Let f : [–1, ∞) → [–1, ∞) is given by f(x) = (x + 1)2 – 1. Show that f is invertible. Also, find the set S = {x: f(x) = f-1(x)}
We have f : [–1, ∞) → [–1, ∞) and f(x) = (x + 1)2 – 1
Recall that a function is invertible only when it is both one-one and onto.
First, we will prove that f is one-one.
Let x1, x2ϵ [–1, ∞) (domain) such that f(x1) = f(x2)
⇒ (x1 + 1)2 – 1 = (x2 + 1)2 – 1
⇒ (x1 + 1)2 = (x2 + 1)2
⇒ x12 + 2x1 + 1 = x22 + 2x2 + 1
⇒ x12 + 2x1 = x22 + 2x2
⇒ x12 – x22 + 2x1 – 2x2 = 0
⇒ (x12 – x22) + 2(x1 – x2) = 0
⇒ (x1 – x2)(x1 + x2) + 2(x1 – x2) = 0
⇒ (x1 – x2)[x1 + x2 + 2] = 0
⇒ x1 – x2 = 0 (as x1, x2ϵ R+)
∴ x1 = x2
So, we have f(x1) = f(x2) ⇒ x1 = x2.
Thus, function f is one-one.
Now, we will prove that f is onto.
Let y ϵ [–1, ∞) (co-domain) such that f(x) = y
⇒ (x + 1)2 – 1 = y
⇒ (x + 1)2 = y + 1
Clearly, for every y ϵ [–1, ∞), there exists x ϵ [–1, ∞) (domain) such that f(x) = y and hence, function f is onto.
Thus, the function f has an inverse.
We have f(x) = y ⇒ x = f-1(y)
But, we found f(x) = y ⇒ 
Hence, 
Thus, f(x) is invertible and 
Now, we need to find the values of x for which f(x) = f-1(x).
We have f(x) = f-1(x)
We can write 
On substituting
, we get
t4 = t
⇒ t4 – t = 0
⇒ t (t3 – 1) = 0
⇒ t (t – 1)(t2 + t + 1) = 0
t2 + t + 1 ≠ 0 because this equation has no real root t.
⇒ t = 0 or t – 1 = 0
⇒ t = 0 or t = 1
Case – I: t = 0
⇒ x + 1 = 0
∴ x = –1
Case – II: t = 1
⇒ x + 1 = 1
∴ x = 0
Thus, S = {0, –1}