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Math Help - Inverse Functions

  1. #1
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    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?

    Thanks in advance!
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  2. #2
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    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)
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  3. #3
    Grand Panjandrum
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    Re: Inverse Functions

    Quote Originally Posted by Blizzardy View Post
    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
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  4. #4
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    Re: Inverse Functions

    Quote Originally Posted by Blizzardy View Post
    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 definition of "inverse function" is that f(f'(x))=x and f'(f(x))= x.

    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.)
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