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Thread: 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
    $\displaystyle f(x) = \frac{1}{x-2}$

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

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

    The o stands for composed with
    $\displaystyle f o f^-^1 $
    $\displaystyle 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
    $\displaystyle f(x) = \frac{1}{x-2}$
    Note that 2 is not in the domain of $\displaystyle f$.
    Try $\displaystyle g(x)=\frac{2x+1}{x}$.

    What is $\displaystyle f\circ g(x)~\&~g\circ f(x)=~?$
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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 $\displaystyle f$.
    Try $\displaystyle g(x)=\frac{2x+1}{x}$.

    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 $\displaystyle f$.
    Try $\displaystyle g(x)=\frac{2x+1}{x}$.
    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 $\displaystyle x=\frac{1}{y-2}$ for $\displaystyle y$.
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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 $\displaystyle x=\frac{1}{y-2}$ for $\displaystyle y$.
    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.

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

    $\displaystyle x(y-2) = 1$

    $\displaystyle xy - 2x = 1$

    $\displaystyle xy = 1 + 2x$

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

    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
    $\displaystyle F(x) ~o~ f^-^1(x)~~and~~F^-^1(x) o F(x)$ I had a little trouble.

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

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

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

    $\displaystyle \frac{x}{1}$

    $\displaystyle = x$
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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.

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

    $\displaystyle x(y-2) = 1$

    $\displaystyle xy - 2x = 1$

    $\displaystyle xy = 1 + 2x$

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

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

    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 $\displaystyle \frac{x(a+ b)}{x(c+ d)}$ but not in $\displaystyle \frac{x+ b}{x(c+ d)}$ or $\displaystyle \frac{x+ b}{x+ d}$.
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