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Math Help - Finding the inverse of a function

  1. #1
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    Finding the inverse of a function

    What is the inverse of f(x) = x/x-1 ?
    this is my attempt to solve it:
    y = x / x-1
    y(x-1) = x
    hence x(x-1) is the inverse.....
    however, substituting with x=5 gives......
    5/4=1.25
    so pluging this is the inverse function should give 5, right?
    lets see:
    1.25(1.25-1) = 1.25*.25= 0.3125. WRONG!!

    can anyone tell me where I went wrong?
    thanks
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  2. #2
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    Re: Finding the inverse of a function

    Expand y(x-1) = x

    This gives yx - y = x. Now you can get all your x terms on the LHS and factor
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    Re: Finding the inverse of a function

    As an alternative, write \displaystyle \frac{x}{x - 1} = \frac{x - 1 + 1}{x - 1} = 1 + \frac{1}{x - 1} so that your original function is

    \displaystyle y = 1 + \frac{1}{x - 1} .

    Now evaluating the inverse function should be much easier.
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    Re: Finding the inverse of a function

    Quote Originally Posted by Prove It View Post
    As an alternative, write \displaystyle \frac{x}{x - 1} = \frac{x - 1 + 1}{x - 1} = 1 + \frac{1}{x - 1} so that your original function is

    \displaystyle y = 1 + \frac{1}{x - 1} .

    Now evaluating the inverse function should be much easier.
    how? why is (x+1-1)/x-1 = 1+ 1/x-1?
    can you please be more clear with your method? thanks
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    Re: Finding the inverse of a function

    Quote Originally Posted by IBstudent View Post
    how? why is (x+1-1)/x-1 = 1+ 1/x-1?
    can you please be more clear with your method? thanks
    My method is perfectly clear if you understand that \displaystyle \frac{a + b}{c} = \frac{a}{c} + \frac{b}{c}, but anyway, \displaystyle \frac{x - 1 + 1}{x - 1} = \frac{x - 1}{x - 1} + \frac{1}{x - 1} = 1 + \frac{1}{x - 1}
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    Re: Finding the inverse of a function

    ahh I see
    Thanks :-)
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    Re: Finding the inverse of a function

    Quote Originally Posted by IBstudent View Post
    how? why is (x+1-1)/x-1 = 1+ 1/x-1?
    can you please be more clear with your method? thanks
    Two comments.

    1. Another way to find the inverse function is to switch the roles of x and y. For example:
    y = \frac{x}{x - 1} \to x = \frac{y}{y - 1}
    Now solve for y and you get your inverse function.

    2. Please be more careful in how you write you problem. For example:
    y = x/x - 1
    is really
    y = (x/x) - 1 = 1 - 1 = 0

    What you meant to write is y = x/(x - 1). Please use parentheses in the future.

    -Dan
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    Re: Finding the inverse of a function

    okay so 1/(x-1) = y-1
    (y-1)(x-1) = 1
    yx-y-x+1 = 1
    y(x-1)=x
    ahhhh I'm confused...:S
    is it x/(x-1) or x^2-x that is the inverse?
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    Re: Finding the inverse of a function

    Quote Originally Posted by IBstudent View Post
    okay so 1/(x-1) = y-1
    (y-1)(x-1) = 1
    yx-y-x+1 = 1
    y(x-1)=x
    ahhhh I'm confused...:S
    is it x/(x-1) or x^2-x that is the inverse?
    It is basic algebra.

    1/(x-1) = y-1

    => x - 1 = 1/(y - 1) => x = .....
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    Re: Finding the inverse of a function

    I know that, so, the inverse of x/(x-1) is x/(x-1) ?!
    thats weird
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    Re: Finding the inverse of a function

    Quote Originally Posted by IBstudent View Post
    I know that, so, the inverse of x/(x-1) is x/(x-1) ?!
    thats weird
    Any function that is symmetric with respect to a reflection about the line y = x will behave that way. Other examples: y = x and y = 1/x. I'm sure there are others, but they're not coming to mind.
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    Re: Finding the inverse of a function

    Quote Originally Posted by IBstudent View Post
    I know that, so, the inverse of x/(x-1) is x/(x-1) ?!
    thats weird
    Not quite and not really.

    If f(x) = \frac{x}{x-1} Note, first that x \ne 1

    Then, f^{-1}(x) = \frac{x}{x-1}, where again, x \ne 1

    Notation is important. Don't just skip it.
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    Re: Finding the inverse of a function

    Quote Originally Posted by Ackbeet View Post
    Any function that is symmetric with respect to a reflection about the line y = x will behave that way. Other examples: y = x and y = 1/x. I'm sure there are others, but they're not coming to mind.
    To expand on this, functions that are their own inverses are called involutes. It is simple to show that the following functions are involutes:

    f(x) = x.
    f(x) = -x + b, where b is a real number.

    f(x) = \frac{ax + b}{cx - a} where a, b and c are real numbers. f(x) = 1/x is included as a special case.

    The question I'll pose for the interested reader is: Are the above functions the only involutes?
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