Originally Posted by

**Blizzardy** The function f and g are defined by

f : x -> (2-3x)/(3-x), x > 3,

g : x -> x^2 - 4x + a, a E R (i.e. a is an element of all real no.), x>1

Define in a similiar manner, the inverse function f ' and show that f^2 (x) = x. Hence determine f^13 in a similiar manner.

I managed to find the inverse of f, f ':

f ' : x -> (2-3x)/(3-x), which is the same as f actually,

then f^2 (x) = f[f(x)] = f[f '(x)] = x

I think this is how it is supposed to be done but why is " f[f '(x)] = x " ?

And how do I determine f^13?

Thanks in advance!