Find where one graph is bigger than the other

[Pi Face]

Homework Statement

My question is part of a bigger one that I'm using in a graphical proof. I have two functions, tanh(ap) and 2a/(1+a^2). The a is the input and p is a constant. I'm trying to find the value for p which is the bridging point between the two functions having 1 intersection and 2 intersections. To do this, I said that we want to find the value for p which makes tanh(ap) greater than 2a(1+a^2) for every point on the interval (0,1) (because the second function's max is at 1, so the two intersections will be on the sides of a=1)

So I have tanhap>2a/(1+a^2)>0, but I'm not sure how to go about solving it. I can't set them equal to each other at a=1 because tanha never reaches 1 so I get an undefined answer. Is there something I can do with limits?

 

[Simon Bridge]

Do I understand that your problem is to find $$p: \tanh(px) > \frac{2x}{1+x^2}: x\in (0,1)$$... what's wrong with putting LHS=RHS and finding the intersection in terms of p as a starting point?

You can gain an understanding of what p does by plotting tanh(px), and 2x(1+x^2), on the same axis, in the interval, for several values of p using a math-script program like matlab, mathematica, or gnu/octave.

But I think I see your problem ... if you had a hard interval like (0,1] then your requirements cannot be met since at x=1, RHS>LHS ... and this is unambiguous. However, you only need LHS>RHS for x values arbitrarily close to 1.

Interestingly, the RHS has a turning point at x=1 ... the slope of the RHS is a maximum at the origin, so any value of p that makes the slope of the LHS the same at the origin will make the LHS > RHS for much of the interval. But I suspect that you can only make the other two intersection points arbitrarily close together as p -> infinity. I suppose a limit formulation would be to find p that satisfies the conditions for an interval (0,z] in the limit z -> 1. Trouble is, p(x) is assymtotic at x=1.