Question

Let f : R $-\left\{-\frac{4}{3}\right\}\to$R be a function defined as f$\left(x\right)$ $=\frac{4x}{3x+4}$ . Show that
f : R $-\left\{-\frac{4}{3}\right\}\to$ Rang (f) is one-one and onto. Hence, find f ^$-$1.

Answer 1

The function $f:\mathbf{R}-\left\{-\frac{4}{3}\right\}\to \mathbf{R}-\left\{\frac{4}{3}\right\}\mathrm{is}\mathrm{given}\mathrm{by}f\left(x\right)=\frac{4x}{3x+4}$.
Injectivity: Let $x,y\in \mathbf{R}-\left\{-\frac{4}{3}\right\}$ be such that
$$\begin{gathered} f\left(x\right)=f\left(y\right) \\ \Rightarrow \frac{4x}{3x+4}=\frac{4y}{3y+4} \\ \Rightarrow 4x\left(3y+4\right)=4y\left(3x+4\right) \\ \Rightarrow 12xy+16x=12xy+16y \\ \Rightarrow 16x=16y \\ \Rightarrow x=y \end{gathered}$$
Hence, f is one-one function.
Surjectivity: Let y be an arbitrary element of $\mathbf{R}-\left\{\frac{4}{3}\right\}$. Then,
f(x) = y
$$\begin{gathered} \Rightarrow \frac{4x}{3x+4}=y \\ \Rightarrow 4x=3xy+4y \\ \Rightarrow 4x-3xy=4y \\ \Rightarrow x=\frac{4y}{4-3y} \end{gathered}$$
As $y\in \mathbf{R}-\left\{\frac{4}{3}\right\},\frac{4y}{4-3y}\in \mathbf{R}$.
Also, $\frac{4y}{4-3y}\ne -\frac{4}{3}$ because $\frac{4y}{4-3y}=-\frac{4}{3}\Rightarrow 12y=-16+12y\Rightarrow 0=-16$, which is not possible.
Thus,
$x=\frac{4y}{4-3y}\in \mathbf{R}-\left\{-\frac{4}{3}\right\}$ such that
$f\left(x\right)=f\left(\frac{4x}{3x+4}\right)=\frac{4\left({\displaystyle \frac{4y}{4-3y}}\right)}{3\left({\displaystyle \frac{4y}{4-3y}}\right)+4}=\frac{16y}{12y+16-12y}=\frac{16y}{16}=y$, so every element in $\mathbf{R}-\left\{\frac{4}{3}\right\}$ has pre-image in $\mathbf{R}-\left\{-\frac{4}{3}\right\}$.
Hence, f is onto.
Now,
$x=\frac{4y}{4-3y}$
Replacing x by $f^{-1}\left(x\right)$ and y by x, we have
$f^{-1}\left(x\right)=\frac{4x}{4-3x}$