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Math Help - Value of m for a function to be its own inverse.

  1. #1
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    Value of m for a function to be its own inverse.

    Given,

    y = f(x) = (3x - 5) / (2x - m)

    Find the value of m such that f(y) = x

    I tried doing it myself, but it just kept getting lengthier & didn't seem to end..

    Any idea how to solve that?

    Thanks for your time.
    Last edited by mr fantastic; March 29th 2011 at 07:25 PM. Reason: Re-titled.
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  2. #2
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    What exactly did you try?
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  3. #3
    Senior Member Sambit's Avatar
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    If f(x) = (3x - 5) / (2x - m), you should know what f(y) is. Equate that with x; and find m. This is not at all a lengthy procedure.
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  4. #4
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    This is what I did: -

    If y = f(x) = (3x - 5) / (2x - m) ----- (1)

    then, f(y) = (3y - 5) / (2y - m) = x [given condition]

    or, [3 {(3x - 5) / (2x - m)} - 5] / [2 {(3x - 5) / (2x - m)}] = x [putting the value of y from eqn. (1)]

    then....

    this is where I came: -

    (-x) + 5m - 15 = (6x^2) - 10x - (2x^2)m + (m^2)x

    This is where I got stuck & couldn't farther anymore steps..
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  5. #5
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    Two comments.

    1. You should have

    \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x, not

    \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}}=x.

    2. Once you carry this correction through, you can equate like powers of x to get the value for m.
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  6. #6
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    Quote Originally Posted by Ackbeet View Post
    Two comments.

    1. You should have

    \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x, not

    \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}}=x.

    2. Once you carry this correction through, you can equate like powers of x to get the value for m.

    Thanks for the correction.

    I did do \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x,, and not \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}}=x.

    Because only after doing \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x, , can you arrive at (-x) + 5m - 15 = (6x^2) - 10x - (2x^2)m + (m^2)x.

    It's just that I forgot to write it here as I'm new to this web.

    However, could you just give a simple example to show how to equate the powers of X on both sides of an equation?

    I know I learned that at Grade 8 or 9.. But I don't remember the exact procedure.

    So if you could just give a simple example of it, that would be very helpful..

    Thanks for your time
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  7. #7
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    Quote Originally Posted by arijit2005 View Post
    Thanks for the correction.

    I did do \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x,, and not \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}}=x.

    Because only after doing \dfrac{3\,\dfrac{3x-5}{2x-m}-5}{2\,\dfrac{3x-5}{2x-m}-m}=x, , can you arrive at (-x) + 5m - 15 = (6x^2) - 10x - (2x^2)m + (m^2)x.

    It's just that I forgot to write it here as I'm new to this web.

    However, could you just give a simple example to show how to equate the powers of X on both sides of an equation?

    I know I learned that at Grade 8 or 9.. But I don't remember the exact procedure.

    So if you could just give a simple example of it, that would be very helpful..

    Thanks for your time
    No problem. In your equation, which looks correct, you can scan the whole equation for the power x^2, and equate all those powers. On the LHS, you don't have any. On the RHS, you have 6-2m. Therefore, 0 = 6-2m.

    Next, you can equate the single powers of x^1. On the LHS, you have -1, and on the RHS, you have -10+m^2. Hence, you have the equation -1 = -10 + m^2.

    Finally, you can equate the constants (powers of x^0). On the LHS, you have 5m-15, and on the RHS, you have 0. Hence, 5m-15 = 0.

    Thankfully, or perhaps the problem was designed with this in view, there is only one value of m that satisfies all three equations, but it does satisfy all three equations. You'd certainly never expect there to be a solution, since over-determined systems of equations often don't have solutions. However, this system does.

    Make sense?
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