Let f : R


Let $f: R-\left\{-\frac{4}{3}\right\} \rightarrow R$ be a function defined as $f(x)=\frac{4 x}{3 x+4}$. Show that

$f: R-\left\{-\frac{4}{3}\right\} \rightarrow$ Rang $(f)$ is one-one and onto. Hence, find $f-1$.


The function $f: \mathbf{R}-\left\{-\frac{4}{3}\right\} \rightarrow \mathbf{R}-\left\{\frac{4}{3}\right\}$ is given by $f(x)=\frac{4 x}{3 x+4}$.

Injectivity: Let $x, y \in \mathbf{R}-\left\{-\frac{4}{3}\right\}$ be such that


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

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

$\Rightarrow 12 x y+16 x=12 x y+16 y$

$\Rightarrow 16 x=16 y$

$\Rightarrow x=y$

Hence, $f$ is one-one function.

Surjectivity: Let $y$ be an arbitrary element of $\mathbf{R}-\left\{\frac{4}{3}\right\}$. Then,


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

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

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

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

As $y \in \mathbf{R}-\left\{\frac{4}{3}\right\}, \frac{4 y}{4-3 y} \in \mathbf{R}$.

Also, $\frac{4 y}{4-3 y} \neq-\frac{4}{3}$ because $\frac{4 y}{4-3 y}=-\frac{4}{3} \Rightarrow 12 y=-16+12 y \Rightarrow 0=-16$, which is not possible.


$x=\frac{4 y}{4-3 y} \in \mathbf{R}-\left\{-\frac{4}{3}\right\}$ such that

$f(x)=f\left(\frac{4 x}{3 x+4}\right)=\frac{4\left(\frac{4 y}{43 y}\right)}{3\left(\frac{4 y}{4 y_{y}}\right)+4}=\frac{16 y}{12 y+16-12 y}=\frac{16 y}{16}=y$, 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.


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

Replacing $x$ by $f^{-1}(x)$ and $y$ by $x$, we have

$f^{-1}(x)=\frac{4 x}{4-3 x}$

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