Two Linear Functions F(x) and G(x) With

Distinct Points of Tangency

By:  Denise Natasha Brewley and Venki Ramachandran


Posed Problem: Find two 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.

Consider two functions of the form,

 f(x)= ax+b


g(x)= cx+d 

We want to find product h(x) = f(x)g(x) such that it is tangent to each function, f(x) and g(x) at two distinct points. We first find the zeros of f(x) and g(x).  So, respectively we have

x1 = -b/a


x2 = -d/c

It follows that

h(x) = f(x)g(x)= (ax+b)(cx+d)


h(x)= acx2+adx+bcx +bd

Now, we take the derivative of h(x).  So,

h'(x) = 2acx+ad+bc

At the point where h(x) is tangent to f(x) and g(x), the slope of f(x) is equal to 'a' and the slope of g(x) is equal to 'c'.  So, we can write

2acx1+ad+bc = c


2acx2+ad+bc = a

If we solve each system for x1 and x2 respectively we obtain,

x1=(c - bc - ad)/2ac


x2=(a - ad - bc)/2ac

Each of these equations can be set equal to the zeros that we found earlier.  So now we can write,

x1=(c - bc - ad)/2ac = -b/a


x2=(a - ad - bc)/2ac = -d/c

One thing to note is that the slope of f(x) and g(x) are negations of each other, that is a = -c.  We could have used any values that we want such that a = -c.  But let us just choose values for the slopes, and it follows that     a= -1 and c = 1.  So if we substitute these values and simplify our x1 and x2 equation above, for each we will obtain b = 1 - d.  So if we now choose a value for d, let say d = 4, then b = -3.  So we can conclude that given     f(x) = -x-3 and g(x) = x+4 the product h(x) = (-x-3)(x+4) yields the following graphs:

You can try this for any values that you like.  The general form of the equations are:

f(x) = ax +b

g(x) = -cx + (1-d)

See if you can explore this on your own.