Q. 165.0( 3 Votes )
Let f : R – → R be a function defined as. Show that f : R – → range(f) is one-one and onto. Hence, find f-1.
We have f : R – → R and
We need to prove f : R – → range(f) is invertible.
First, we will prove that f is one-one.
Let x1, x2ϵ A (domain) such that f(x1) = f(x2)
⇒ (4x1)(3x2 + 4) = (3x1 + 4)(4x2)
⇒ 12x1x2 + 16x1 = 12x1x2 + 16x2
⇒ 16x1 = 16x2
∴ 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 range(f) (co-domain) such that f(x) = y
⇒ 4x = 3xy + 4y
⇒ 4x – 3xy = 4y
⇒ x(4 – 3y) = 4y
Clearly, for every y ϵ range(f), there exists x ϵ A (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 ⇒
Thus, f(x) is invertible and
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