Q. 19

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)}

Answer :

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}


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