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TANGET PRODUCT BETWEEN TWO POLYNOMIAL FUNCTIONS

By Dario Gonzalez

BEFORE FACING THE PROBLEMS

For this assignment I have decided to solve problems number 3 and 4, which are the followings:

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.

Problem 4: Repeat problem 3 above where f(x) and g(x) are quadratic functions and each function, f(x) and g(x) is tangent to h(x) in two different points. That is, h(x) is a fourth degree equation and each of the second degree equations, f(x) and g(x) is tangent to h(x) in two points.

However, before to find a solution for each problem, I would prefer to discuss some issues related to the conditions of the problems.  This discussion will serve as a background and will provide important principles to solve problems 3 and 4.

 

1) General characteristics of the product of two functions:

The product of two functions f(x) and g(x) is a function h(x) such that if f(x1) = y1 and g(x2) = y2, then h(x1) = y1y2.  In other words, when we multiply two functions, we are multiplying their images for each x Î the intersection between the domains of f(x) and g(x).  Thus, graphically, the product function h(x) is a curve that passes by the point (x1 , y1y2) where f(x1) = y1 and g(x2) = y2.  This is shown in figure 1 below.

Figure 1

 

According to this graphic reasoning, we can conclude an important characteristic of the product function.  If h(x) is the product of the functions f(x) and g(x) then the graph of h(x) will intersect to one of the factor functions when the image of one of them is 0 or 1.  Indeed, if f(x1) = 0 then h(x1) = 0 because f(x1)g(x1) = 0; so, h(x) intersect to f(x) at (x1 , 0).  Similarly, if f(x1) = 1 then h(x1) = g(x1) because f(x1)g(x1) = g(x1); so, h(x) intersect to g(x) at (x1 , g(x1)).  This characteristic is exemplified in figure 2 below.

Figure 2

 

Thus, we can summarize this characteristic of the product function in the following statement:

Statement 1: Let f(x) and g(x) be two continuous functions in the interval [a , b], and let h(x) be the product between f(x) and g(x), then:

·         If f(x1) = 0 for some  then h(x1) = f(x1) = 0.

·         If f(x1) = 1 for some  then h(x1) = g(x1).

 

2) Tangent and no tangent intersections:

In the figure 3 that is presented bellow these lines shows the appearance of tangent intersections between two graphs, and the appearance of a secant intersection between two graphs.

Figure 3 (a)

Figure 3 (b)

 

From the figure 3 (a), we can realize an important characteristic of the tangent intersections; they do not implicate that the graphs have to change their relative order between them.  That is, if two graphs intersect each other tangentially, and the intersection occurs when x = x1 then the graph that had biggest images before x = x1 is going to have the biggest images after x = x1.  Whereas, from figure 3 (b), we can conclude when the intersection is no tangential, and the intersection occurs when x = x1 then the graph that had biggest images before x = x1 is going to have the smallest images after x = x1, and vice versa.  We can summarize this idea in the following statement:

Statement 2: Let f(x) and g(x) be two continuous functions in the interval [x1 – h , x1 + h] where f(x1) = g(x1), and h is taken such that x = x1 is the only solution of the equation f(x) = g(x) in [x1 – h , x1 + h].  Thus, if either f(x) > g(x)  in [x1 – h , x1 + h] or f(x) < g(x)  in [x1 – h , x1 + h], then the intersection in x = x1 is tangential.

 

3) General characteristic of the problems:

This part of the discussion could serve as a first approach to the resolution of problems 3 and 4 since here I will present some general conditions and ideas that it is necessary to review before we face the details in each problem.

First of all, both problems involve finding two polynomial functions f(x) and g(x), with the same degree, whose product h(x) = f(x)g(x) is tangent to each of f(x) and g(x).  According to statement 2, h(x) is going to be tangent to each of f(x) and g(x) in one of two cases:

·         Case (i): h(x) > f(x) and h(x) > g(x)  such that x is different from the intersection points.

·         Case (ii): h(x) < f(x) and h(x) < g(x)  such that x is different from the intersection points.

The idea is shown in the figure 4 bellow:

Figure 4

 

However, both problems involve polynomial functions what impedes case (i) holds.  Indeed, if we remember that h(x) = f(x)g(x), then we observe that, in order to satisfy case (i), it should happen either f(x) > 1 and g(x) > 1 or f(x) < -1 and g(x) < -1, which is impossible if the factor functions are odd-degree polynomial functions because every odd-degree polynomial function has images between –1 and 1 for some interval I.

At the same time, if the two factor functions are even-degree polynomials the case (i) does not hold given the conditions presented in problems 3 and 4.  In general, both problems involve that the product h(x) must be tangent to each of f(x) and g(x) in a number of points that should be equal to the degree of f(x) and g(x).  In other words, if f(x) and g(x) are polynomials of degree 2 then h(x) = f(x)g(x) must be tangent to each of f(x) and g(x) in two different points.  According to this condition, we will have two situations with even-degree polynomial functions.

First, case (i) could implicate that f(x) > 1 and g(x) > 1, but given that f(x) and g(x) are degree 2n, h(x) will be tangent to each of f(x) and g(x) in only n different points, which would be for f(x) when g(x) = 1 and for g(x) when f(x) = 1.  This situation is shown in the figure 5 bellow.

Figure 5

 

Second, case (i) could also implicate that f(x) < -1 and g(x) < -1, then h(x) > 1 so there will be no intersections between h(x) and the functions f(x) and g(x).  Thus, the case (i) does not hold for even-degree polynomial functions either which makes us to obtain crucial conclusions that I summarize in the following statements.

Statement 3: Let f(x) and g(x) be polynomial functions of degree n, and let h(x) = f(x)g(x) such that h(x) is tangent to each of f(x) and g(x) in n different points, then h(x) < f(x) and h(x) < g(x)  such that x is different from the intersection points.  This condition involves exactly three relationships between images of f(x) and images of g(x):

·         Relationship 1: if f(x) < 0 then g(x) > 1

·         Relationship 2: if f(x) > 1 then g(x) < 0

·         Relationship 3: if 0 < f(x) < 1 then 0 < g(x) < 1

Statement 4: Let f(x) and g(x) be polynomial functions such that  and , and let h(x) = f(x)g(x) then, in order to satisfy case (ii) and given that h(x) is even-degree polynomial function, .

 

SOLVING THE PROBLEMS:

After having the review presented above, we now can face problems 3 and 4 to find the solution for each one of them.

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.

Suppose a linear function .  According to statement 1, the points where f(x) = 0 and f(x) = 1 are crucial because they determine the possible (note that I wrote “possible” in italic letters) intersections between the function product h(x) and the factor functions f(x) and g(x).

These points are:

 then

 then

 

Now, the figure 6 shows the hypothetic graph of f(x) where were marked the points  and .  Also, figure 6 shows the regions where the other linear function should pass through in order to satisfy relationships 1, 2 and 3 presented in statement 3.

Figure 6

 

According to this graphic argument, the other linear function g(x) must satisfy  and .  First, in order to determinate g(x), we could calculate the slope of g(x):

 

Finally, since we know that  is a root of g(x), we can write its equation as follow:

 

Note that the equation of g(x) has been written in terms of the equation of f(x), and it is not only because this expression evinces the relationship between the equation of f(x) and g(x), but it also suggests an interesting conjecture that would permit to face a problem similar to problem 3 and 4, when the functions involved in it are polynomials of degree n > 2.  This conjecture will be presented again later.

Thus, two linear functions f(x) and g(x) whose product h(x) is tangent to each of f(x) and g(x) in one point for each must exhibit the following relationship: if  then .  For example, if we consider  then .  Thus, h(x) = f(x)g(x) will tangent to f(x) in x = 1.75 and to g(x) in x = 3.  The figure 7 shows the graph of f(x), g(x) and h(x):

Figure 7

 

Important: Note that the tangent points represent the roots of f(x) and g(x).  In other words, h(x) is tangent to f(x) when f(x) = 0, and h(x) is tangent to g(x) when g(x) = 0.

 

Problem 4: Repeat problem 3 above where f(x) and g(x) are quadratic functions and each function, f(x) and g(x) is tangent to h(x) in two different points. That is, h(x) is a fourth degree equation and each of the second degree equations, f(x) and g(x) is tangent to h(x) in two points.

It is a fact that one of these two quadratic functions, let’s say f(x), will present images > 1.  Indeed, statement 4 states that one of these quadratic functions should have positive orientation; that is, if  we can consider that .  Moreover, according to what was discussed in the part 3 “General characteristic of the problems” f(x) should also present images between 0 and 1, which implicates that the equation f(x) = 1 will have two solutions; in other words, there are two values of x, let’s say x1 and x2 such that f(x1) = f(x2) = 1.  Therefore, given the positive orientation of f(x), f(x) > 1 either " x Î ]-¥ , x1] or " x Î [x2 , +¥[ which implicates that g(x) < 0 on the same intervals, and f(x1) = f(x2) = 1 whereas g(x1) = g(x2) = 0.  Also, given the symmetry of every parabola, it easy to note that f(x) and g(x) will have their vertices for the same .  Given the anterior lines, let’s say that the graph of f(x) and g(x) are as the ones shown in figure 8 that appears bellow:

Figure 8

 

Remember that figure 8 represent a tentative graph of f(x) and g(x).  So, according to statement 1, x1 and x2 will be intersection points between g(x) and the product function h(x); that is, g(x) already has the two intersections requested by the problem.  In contrast, if we consider figure 8 f(x) have no intersection points with h(x) yet.  Therefore, it is clear that the graphs shown in figure 8 are not the exactly representation of f(x) and g(x), but these graphs provide the following reasoning: Having intersection points between f(x) and h(x) implicates that these intersections should occur when f(x) = 0.

According to statement 1, we would have two intersections between f(x) and h(x) if one of the following conditions is satisfied for some x3, x4 Î ]x1 , x2[ such that x3 x4:

1)      f(x3) = 0 and g(x4) = 1.

2)      f(x3) = 0 and f(x4) = 0.

3)      g(x3) = 1 and g(x4) = 1.

If we observe figure 8, condition 1 would implicate that the vertex of f(x) should be located at (x3 , 0), and the vertex of g(x) should be located at (x4 , 1).  However, both vertices have the same x-coordinate then , but it is contradictory with our assumption that .  Thus, condition 1 does not hold.

Conditions 2 would implicate that f(x) < 0 , which in turn implicates that g(x) > 1  according to relationship 1 of statement 3.  Latter is crucial because it involves that the solution for f(x) = 0 and g(x) = 1 are the same; in other words, conditions 2 and 3 implicate each other.  Thus, if f(x3) = 0 and f(x4) = 0 then g(x3) = 1 and g(x4) = 1.

In conclusion, it is reasonable to visualize the graph of f(x) as it is shown in figure 9.  Also, figure 9 shows the regions through which g(x) should passes according to relationships 1, 2 and 3 in statement 3.

Figure 9

 

Thus, suppose that  with  then, according to figure 9, x3 and x4 are roots of f(x), so we can rewrite the equation as .  Moreover, given that f(x1) = 1, we can write the following expression:

                          (expression 1)

 

Now consider the blue points by which g(x) should passes given in figure 9, we can suggest the axes translation  and  to write the equation of g(x).  In the  axes-system, x3 and x4 are roots of the graph of g(x).  So, the graph of g(x) in the  axes-system would have the following equation:

 

Then, by replacing the relationships  and , we obtain:

 

Soon, from the figure 9, we know that g(x1) = 0.  Therefore, we obtain an expression as follow:

                         

                        (expression 2)

 

Finally, taking expressions 1 and 2 we can write:

 

In conclusion, if  then the equation of g(x) should be .  In other words, Let f(x) and g(x) be quadratic functions such that each one has two roots, and let h(x) = f(x)g(x) and tangent to each of f(x) and g(x) in two different points then , and the tangent intersections with h(x) occur in the roots of f(x) and g(x).

For instance, if  then .  Thus, f(x), g(x) and h(x) = f(x)g(x) will have the following graphs shown by figure 10 bellow:

Figure 10

 

 

CONCLUSION

After this long discussion about problems 3 and 4, it is possible to establish some conclusion that permit us to solve similar problems when the polynomials presented are degree n > 2.  These conclusions can be summarized as follow:

Conclusion 1: The tangent intersections between the product function and the factor functions occur in the roots of the latter.  So, since the number of intersections is equal to degree of each factor function then each factor function must have as many real roots as its degree.

Conclusion 2: If h(x) = f(x)g(x) is tangent to each of f(x) and g(x) in as many points as the degrees of each one then there exist the following relationship between f(x) and g(x):

 

Thus, it is possible to solve problems similar to problems 3 and 4.  For example, consider the 5-degree function  whose graph is shown in figure 11 that appears bellow:

Figure 11

 

Note that there are three real solutions for f(x) = 1.  This is a consequence of conclusion 1 since if g(x) is going to have 5 real roots then according to relationship 1 and 2 in statement 3, f(x) must be equal to 1 for the same values of x where g(x) =0.  So, 5 roots in g(x) implicates 5 “ones” in f(x).

Finally, according to conclusions 1 and 2, we can find the equation of g(x) by the relationship .  Thus, the graph of f(x), g(x) and h(x) will be as follow below:

Figure 12

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