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 solution is available in R.
So $f^{-1}${- 3} = Φ .