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

(N/A) For one-one,let $f(x_1) = f(x_2)$.
$\Rightarrow \frac{x_1}{x_1+2} = \frac{x_2}{x_2+2}$
$\Rightarrow x_1(x_2+2) = x_2(x_1+2)$
$\Rightarrow x_1x_2 + 2x_1 = x_1x_2 + 2x_2$
$\Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2$.
Thus,$f$ is one-one.
For the inverse,let $y = f(x) = \frac{x}{x+2}$.
$y(x+2) = x \Rightarrow xy + 2y = x \Rightarrow 2y = x(1-y) \Rightarrow x = \frac{2y}{1-y}$.
Since $f$ is onto its range,the inverse function $f^{-1}: \text{Range } f \rightarrow [-1,1]$ is defined by $f^{-1}(y) = \frac{2y}{1-y}$.