Let A = {x ϵ R | –1 ≤ x ≤ 1} and let f : A → A, g : A → A be two functions defined by f(x) = x2 and g(x) = sin πx/2. Show that g-1 exists but f-1 does not exist. Also, find g-1.
We have f : A → A where A = {x ϵ R | –1 ≤ x ≤ 1} defined by f(x) = x2.
Recall that a function is invertible only when it is both one-one and onto.
First, we will check if f is one-one.
Let x1, x2ϵ A (domain) such that f(x1) = f(x2)
⇒ x12 = x22
⇒ x12 – x22 = 0
⇒ (x1 – x2)(x1 + x2) = 0
⇒ x1 – x2 = 0 or x1 + x2 = 0
∴ x1 = ±x2
So, we have f(x1) = f(x2) ⇒ x1 = ±x2.
This means that two different elements of the domain are mapped to the same element by the function f.
For example, consider f(–1) and f(1).
We have f(–1) = (–1)2 = 1 and f(1) = 12 = 1 = f(–1)
Thus, f is not one-one and hence f-1 doesn’t exist.
Now, let us consider g : A → A defined by g(x) = sin
First, we will prove that g is one-one.
Let x1, x2ϵ A (domain) such that g(x1) = g(x2)
(in the given range)
∴ x1 = x2
So, we have g(x1) = g(x2) ⇒ x1 = x2.
Thus, function g is one-one.
Let y ϵ A (co-domain) such that g(x) = y
Clearly, for every y ϵ A, there exists x ϵ A (domain) such that g(x) = y and hence, function g is onto.
Thus, the function g has an inverse.
We have g(x) = y ⇒ x = g-1(y)
But, we found g(x) = y ⇒
Hence,
Thus, g(x) is invertible and