Question

Consider $f:R-\{-\frac{4}{3}\}\to R-\left\{\frac{4}{3}\right\}$ given by $f\left(x\right)=\frac{4x+3}{3x+4}$. Show that $f$ is bilective. Find the inverse of $f$ and hence find $f^{-1}\left(0\right)$ and $x$ such that $f^{-1}\left(x\right)=2$.

Answer 1

One - One :

Let $x_1.x_2\in R-\{-\frac{4}{3}\}$

Now, $f\left(x_1\right)=f\left(x_2\right)\Rightarrow \frac{4x_1+3}{3x_1+4}=\frac{4x_2+3}{3x_2+4}$

$\Rightarrow 12x_1{x}_2+16x_1+9x_2+12$

$=12x_1{x}_2+9x_1+16x_2+12$

$\Rightarrow (16-9)x_1=(16-9)x_2$

$\Rightarrow x_1=x_2$

Thus, $f$ is one-one function.

Onto:

Let $y=\frac{4x+3}{3x+4}$

$\Rightarrow 3xy+4y=4x+3$

$\Rightarrow x(3y-4)=3-4y$

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

$\Rightarrow \forall y\in$ codomain $\exists x\in \text{domain}[\because x\ne \frac{4}{3}]$

Thus, $f$ is onto function.

Hence, $f$ is one-one onto function.

Now, $f^{-1}\left(x\right)=\frac{3-4x}{3x-4}$

$f^{-1}\left(0\right)=\frac{3}{-4}$

$f^{-1}\left(x\right)=2$

$\Rightarrow \frac{3-4x}{3x-4}=2$

$\Rightarrow 3-4x=6x-8$

$\Rightarrow 10x=11$

$\Rightarrow x=1.1$