Show that the function f in A= R-\left \{ \frac{2}{3} \right \}\, defined\: as\: f\left ( x \right )= \frac{4x+3}{6x-4}  is one-one and onto. Hence .find f^{-1}\cdot

 

 

 

 
 
 
 
 

Answers (1)

Let \alpha ,\beta \; \epsilon A\; \; and\; \; f\left ( \alpha \right )= f\left ( \beta \right )
\therefore \frac{4\alpha +3}{6\alpha -4}= \frac{4\beta +3}{6\beta -4}
\Rightarrow 24\alpha \beta -16\alpha +18\beta -12= 24\alpha \beta -16\beta +18\alpha -12
\Rightarrow \alpha = \beta  so f is one-one
Let y\epsilon A  such that  y= \frac{4x+3}{6x-4}
\Rightarrow 6xy-4y= 4x+3\; \Rightarrow\; 6xy-4x= 4y+3
\Rightarrow x= \frac{4y+3}{6y-4}\: \epsilon A  for all y\, \epsilon \, A  so f is onto
\therefore f^{-1}\left ( y \right )= \frac{4y+3}{6y-4}\; or\; f^{-1}\left ( x \right )= \frac{4x+3}{6x-4}

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