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) = \(\frac{4x}{4-3y}\)