Show that $f:[-1,1] ightarrow mathbf{R}$, given by

Solution:

$f:[-1,1] \rightarrow \mathrm{R}$ is given as $f(x)=\frac{x}{(x+2)}$.

Let $f(x)=f(y)$.

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

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

$\Rightarrow 2 x=2 y$

$\Rightarrow x=y$

∴ f is a one-one function.

It is clear that $f:[-1,1] \rightarrow$ Range $f$ is onto.

$\therefore f:[-1,1] \rightarrow$ Range $f$ is one-one and onto and therefore, the inverse of the function:

$f:[-1,1] \rightarrow$ Range $f$ exists.

Let $g$ : Range $f \rightarrow[-1,1]$ be the inverse of $f$.

Let y be an arbitrary element of range f.

Since $f:[-1,1] \rightarrow$ Range $f$ is onto, we have:

$y=f(x)$ for same $x \in[-1,1]$

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

$\Rightarrow x y+2 y=x$

$\Rightarrow x(1-y)=2 y$

 

$\Rightarrow x=\frac{2 y}{1-y}, y \neq 1$

Now, let us define $g$ : Range $f \rightarrow[-1,1]$ as

$g(y)=\frac{2 y}{1-y}, y \neq 1$