Exploring h(x) = f(x)g(x) when f and g are linear
Goal: Find 2 linear functions f(x) and g(x) such that their product h(x) = f(x)g(x) is tangent to each of f(x) and g(x) at two distinct points. Discuss and illustrate the method and results.
As with any open ended problem of this type, our first thought may be to plug in some numbers and look at what happens. Because of that, I made a first attempt of:
Clearly from this result, guessing and checking will likely be a futile effort.
My first attempts included setting up a system of equations and attempting to plug in values that worked. While one attempt was successful, further attempts were unsuccessful, showing the futility of these efforts. Because of that I decided to take a more computational approach (with the help of the CAS Maple).
I started with the following system of equations:
Now since I know at some fixed x, f(x) and h(x) intersect and are tangent, and similarly for some fixed y for h(y) and g(y), I have the following system of equations, where equations (3) and (4) come from the fact that f'(x) = h'(x) and g'(y) = h'(y).
Now I began to play with Maple. First I solved equation (3) for x and plugged that into equation (1) elimating x. Then I solved that equation for a. Similarly, I solved equation (4) for y and plugged it into equation (2) eliminating y. Then I solved this equation for a. After solving both equations for a, I set the two equations equal (hence eliminating a). After doing this, I solve the equation for d, during which the variable c is eliminated, giving me:
Now that I have the relationship d = 1-b, I plug that into equations (1) and (3). Again I solve the new equation (3) for x and plug it into equation (1) and solve for a. In this case I get the relationship:
a = -c
a - c = 0
Now we have a set of two parameters.
a - c = 0
b + d = 1
When playing with these we find that any pair of numbers that satisfy the parameters will work as seen with the animations below.
Animation 1(a and c vary):
Animation 2 (b and d vary):
Animation 3 (all 4 vary):
Another set of arguments on this problem can be found here. The author of this article uses a different technique to obtain the same set of equations a - c = 0 and b + d = 1. They point out that f(x) = 0 when g(x) = 1 and g(x) = 0 when f(x) = 1. However, my animations go one step further to show that f(x) + g(x) = 1 for all x (not just when one or the other equals 0 or 1). Another observation that can be seen from this article is that the point of intersection of the f(x) and g(x) is always at y = 1/2.