If f : R → R be defined by f(x) = x2 + 1, then find f-1{17} and f-1{

1.	If f : R → R be defined by f(x) = x2 + 1, then find f-1{17} and f-1{–3}.
Answer» 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} = ∅

If f : R → R be defined by f(x) = x2 + 1, then find f-1{17} and f-1{–3}.

Answer»

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} = ∅