Math 6680 – Jackie Gammaro

Assignment #1 - Problem #3

 

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.  Problem Solve - Trial and Error

 

After a few tries, I let f(x) = 2x + 4 and g(x) = -2x - 3, thus their product h(x) =(2x + 4)(-2x - 3).

The following is a graph of f(x), g(x), and h(x).

   

                                                           

 

II.  What are the equations?

 

After this little success I tried to generalize.  I noticed the slopes of the two linear functions were opposites of each other and that their was a relationship between the y-intercepts of f(x) and g(x).

Thus I let f(x) = mx + b,  g(x) = -mx – b + 1, and h(x) = (mx + b)( -mx – b + 1).  Below is the graph of these functions.  Note: Graphing Calculator 3.5 chooses specific values for m and b.  In this case    m = 1 and b =1.

                                                       

 

III.  What are the points of intersection?

 I noticed the intersections of f(x) with h(x) and of g(x) with h(x) lie on the y-axis, thus when the functions equal zero.   How can I show this algebraically?

 

If f(x) = mx + b,  g(x) = -mx – b + 1, and h(x) = (mx + b)( -mx – b + 1), to find the points of intersection, determine when f(x) = h(x) and when g(x) = h(x).

 

f(x) = h(x)                                                                    g(x) = h(x)                                                            

       mx + b = (mx +b)(-mx –b+1)                                    -mx-b+1 = (mx+b)(-mx-b+1)

                 1 = -mx –b+1                                                               1 = mx + b

                           0 = -mx -b                                                                1-b = mx

            x =                                                                         x =

 

  f =                                     

 

                                       

 

 

 

Thus, the points of intersection are and .

 

IV.  Do the points of intersection always lie on the x-axis?

 

I also wondered, and If they do, then WHY? 

 

1.     First I examined the graphs of f(x –k), g(x –k) and h(x – k) and then determined algebraically, that horizontal translations of the functions dont change the points of intersection off the x-axis.

2.     I then examined the graphs of f(x) + k, g(x) + k and h(x) + k. Just from viewing the graphs in Graphing Calculator, I determined the stipulation that h(x) has to be tangent to f(x) and g(x) at the points of intersection was not visible. However, points of intersection do lie on the x-axis.  In this case m, b, k =1.

                                                           

 

 

 

3.     At this point I wondered, WHY the x-axis?

I studied the graphs, I studied my algebra, and it wasnt clicking.  Then the lightbulb went off!

 

I had to determine when f(x) = h(x) and g(x) = h(x) and I know h(x) = f(x)*g(x).

 

So I let f(x) = a and g(x) = b, and assuming x , substituted into the previous equations which results to

           

            f(x) = h(x)                    and                 g(x) = h(x)

 

                 a = ab                                                  b = ab

 

                       a = 0 or b = 1 or a,b=0                            b = 0 or a = 1 or a,b=0

 

 

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