If you have 2 functions, f(x) and g(x), you would find f(g(x)) and g(f(x)). If, after you simplify both and you end up with just "x" as the answer, they are inverses of each other. For example, f(x) = 3x - 2 and g(x) = (x+2)/3
f(g(x)) takes the g function and plugs it into the x in the f function, like this:
![f(g(x))=3[ \frac{(x+2)}{3} ]-2](https://tex.z-dn.net/?f=f%28g%28x%29%29%3D3%5B%20%5Cfrac%7B%28x%2B2%29%7D%7B3%7D%20%5D-2%20)
cancel out the 3's to get
f(g(x))=(x+2)-2 which is just x. Now do it the other way: g(f(x))
![g(f(x))= \frac{[(3x-2)+2]}{3}](https://tex.z-dn.net/?f=g%28f%28x%29%29%3D%20%5Cfrac%7B%5B%283x-2%29%2B2%5D%7D%7B3%7D%20)
3x-2+2 is 3x, so we have
See?
f(g(x)) takes the g function and plugs it into the x in the f function, like this:
cancel out the 3's to get
f(g(x))=(x+2)-2 which is just x. Now do it the other way: g(f(x))
3x-2+2 is 3x, so we have
See?