f(x) = -3$: \[ x^2 + 1 = -3 \implies x^2 = -4, \] which has no real solutions. Hence, the pre-image is $\phi$.