Tangents to Functions

by

Susan Sexton

 

 

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. Discuss and illustrate the method and the results.

 

I first tried graphing two different functions and their product.

 

 

 

Looking at the result some modifications needed to be made . . .

First one of the lines needed to have a negative slope this would create a different product function.

 

 

This appears to be better but the linear functions again need to be modified . . .

Hmmm . . .

Maybe some algebra (and calculus) may help.

 

I will first generalize my two functions:

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

so h(x)=(ax+b)(cx+d)

 

By distributing h(x) new form is:

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

h(x) = acx2 + (ad+bc)x + bd

 

Finding the derivative should help in finding the points tangent to h(x) at any given x. 

 

h¢(x)=2acx + ad +bc

 

Since the derivative is the slope of the line tangent to the function at x, I need to find x-values.  Therefore I will find the roots of f(x) and g(x) which are x1 = -b/a and x2 = -d/c respectively.

 

By substituting the x values into the derivative this will yield the slope of the line tangent to h(x) at x.

 

h¢(x1)= 2acx1 + ad +bc         and            h¢(x2)=2acx2+ ad +bc

h¢(-b/a)=2ac(-b/a) + ad +bc             h¢(-d/c)=2ac(-d/c) +ad +bc

   =-2bc + ad + bc                                      =-2ad + ad + bc

            =ad – bc                                                  =-ad + bc

 

Setting these equal to the slopes of the given functions, f(x) and g(x) we get:

a = ad – bc          and  c = -ad + bc

 

This means that the slopes of the tangent functions are the same but with opposite signs.

So now I adjust my original functions to reflect this:

 

 

This looks a lot better!  Now to work on the y-intercepts.

Back to the algebra (and calculus) . . .

 

Finding that

a = ad – bc and   c = -ad + bc

 

lead to the conclusion that a = -c.

 

I need to find a relationship between b and d (the two y-intercepts).

 

Using substitution I get:

a = ad – b(-a)

1 = d + b

 

So the sum of the y-intercepts is 1.

 

Let me again work on the original linear functions:

 

 

That looks pretty good, but let me zoom in a little to see if those lines are indeed tangent.

 

I think that confirms it!

 

So this may work for any two linear functions . . .

 

If f(x) = 3x + 5 then what are the functions g(x) and h(x) and their respective graphs?

(Answer)

 

If using strictly algebra (no calculus involved) is there another approach?

(Answer)

 

 

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