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


Hence,


Thus, f(x) is invertible and


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