Question

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 $f^{-1}$.

Answer 1

The function f is one-one.

Reason: Let $x_1,x_2ϵA$ be such that $f\left(x_1\right)=f\left(x_2\right)$

$\frac{4x_1+3}{6x_1-4}=\frac{4x_2+3}{6x_2-4}$

$24x_1{x}_2-16x_1+18x_2-12=24x_1{x}_2+18x_1-16x_2-12$

$\Rightarrow 34x_2=34x_1\Rightarrow x_2=x_1\Rightarrow$ f is one-one.

The function f is onto:

We shall find the range of the function f.

For the range of f, let y = f(x)

$y=\frac{4x+3}{6x-4}\Rightarrow 6xy-4y=4x+3$

$\Rightarrow (6y-4)x=4y+3\Rightarrow x=\frac{4x+3}{6x-4}$ but $xϵA$

$6y-4\ne 0\Rightarrow y\ne \frac{2}{3}$

Range of $f$ is $R-\frac{2}{3}=A$

Range of $f=$Co-domain of $f$

$f$ is onto function.

Thus, the function f is one-one and onto, f is invertible.

To find $f^{-1}$:

Let $f\left(x\right)=y$

$y=\frac{4x+3}{6x-4}$

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

$(6y-4)x=4y+3$

$x=\frac{4y+3}{6y-4}$

Now, $f\left(x\right)=y$ and f is inversible.

$f^{-1}\left(y\right)=x\Rightarrow f^{-1}\left(y\right)=\frac{4y+3}{6y-4}$

Thus, the function $f^{-1}:A\to A$ is given by $f^{-1}\left(y\right)=\frac{4y+3}{6y-4}$

i.e., $f^{-1}=\frac{4y+3}{6y-4}$

So, $f^{-1}=f$