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Math Help - inverse of x-1/x

  1. #1
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    Cool inverse of x-1/x

    Its been quite a few years since my calculus days so I'm probably missing something obvious. Somebody can probably do this in a couple seconds...

    f(x)=y-(1/y)

    solve for y.

    many thanks
    Last edited by scottboston; May 15th 2010 at 08:06 PM.
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  2. #2
    MHF Contributor matheagle's Avatar
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    first of all you have a typo

    you probably mean f(y)=y-1/y

    in any case to solve y=x-1/x, you multiply by x giving you a quadractic

    xy=x^2-1 so use the quadratic equation and solve for x.
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  3. #3
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by scottboston View Post
    Its been quite a few years since my calculus days so I'm probably missing something obvious. Somebody can do this a a couple seconds...

    f(x)=y-(1/y)

    solve for y.

    many thanks
    Not sure if you were looking for a more elegant technique here, but the quadratic formula should do the job:

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

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

    xy = x^2-1

    x^2-yx - 1 = 0

    Apply quadratic formula:

    x = \frac{y\pm \sqrt{y^2+4}}{2}

    Good luck!
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  4. #4
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    Thanks for the replies

    Something is up using the quadratic here because I'm getting 0=4 when I try to use the result <br /> <br />
0 = (x + \frac{y + \sqrt{y^2+4}}{2})(x + \frac{y - \sqrt{y^2+4}}{2})<br />

    Any thoughts? thanks
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  5. #5
    Junior Member autumn's Avatar
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    Quote Originally Posted by scottboston View Post
    Thanks for the replies

    Something is up using the quadratic here because I'm getting 0=4 when I try to use the result <br /> <br />
0 = (x + \frac{y + \sqrt{y^2+4}}{2})(x + \frac{y - \sqrt{y^2+4}}{2})<br />

    Any thoughts? thanks
    why are you taking x plus these roots?
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  6. #6
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    I'm just trying to solve one of them at zero

    Souldn't I be trying to solve this?
    <br />
x = \frac{y + \sqrt{y^2+4}}{2}<br />

    And then this?
    <br />
x = \frac{y - \sqrt{y^2+4}}{2}<br />


    I know I'm missing something here...something tells me I'll be doing and infinite loop of quadratic equations.
    Last edited by scottboston; May 15th 2010 at 11:47 PM.
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  7. #7
    MHF Contributor matheagle's Avatar
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    try subtracting
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  8. #8
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    ok

    <br /> <br />
0 = (x - \frac{y + \sqrt{y^2+4}}{2})(x - \frac{y - \sqrt{y^2+4}}{2})<br />

    try the first one....

    <br />
0 = x - \frac{y + \sqrt{y^2+4}}{2}<br />

    <br />
-2x = -y + \sqrt{y^2+4}<br />

    This is where I end. I don't think I want any xy multiple so I'm not sure what to do with the square root. How am I doing?
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