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 bijective.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

Let $A=R-\{-\frac{4}{3}\}\to R-\left\{\frac{4}{3}\right\}$.So,$f:A\to B$ given by $f\left(x\right)=\frac{4x+3}{3x+4}$.

Let $y=f\left(x\right)=\frac{4x+3}{3x+4},yϵB\Rightarrow 3xy+4y=4x+3\Rightarrow x(3y-4)=3-4y$

$\Rightarrow x=\frac{3-4y}{3y-4}$.Consider $g:B\to A$ defined as $g\left(y\right)=\frac{3-4y}{3y-4}$.

By (i) and (ii),$gof=I_{A}andfog=I_B$.Hence f is bijective.Also,$f^{-1}=\frac{3-4y}{3y-4}$

Now,$f^0=\frac{3-4\times 0}{3\times 0-4}=-\frac{3}{4}$

Also,$f^x=\frac{3-4x}{3x-4}=2\Rightarrow 3-4x=6x-8\therefore x=\frac{11}{10}$