# When is f negative?

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CARA on 2 Jun 2015
Commented: Roger Stafford on 3 Jun 2015
Hello, I have the following function
f(x)= b * c^(1−1/a) * [((1-x)/x)^(1/a) * (a-1) - ((1-x)/x)^((1/a)-1)] + a
where the parameters b,c and a are all in (0,1) and the variable x is also in (0,1). I want to know the parametric assumptions under which this is always negative.
How do I solve it?

Salaheddin Hosseinzadeh on 2 Jun 2015
Hi,
I can suggest a nasty & a bit dumb solution which would be making 4 for loops. Or maybe 3 would be enough. Make it 4 and keep it simple.
Then create a b c and x values, calculate the f value for each and every combination of these parameters.
Make sure you're saving them in a fashion that you can later discriminate the range of input values.
It may not be a fast and efficient solution but it is fast for PC and it should work.
CARA on 3 Jun 2015
Thanks! I will definitely try it if I fail to solve it analytically.

Roger Stafford on 2 Jun 2015
If a > 0, there are no possible combinations of the parameters a, b, and c which lie in (0,1) that will make f(x) always be negative for all x in (0,1). That is because as x approaches 1 the quantity
[((1-x)/x)^(1/a) * (a-1) - ((1-x)/x)^((1/a)-1)]
will approach arbitrarily close to zero and with 'a' added onto f(x) (at the right end,) it must necessarily at some point become positive.
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Roger Stafford on 3 Jun 2015
Cara, when you stated, "I want to know the parametric assumptions under which this is always negative," I understood that to mean that you were looking for fixed parameter values that possessed the property that f(x) would be negative for all x values from 0 to 1 for the same parameters. There are no such fixed parameter values that can do this.
Even if you are allowing the parameter values to vary along with x, if x is any number above about 0.7822, there exists no set of parameters in (0,1) for which f(x) can be made negative for that x.