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Question
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 = 34x1⇒x1= 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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