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Math Help - Distribution Maximum

  1. #1
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    Distribution Maximum

    need help plz
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  2. #2
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    Hello, Harry!

    Just take the derivative, equate to zero, and solve . . .


    Assuming x > 0, we have: .f(x) .= .2axe^{-ax}


    Differentiate and equate to zero:

    . . f'(x) .= .2axe^{-2ax}(-2ax) + 2ae^{-ax} .= .0

    Divide by 2ae^{-ax}: .-2ax + 1 .= .0

    And we have: .2ax .= .1 . . . . x = 1/2a

    . . . . . . . . . . . . . . .__
    Therefore: .x .= .1/√2a

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