સાબિત કરો કે $f:[-1,1] \rightarrow R ,$ $f(x)=\frac{x}{(x+2)}$ દ્વારા વ્યાખ્યાયિત વિધેય એક-એક છે. વિધેય $f:[-1,1] \rightarrow,$ નો વિસ્તાર$ f(x)=\frac{x}{(x+2)}$ તો $f$ નું પ્રતિવિધેય શોધો.

સૂચન : $f$ ના વિસ્તારમાં આવેલ $y$ ને સંગત કોઈક $x \in [ - 1,1]$ માટે $y=f(x)=\frac{x}{x+2}$, એટલે કે, $x = \frac{{2y}}{{(1 - y)}}$,

$f :[-1,1] \rightarrow R$ is given as $(x)=X(X+2)$

For one - one

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

$\Rightarrow(X+2)=(Y+2)$

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

$\Rightarrow 2 x=2 y$

$\Rightarrow x=y$

$\therefore f$ 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 some $x \in[-1,1]$

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

$\Rightarrow x y+2 y=x$

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

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

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

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

Now,

$(g o f)(x)=g(f(x))=g\left(\frac{x}{(x+2)}\right)$ $=2\left(\frac{2\left(\frac{x}{x+2}\right)}{1-\left(\frac{x}{x+2}\right)}\right)$ $=\frac{2 x}{x+2-x}=\frac{2 x}{2}=x$

and

$(f o g)(y)=f(g(y))=f\left(\frac{2 y}{1-y}\right)$ $=\frac{\frac{2 y}{1-y}}{\frac{2 y}{1-y}+2}$ $=\frac{2 y}{2 y+2-2 y}=\frac{2 y}{2}=$

$\therefore $ $g o f=x=I_{[-1,1]}$ and $fog =y=$ $I_{Range}$

$\therefore $  $-f^{-1}=g$

$\Rightarrow f^{-1}(y)=\frac{2 y}{1-y}$,  $y \neq 1$