Question

prove that function "f" defined as $f:R-(-2)\to R-1,f\left(x\right)=\frac{x}{x+2}$ is one - one and onto. Also find $f^{-1}$

Answer 1

$f\left(x\right)=\frac{x}{x+2}$

$x+2\ne 0$

Domain of function $=R-\{-2\}$

$\Rightarrow x\ne -2$

Therefore,

$f\left(y\right)=\frac{y}{y+2}$

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

$\frac{x}{x+2}=\frac{y}{y+2}$

$x(y+2)=y(x+2)$

$xy+2x=xy+2y$

$\Rightarrow x=y$

Hence the function is one-one.

Now, let $y=f\left(x\right)$

$\Rightarrow y=\frac{x}{x+2}$

$\Rightarrow y(x+2)=x$

$\Rightarrow xy+2y=x$

$\Rightarrow x-xy=2y$

$\Rightarrow x=\frac{2y}{1-y}=f\left(y\right).....\left(i\right)$

$1-y\ne 0$

$\Rightarrow y\ne 1$

Hence range of the function is $R-\left\{1\right\}$.

On replacing $y$ with $x$ in equation $\left(i\right)$, we get inverse of $f\left(x\right)$.

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