**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 *

*and *

*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

*x _{1 }= -b/a *

*and *

*x _{2} =
-d/c*

It follows that

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

*and*

*h(x)= acx ^{2}+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

*2acx _{1}+ad+bc = c*

*and*

*2acx _{2}+ad+bc = a*

If we solve each system for *x _{1}*
and

*x _{1}=(c - bc - ad)/2ac *

*and *

*x _{2}*

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

*x _{1}=(c - bc - ad)/2ac = -b/a *

*and *

*x _{2}*

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 *x _{1 }*

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.

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