Find the inverse of a function.

Question

$$ g:[-1,1] \to \mathbb R\\ g(x)={\frac{x}{x+2}} $$

$f:[-1,1] \to$ range of f. Find the inverse of $ f.$

$\forall y\in \text{range of }g$ there exist some ${\frac{2y}{1-y}}\in [-1,1]$ such that $g({\frac{2y}{1-y}})=y$

I'm getting problem that when I am replacing $x$ by ${\frac{2y}{1-y}}$ I'm not getting $y$.

Kindly help me out.

Answer 1

You have $$g\left(\frac{2y}{1-y}\right)=\frac{\frac{2y}{1-y}}{\frac{2y}{1-y}+2}=\frac{\frac{2y}{1-y}}{\frac{2y+2(1-y)}{1-y}}=\frac{2y}{2y+2(1-y)}=\frac{2y}{2}=y.$$

Answer 2

Must be some calculation error. The inverse is actually: $\frac{2y}{1-y}$. If you substitute this and check, then you will find it satisfying the equation :$g({\frac{2y}{y-1}})=y$