Question

Show that $f:[-1,1]\to R$, given by $f\left(x\right)=\frac{x}{(x+2)},x\ne -2$, is one-one. Find the inverse of the function $f:[-1,1]\to$ Range f.

Answer 1

Function $f:[-1,1]\to R$ is given $f\left(x\right)=\frac{x}{(x+2)}$
Let f(x)=f(y)
$\Rightarrow \frac{x}{x+2}=\frac{y}{y+2}\Rightarrow xy+2x=xy+2y\Rightarrow 2x=2y\Rightarrow x=y$
Therefore, f is a one-one function.
Let $y=\frac{x}{x+2}\Rightarrow x=xy+2y\Rightarrow x=\frac{2y}{1-y}$
So, for every y except 1 in the range there exists x in the domain such that f(x)=y, Hence, function f is onto.
Therefore, $f:[-1,1]\to$ Range f , is one-one and onto and therefore, the inverse of the function $f:[-1,1]\to$ Range f exists.
Let y be an arbitrary element of range f.
Since, $f:[-1,1]\to$ Range f is onto, we have y =f(x)for some $x\in [-1,1]$
$\Rightarrow y=\frac{x}{x+2}\Rightarrow xy+2y=x\Rightarrow x(1-y)=2y\Rightarrow x=\frac{2y}{1-y},y\ne 1$
Now, let us define g:Range $f\to [-1,1]asg\left(y\right)=\frac{2y}{1-y},y\ne 1$
Now, $\left(gof\right)\left(x\right)=g\left(f\right(x\left)\right)=g\left(\frac{x}{x+2}\right)=\frac{2\left(\frac{x}{x+2}\right)}{1-\frac{x}{x+2}}=\frac{2x}{x+2-x}=\frac{2x}{2}=x\left(fog\right)\left(y\right)=f\left(g\right(y\left)\right)=f\left(\frac{2y}{1-y}\right)=\frac{\frac{2y}{1-y}}{\frac{2y}{1-y}+2}=\frac{2y}{2y+2-2y}=\frac{2y}{2}=y$
Therefore, gof $=fog=I_R$, Therefore, $f^{-1}=g$
Therefore, $f^{-1}\left(y\right)=\frac{2y}{1-y},y\ne 1$