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Show that the function f in A=R-{2/3} defined as - Mathematics

Show that the function f in `A=R-{2/3} ` defined as `f(x)=(4x+3)/(6x-4)` is one-one and onto hence find f-1

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Solution

The given function is

`f(x)=(4x+3)/(6x-4)`

`Let f(x_1)=f(x_2)`

`(4x_1+3)/(6x_1-4)=(4x_2+3)/(6x_2-4)`

⇒ 24x1x2 − 16x­1 + 18x2 − 12 = 24x1x2 + 18x1 − 16x2 − 12

⇒18x2 + 16x2 = 18x1 + 16x1

⇒34x2 = 34x1x1= x2

Therefore f(x) is one − one.

Since, `(4x+3)/(6x-4) ` is a real number, therefore, for every y in the co–domain (f), there exists a number x in `R-{2/3}` such that

`f(x)=y=(4x+3)/(6x-4)`

Therefore, f(x) is onto

Now let `y=(4x+3)/(6x-4)`

`6xy-4y=4x+3`

`x=(4y+3)/(6y-4)`

`f^(-1)(x)=(4x+3)/(6x-4)`

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