It looks promising
as we now
appear to have g(x) and h(x) tangent at two points. Perhaps a
f(x) will create two tangents there. Let's
try f(x) = x2-1.
It looks like success! Let's Prove it.
f '(x) = 2x, g'(x) = -2x, and h'(x) = (fg)'(x) =2x(-x2+2)-2x(x2-1) = -4x3+6x. We'll need these to find the slopes of the tangent lines.
Inspired by this success and some intuition gained from the graphs, lets see if we can algebraically find a family of solutions.
If we let f(x) = (x-a)(x+a) and g(x) = -(x-b)(x+b), then h(x) = -(x-a)(x+a)(x-b)(x+b), and we are guaranteed that f and g will always intersect h at (a,0), (-a,0) and (b,0), (-b,0) respectively. Thus we need only find a and b such that f '(a) = h'(a) and g'(b) = h'(b). This is actually easier than it seems.
Since f(x) = x2 - a2 , g(x) = x2 - b2, and h(x) = -x4 + (a2+b2)x2 + a2b2,
f '(x) = 2x, g'(x) = -2x, and h'(x) = -4x3 + 2xa2 +2b2x.
A little algebra with f '(a) = h'(a) gives, ab2-a3-a=0 and, ba2-b3-b=0 from g'(b)=h'(b).If we assume that a and b are nonzero, we get the hyperbolas b2 - a2 = 1 and a2 - b2 = 1.