Given f : R → R and f(x) = x² + 1.

We need to find f^-1{17} and f^-1{–3}.

Let f^-1{17} = x

⇒ f(x) = 17

⇒ x² + 1 = 17

⇒ x² – 16 = 0

⇒ (x – 4)(x + 4) = 0

∴ x = ±4

Clearly, both –4 and 4 are elements of the domain R.

Thus, f^-1{17} = {–4, 4}

Now, let f^-1{–3} = x

⇒ f(x) = –3

⇒ x² + 1 = –3

⇒ x² = –4

However, the domain of f is R and for every real number x, the value of x² is non-negative.

Hence, there exists no real x for which x² = –4.

Thus, f^-1{–3} = ∅