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

Hence,

Thus, f(x) is invertible and

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