A Relation of Linear Functions and Their Product

By: Stephen
Bismarck

Find two ** linear **functions

**h(x) = f(x).g(x)**

I knew that the two functions had to have opposite slopes (one
positive and the other negative).
So I started off with

f(x) = nx +
n and g(x) = -nx + n

h(x)
= (nx + n)(-nx + n)

I did this because I want to see if all the constants and
coefficients were the same.

This for n = 1 and I observed
that the bigger n got the further away the vertex of the parabola got from the intersection
of the lines.

This is for n = 2. So from observing that I thought about values smaller than n = 1 and this is where I got my first solution.

This is for n = .5 but how do I know it’s a solution. Well since we want each line to be tangent to the product of the lines, each must only touch the parabola. So I set each linear equation equal to the product of the lines.

As you can see from the
picture and if you plug the values for x back into the equation, the distinct
points are (1,0) and (-1,0).

So my first solution is

For the second solution I tried to keep f(x) constant and change
g(x). I decided to pick f(x) = x
and g(x) = -nx + n so h(x) = x(-nx +n).

I found this graph for when n = 1. It looks like the tangents are at (0,0) and (1,0). I decided to use some calculus to
determine if they are tangents.

I took the derivative of the product of the two lines, h(x) =
x(-x+1), and got h’(x) = -2x + 1 so I plugged in the x values of 0 and 1
from the points that I thought might be where the tangents are.

h’(0) = 1 and h’(1) = -1

So to find the equation of the tangent line I plugged in the slope
and a point on the graph of h(x) and got y = x and y = -x +1. This supports my hypothesis that f(x) =
x and g(x) = -x + 1 are solutions.

The two solutions that were found to this problem were

1.

2.
f(x) = x g(x) = -x +
1 and h(x) = x(-x + 1)

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1.

2.