Assignment 1:

Finding Two Linear Functions Such that Their Product is Tangent to Each Linear Function at Two Distinct Points

by

Ángel M. Carreras Jusino

Goal:

• Find two linear functions f(x) and g(x) such that their product h(x) = f(x)∙g(x) is tangent to each f(x) and g(x) at two distinct points.

Let f(x) and g(x) be linear functions. So we can write them of the form f(x) = ax + b and g(x) = cx + d, where a, b, c, d ∈ ℜ.

Now if h(x) = f(x)∙g(x), then h(x) = (ax + b)(cx + d) = acx2 + (ad + cb)x + bd .

If we look at the following image one possible case is when the linear functions are concurrent with the axis of symmetry of the parabola h(x) resulting from their product.

For an equation of a parabola written in the form y = Ax2 + Bx + C its axis of symmetry is given by x = -B2A.

So for the case of h(x) we have that the axis of symmetry is given by

Now because f(x) and g(x) are concurrent with this line the x coordinate of the intersection of f(x) and g(x) is going to be the value of the axis of symmetry.

Therefore,

So the slope of one of the linear functions is the additive inverse of the other, hence we can rewrite f(x) and g(x) as

f(x) = ax + b

g(x) = -ax +d

Now without loss of generalization we let the axis of symmetry be the y-axis, therefore f(x) and g(x) intersects at their y-intercept. Consequently b = d.

Summarizing what we have until now and their implications we have

f(x) = ax + b

g(x) = -ax +b

h(x) = f(x)∙g(x) = -a2x2 + b2

Because h(x) is tangent to f(x) then h'(x) = f '(x) for the point of tangency.

h'(x) = f '(x)

-2a2x = a

x = -12a

Substituting this value in f(x) we have that f(-12a) = a(-12a) + b = -½ + b.

Now the point (-12a , -½ + b) must lie on h(x), therefore

As a result b = ½, the y-intercept of the linear equations f(x) and g(x). Therefore

f(x) = ax + ½

g(x) = -ax + ½

h(x) = -a2x2 + ¼

Consequently the parameter a can be any real number, different from zero, in order to the product of f(x) and g(x) be tangent to each of them at two distinct points.

Finally, lets look an animation of this result.