**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) andg(x) such that their producth(x) =f(x)∙g(x) is tangent to eachf(x) andg(x) at two distinct points.

Let

f(x) andg(x) be linear functions. So we can write them of the formf(x) =ax+bandg(x)= cx+d, wherea,b,c,d∈ ℜ.Now if

h(x) =f(x)∙g(x), thenh(x) = (ax+b)∙(cx+d) =acx^{2 }+ (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=Ax^{2}+Bx+Cits axis of symmetry is given byx=^{-B}⁄_{2A}.So for the case of

h(x) we have that the axis of symmetry is given byNow because

f(x) andg(x) are concurrent with this line thexcoordinate of the intersection off(x) andg(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) andg(x) as

f(x) =ax+b

g(x) = -ax+dNow without loss of generalization we let the axis of symmetry be the

y-axis, thereforef(x) andg(x) intersects at theiry-intercept. Consequentlyb=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) = -a^{2}x^{2}+b^{2}Because

h(x) is tangent tof(x) thenh'(x) =f'(x) for the point of tangency.

h'(x) =f'(x)-2

a^{2}x=a

x=^{-1}⁄_{2a}Substituting this value in

f(x) we have thatf(^{-1}⁄_{2a}) =a(^{-1}⁄_{2a}) +b= -½ +b.Now the point (

^{-1}⁄_{2a}, -½ +b) must lie onh(x), thereforeAs a result

b= ½, they-intercept of the linear equationsf(x) andg(x). Therefore

f(x) =ax+ ½

g(x) = -ax+ ½

h(x) = -a^{2}x^{2}+ ¼Consequently the parameter

acan be any real number, different from zero, in order to the product off(x) andg(x) be tangent to each of them at two distinct points.Finally, lets look an animation of this result.