Question

If f : R → R be defined by f(x) = x² + 1, then find f⁻¹ [17] and f⁻¹ [−3].

 

Answer 1

If f : A → B is such that y ∈ B, then \[f^{- 1}\] { y }={x ∈ A: f (x) = y}.
In other words, f ^-1{ y} is the set of pre - images of  y.
Let

\[f^{- 1}\] {17} = x .
Then, f (x) =17 .
⇒ x² +1 = 17
⇒ x² = 17 -1 = 16
⇒ x = ± 4
∴ \[f^{- 1}\] {17} = { -4,4} 

Again,

let

\[f^{- 1}\] { -3} = x .
Then, f (x) =-3
⇒ x² + 1 = -3
⇒ x² =- 3 - 1 = -4 

⇒ \[x = \sqrt{- 4}\]

Clearly, no soluti on is available in R.
So

\[f^{- 1}\] {- 3} = Φ .