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Math Help - Minimum Value

  1. #1
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    Minimum Value

    Not sure if this is in calculus section as I am not sure of approach.

    Given

    f(x) = (a+b+x)/(3abx) , x>0 and a ,b are positive real numbers

    i)Show that the minimum value of f(x) occurs when x = (a + b)/2

    Thanks. No idea how to approach. Tried differentiation but ended with

    f'(x) = -(a+b)/(3abx^2) which does not have turning point.
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  2. #2
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    Quote Originally Posted by Lukybear View Post
    Not sure if this is in calculus section as I am not sure of approach.

    Given

    f(x) = (a+b+x)/(3abx) , x>0 and a ,b are positive real numbers

    i)Show that the minimum value of f(x) occurs when x = (a + b)/2

    Thanks. No idea how to approach. Tried differentiation but ended with

    f'(x) = -(a+b)/(3abx^2) which does not have turning point.
    You have differentiated this correctly: the given function does not have a turning point or a minimum value.

    The likeliest explanation is that there is a mistake in the question (or you have copied it incorrectly). If the correct version is f(x) = (a+b+x)^2/(3abx) then there is a minimum at x = (a + b)/2.
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  3. #3
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    Ah yes. That would be most likely explanation as it is proving inequalities. Thanks.
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