I have some problem understanding how to calculate the inverse of a function. I have an example below:
Calculate the following:
$(g\prime)^{-1}(y)$.
The $y$-value is:
$y(s)=g\prime=2s-1 $.
Solved with respect to $s$:
$s=\frac{y(s)+1}{2}$.
This gives the following result:
$(g\prime)^{-1}(y)=s=\frac{y(s)+1}{2}$.
My problem is that I don't understand how we can write that $(g\prime)^{-1}(y)=s$. Can anyone help me to understand that? David
If you have a function $$g:\quad{\mathbb R}\to{\mathbb R},\qquad s\mapsto y:=2s-1\tag{1}$$ then the inverse function $g^{-1}$ is obtained by solving algebraically the equation $y=2s-1$ for $s$ in terms of $y$ (if possible). In the case at hand one obtains $s={1\over2}(y+1)$. Therefore the flow diagram for $g^{-1}$ appears as $$g^{-1}:\quad{\mathbb R}\to{\mathbb R},\qquad y\mapsto s:={1\over2}(y-1)\ .\tag{2}$$ Another matter are the derivatives of $g$ and $g^{-1}$. Differentiating $(1)$ and $(2)$ according to the rules we obtain $$g'(s)\equiv2,\qquad (g^{-1})'(y)\equiv{1\over2}\ .$$ There is also a general law about $(g^{-1})'$. It says that $$(g^{-1})'(y)={1\over g'\bigl(g^{-1}(y)\bigr)}\ .$$ You can verify yourself that it is fulfilled in the example at hand.