If f : R <span la

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} =

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