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Math Help - Inverse Function

  1. #1
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    Inverse Function

    I got to calculate the inverse function of a normal function.

    The function is given by: Y=x-0.5x^2.
    So I need the function which ginve x=..........
    I am in doubt if there is an inverse function of this function.
    I hope someone could provide me the correct answer.

    Thanx in advance
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  2. #2
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    Quote Originally Posted by Divius View Post
    I got to calculate the inverse function of a normal function.

    The function is given by: Y=x-0.5x^2.
    So I need the function which ginve x=..........
    I am in doubt if there is an inverse function of this function.
    I hope someone could provide me the correct answer.

    Thanx in advance
    So you have to solve x = y - \frac{y^2}{2} for y.

    Option 1: Complete the square.

    x = - \frac{1}{2} \left( y^2 - 2y \right) = - \frac{1}{2} \left( [y - 1]^2 - 1 \right)

    \Rightarrow -2x = [y - 1]^2 - 1 \Rightarrow 1 - 2x = [y - 1]^2

    \Rightarrow \pm \sqrt{1 - 2x} = y - 1

    \Rightarrow y = 1 \pm \sqrt{1 - 2x}.


    Option 2: Use the quadratic formula.

    2x = 2y - y^2 \Rightarrow y^2 - 2y + 2x = 0.

    a = 1, b = -2, c = 2x:

    y = \frac{2 \pm \sqrt{4 - 8x}}{2} = \frac{2 \pm 2 \sqrt{1 - 2x}}{2} = 1 \pm \sqrt{1 - 2x}.


    In each case, the domain of y = x - \frac{x^2}{2} will determine the range of the inverse and hence which of the \pm \sqrt{1 - 2x} solutions is wanted.
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  3. #3
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    Thank you very much for the clear answer to my question!
    However, I calculated a mean value for y by repeating the
    <br />
y = x - \frac{x^2}{2}<br />

    3 times for different x values. Next I divided the sum of the values by 3 to determine the mean y value. (in my example the mean y value is 0.007667)

    I need to find the x value which is equal to this mean y value. So given a value of y I have to determine which x value corresponds with this value.
    Last edited by Divius; March 10th 2008 at 08:07 AM.
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  4. #4
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    Quote Originally Posted by Divius View Post
    Thank you very much for the clear answer to my question!
    However, I calculated a mean value for y by repeating the
    <br />
y = x - \frac{x^2}{2}<br />

    3 times for different x values. Next I divided the sum of the values by 3 to determine the mean y value. (in my example the mean y value is 0.007667)

    I need to find the x value which is equal to this mean y value. So given a value of y I have to determine which x value corresponds with this value.
    So you want to solve <br />
0.007667 = x - \frac{x^2}{2}<br />
.

    Re-arrange into a quadratic = 0.

    Or just substitute y = 0.007667 into <br />
x = 1 \pm \sqrt{1 - 2y}<br />
.
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  5. #5
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    This is what I was looking for, sorry I didn't used your first post as it makes it clear. However, I don't understand this step:
    <br />
y = \frac{2 \pm \sqrt{4 - 8x}}{2} = \frac{2 \pm 2 \sqrt{1 - 2x}}{2}
    Last edited by Divius; March 10th 2008 at 01:38 PM.
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