# Thread: Inverse Functions

1. ## Inverse Functions

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?

2. ## Re: Inverse Functions

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)

3. ## Re: Inverse Functions

Originally Posted by Blizzardy
I think this is how it is supposed to be done but why is " f[f '(x)] = x " ?
By definition! Hence your will have constructed f' from f so that this is the case.

CB

4. ## Re: Inverse Functions

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?

Since the inverse function of f is f itself, $f^2(x)= f(f(x))= x$.
Then $f^{13}(x)= f^{11}(f^2(x))= f^11(x)= f^9(f^2(x))= f^9(x)= ...$.
(Often " $f^2(x)$" is used to mean f(x) times f(x), the ordinary product of f with itself, but since they tell you here that " $f^2(x)= x$" it clearly means the composition of f with itself and so $f^{13}$ is the repeated composition.)