The composition of function f and f ' is
f[ f '(x)] = f[(2-3x)/(3-x)] = {2 - 3[(2-3x)/(3-x)]} / {3 -[(2-3x)/(3-x)]}
when simplified gives f[ f '(x)] = x
As you did f^2(x) = x similarly f^3(x) = f(x) ; f^4(x) = x etc
Hence f^13(x) = f(x)
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!
The composition of function f and f ' is
f[ f '(x)] = f[(2-3x)/(3-x)] = {2 - 3[(2-3x)/(3-x)]} / {3 -[(2-3x)/(3-x)]}
when simplified gives f[ f '(x)] = x
As you did f^2(x) = x similarly f^3(x) = f(x) ; f^4(x) = x etc
Hence f^13(x) = f(x)
The definition of "inverse function" is that f(f'(x))=x and f'(f(x))= x.
Since the inverse function of f is f itself, .
Then .
(Often " " is used to mean f(x) times f(x), the ordinary product of f with itself, but since they tell you here that " " it clearly means the composition of f with itself and so is the repeated composition.)