When is f negative?

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?

Thanks in advance, Cara

Answer 1

[Salaheddin Hosseinzadeh]

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]

Thanks! I will definitely try it if I fail to solve it analytically.

Answer 2

[Roger Stafford]

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.

[CARA]

Thank you for your answer! However I should be able to have different parametric constraints for different values of x. Correct?

[Roger Stafford]

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.