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Show that the function f on \(A=R-\{\frac{-4}{3}\}\)into itself, defined by \(f(x)=\frac{4x}{(3x+4)}\) is one-one and onto. Hence, find f-1.

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To Show: that f is one-one and onto

To Find: Inverse of f

[NOTE: Any functions is invertible if and only if it is bijective functions (i.e. one-one and onto)]

one-one function: A function f : A → B is said to be a one-one function or injective mapping if different elements of A have different images in B. Thus for x1, x2 ∈ A & f(x1), f(x2) ∈ B, f(x1) = f(x2) ↔ x1= x2 or x1 x2 ↔ f(x1) ≠ f(x2)

onto function: If range = co-domain then f(x) is onto functions.

So, We need to prove that the given function is one-one and onto

Let x1, x2 ∈ Q and f(x) = \(\frac{4x}{(3x+4)}\).So f(x1) = f(x2) → \(\frac{(4x_1)}{(3x_1+4)}=\frac{(4x_2)}{(3x_2+4)}\) → = on solving we get x1=x2

So f(x1) = f(x2) ↔ x1= x2, f(x) is one-one

Given co-domain of f(x) is R except 3x+4=0.

Let y = f(x) = \(\frac{(4x)}{(4+3y)}\) So x = \(\frac{4x}{4-3y}\) [Range of f(x) = Domain of y]

So Domain of y is R = Range of f(x)

Hence, Range of f(x) = co-domain of f(x) = R except 3x+4=0

So, f(x) is onto function

As it is bijective function. So it is invertible

Invers of f(x) is f-1(y) = \(\frac{4x}{4-3y}\)

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