Question 166996
Isaac, it's not a power but it stands for the inverse function. 
For example, if I have a function {{{f(x)}}} such that
{{{f(1)=5}}}
then the inverse function, {{{f^(-1)(x)}}} would be such that when I plug in x=5, I would get back y=1. 
{{{f^(-1)(5)=1}}}
.
.
.
1.To find the inverse, first change to x,y notation.
{{{y=2*x-3}}}
Then interchange the positions of x and y and solve for y.
{{{highlight(x)=sqrt(2*highlight(y)-3)}}}
{{{x=2y-3}}}
{{{2y=x+3}}}
{{{y=(x+3)/2}}}}
The solution for the new y is the inverse function.
{{{f^(-1)(x)=(x+3)/2}}}
.
.
.
Let's try it. 
{{{f(x)=2x-3}}}
{{{f(1)=2(1)-3=2-3=-1}}}
Now when I plug in x=-1 into the inverse function, 
I should get back y=1.
{{{f^(-1)(x)=(x+3)/2}}}
{{{f^(-1)(-1)=(-1+3)/2}}}
{{{f^(-1)(-1)=2/2}}}
{{{f^(-1)(-1)=1}}}
.
.
.
You interested at the value of the inverse function for x=0.
{{{f^(-1)(x)=(x+3)/2}}}
{{{f^(-1)(0)=(0+3)/2}}}
{{{f^(-1)(0)=3/2}}}
You could have also solved it by asking what x makes my original function equal to zero.
{{{2x-3=0}}}
{{{2x=3}}}
{{{x=3/2}}}
The inverse functions interchange x and y values.
Hopefully that makes sense. Good luck!