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)}

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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