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Math Help - The inverse of a one to one function

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    Senior Member vaironxxrd's Avatar
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    The inverse of a one to one function

    Hello forum,

    I personally believe, though I'm not sure, that this topic should be in the Algebra forum. Though, because I'm taking this from a Precalculus class I wasn't entirely sure whether to place it there.

    Find the inverse of
    f(x) = \frac{1}{x-2}

    To my understanding I must find f^-^1
    I have the answers, but I'm interested in knowing how to find the inverse

    F^-^1(x) = How can I find this?

    The o stands for composed with
    f o f^-^1
    f(f^-^1(x)) = \frac{1}{f^-^1(x)-2}
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    Re: The inverse of a one to one function

    Quote Originally Posted by vaironxxrd View Post
    Find the inverse of
    f(x) = \frac{1}{x-2}
    Note that 2 is not in the domain of f.
    Try g(x)=\frac{2x+1}{x}.

    What is f\circ g(x)~\&~g\circ f(x)=~?
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    Senior Member vaironxxrd's Avatar
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    Re: The inverse of a one to one function

    [QUOTE=Plato;749605]Note that 2 is not in the domain of f.
    Try g(x)=\frac{2x+1}{x}.

    Hello Plato,

    I remember you are always actively helping me. I dont know if I portray the following well when I ask for help. But I'm eager to learn the process, the concept.

    I'm on a rush to school I will solve it as soon as I have a break. Thanks for replying ^_^
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    Re: The inverse of a one to one function

    [QUOTE=vaironxxrd;749609]
    Quote Originally Posted by Plato View Post
    Note that 2 is not in the domain of f.
    Try g(x)=\frac{2x+1}{x}.
    I dont know if I portray the following well when I ask for help. But I'm eager to learn the process, the concept.
    Here is the process.
    Solve x=\frac{1}{y-2} for y.
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    Senior Member vaironxxrd's Avatar
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    Re: The inverse of a one to one function

    [QUOTE=Plato;749610]
    Quote Originally Posted by vaironxxrd View Post
    Here is the process.
    Solve x=\frac{1}{y-2} for y.
    Lets see if I can get it right this time. I forgot some of the rules when multiplying fractions with variables, fear not, there are many resources.

    x = \frac{1}{y-2}

    x(y-2) = 1

    xy - 2x = 1

    xy = 1 + 2x

    y = \frac{1+2x}{x}

    Wouldn't the two x cancel out?
    My best answer is because we are adding 1 to it and therefore the values will change
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    Senior Member vaironxxrd's Avatar
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    Re: The inverse of a one to one function

    When proving that
    F(x) ~o~ f^-^1(x)~~and~~F^-^1(x) o F(x) I had a little trouble.

    F(F^-^1(x)) = \frac{1}{\frac{1+2x}{x}-2}

    \frac{1}{\frac{1+2x}{x}-2}*x

    \frac{x}{1+2x-2x}

    \frac{x}{1}

     = x
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    Re: The inverse of a one to one function

    [QUOTE=vaironxxrd;749802]
    Quote Originally Posted by Plato View Post

    Lets see if I can get it right this time. I forgot some of the rules when multiplying fractions with variables, fear not, there are many resources.

    x = \frac{1}{y-2}

    x(y-2) = 1

    xy - 2x = 1

    xy = 1 + 2x

    y = \frac{1+2x}{x}

    Wouldn't the two x cancel out?
    What is \frac{1+ 2(3)}{3}? Is that the same thing as \frac{1+2}{1}?

    My best answer is because we are adding 1 to it and therefore the values will change
    I'm not sure I can makes sense out of that sentence! Your best answer to what question? The question "Wouldn't the two x cancel out?" require a "yes" or "no" answer, not "because".

    In any case, the only time something in numerator and denominator cancel is when they are multipied by the rest of the numerator and denominator: you can cancel the "x"s in \frac{x(a+ b)}{x(c+ d)} but not in \frac{x+  b}{x(c+ d)} or \frac{x+ b}{x+ d}.
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