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    On page 27 of the Pearson Edexcel C3 textbook example 13c says "find values of x such that f(x)=f^-1(x)" how did they simplify that to:

    f(x) = f^-1(x)
    f(x) = x

    How can f^-1(x) just become x?
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    (Original post by mp_x)
    On page 27 of the Pearson Edexcel C3 textbook example 13c says "find values of x such that f(x)=f^-1(x)" how did they simplify that to:

    f(x) = f^-1(x)
    f(x) = x

    How can f^-1(x) just become x?
    f(x) = x is a function that satisfies f(x) = f^-1(x).

    http://www.thestudentroom.co.uk/show....php?t=3540817

    Try reading my post in this thread.


    Or thinking of it graphically, the inverse fuction is a reflection of the function in the line y = x. So only the function y = x could be the inverse of itself.
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    (Original post by mp_x)
    On page 27 of the Pearson Edexcel C3 textbook example 13c says "find values of x such that f(x)=f^-1(x)" how did they simplify that to:

    f(x) = f^-1(x)
    f(x) = x

    How can f^-1(x) just become x?
    You know that f(x) and f^{-1} (x) are simply reflections of one another in the line y=x. So when they give you the equation f(x) = f^{-1}(x) you know that the solutions to this equation lies on the line y=x allowing you to simplify your equation as f(x) = x = f^{-1}x.

    Examples:

    e^x = \ln x, this one is hard to solve by itself - so let's use the fact that they are inverse functions of one another and satisfy e^x = x = \ln x which makes life slightly easier (although not very much so, unless you want to use the Lambert W function).

    Here's another one, solve: x^2 = \sqrt{x} for x>0 - they are inverses of one another and hence satisfy x^2 = x so x^2 = x \iff x^2 - x = 0 \iff x(x-1) = 0.
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    (Original post by notnek)
    f(x) = x is the only function that satisfies f(x) = f^-1(x).

    http://www.thestudentroom.co.uk/show....php?t=3540817

    Try reading my post in this thread.


    Or thinking of it graphically, the inverse fuction is a reflection of the function in the line y = x. So only the function y = x could be the inverse of itself.
    (Original post by Zacken)
    You know that f(x) and f^{-1} (x) are simply reflections of one another in the line y=x. So when they give you the equation f(x) = f^{-1}(x) you know that the solutions to this equation lies on the line y=x allowing you to simplify your equation as f(x) = x = f^{-1}x.

    Examples:

    e^x = \ln x, this one is hard to solve by itself - so let's use the fact that they are inverse functions of one another and satisfy e^x = x = \ln x which makes life slightly easier (although not very much so, unless you want to use the Lambert W function).

    Here's another one, solve: x^2 = \sqrt{x} for x>0 - they are inverses of one another and hence satisfy x^2 = x so x^2 = x \iff x^2 - x = 0 \iff x(x-1) = 0.
    I'm still confused sorry
    I get that f^-1f(x) = x because you're basically doing something then undoing it but I just don't get how f^-1(x) = x
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    (Original post by Zacken)
    You know that f(x) and f^{-1} (x) are simply reflections of one another in the line y=x. So when they give you the equation f(x) = f^{-1}(x) you know that the solutions to this equation lies on the line y=x allowing you to simplify your equation as f(x) = x = f^{-1}x.

    Examples:

    e^x = \ln x, this one is hard to solve by itself - so let's use the fact that they are inverse functions of one another and satisfy e^x = x = \ln x which makes life slightly easier (although not very much so, unless you want to use the Lambert W function).

    Here's another one, solve: x^2 = \sqrt{x} for x>0 - they are inverses of one another and hence satisfy x^2 = x so x^2 = x \iff x^2 - x = 0 \iff x(x-1) = 0.
    I now understand this graphically but not in principle as it doesn't follow the principle in my reply above
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    (Original post by mp_x)
    I'm still confused sorry
    I get that f^-1f(x) = x because you're basically doing something then undoing it but I just don't get how f^-1(x) = x
    The inverse of a function f(x) maps the output of the function back to the input.

    So if f(a) = b then f^{-1}(b) = a.


    Try it with f(x) = x + 1 and f^{-1}(x) = x - 1 :

    f(a) = a + 1

    f^{-1}(a + 1) = a


    Now if f = f^{-1} then it must follow that if f(a) = b then f(b) = a. So this function must be able to "undo" its mapping.

    Let's try a function f(x) = x+1. This clearly won't work because the mapping will always add 1 to the input.

    A function can undo it's mapping if it maps any element to itself i.e. the function f(x)=x.


    If you're still not sure, please expain the specific part of my post that confuses you.
 
 
 
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