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 $\phi$.