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I am trying to use Mathematica to quickly determine the sign of a derivative. When I use the function Sign, Mathematica informs me the answer depends on the sign of a numerator, which is rather long. Following some advice on using $Assumptions, I have now tried the following, but without the desired effect; I would have thought that Mathematica would have replaced Sign[ ] with -1 or 1 when it has been input the restrictions:

$Assumptions = 0 <= a <= 1 && 0 <= b <= 1 && c == 1 

Sign[(b c (-I \[Pi] + Log[-b + (-1 + b) a] - Log[-b + (1 + b) a]) + c (-4 Log[a] + I (-2 + b) (\[Pi] + I Log[-b + (-1 + b) a]) + (2 + b) Log[-b + (1 + b) a]) a)/(b^2 (-1 + a)^3)]
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  • $\begingroup$ Is my code correct? $\endgroup$
    – user120911
    Aug 2, 2018 at 15:29
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    $\begingroup$ The Sign[0] is 0, Sign[Indeterminate] is Indeterminate, and Sign[z] is z/Abs[z] if z is a nonzero complex number. That last two can occur in the specified domain. $\endgroup$
    – Michael E2
    Aug 3, 2018 at 12:30
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    $\begingroup$ Also, generally you have to apply a function that makes use of $Assumptions (such as Simplify) in order for the assumptions to have an effect. I don't think Sign uses $Assumptions. $\endgroup$
    – Michael E2
    Aug 3, 2018 at 12:32

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You could use $Assumptions=... to define global attributes of variables!

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