EMT 668 Assignment 1
Activity 3

(WRITE UP)

Activity 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 result.

In this activity I have attempted to find a solution to the problem using Algebra Xpresser as the tool. That is I start by making a conjucture, then I test this on Algebra Xpresser , modify the conjecture and test again. I have followed this route as I would like to convince myself, and the reader, that the computer can be a valuable tool in exploring Mathematical properties. Stated more strongly - I would like to suggest that the computer can be used as the tool of exploration.

Starting off:

Based on the experience of EMT 668 Assignment 1 (activity 2) we can make a few observations that will help in making the first stab at this problem.

(i) When h(x) = f(x).g(x) with f(x) and g(x) are both linear functions, h(x) will be a second degree equation (a parabola),

(ii) The x intercepts of f(x) and g(x) will also be the x intercepts h(x).

(iii) h(x) will open up if f(x) and g(x) have the same gradient, but will open down if f(x) and g(x) opposite gradients, and

(iv) we expect that f(x) and g(x) will each cut h(x) at most twice or at least once. This is quite an obvious statement and it is in fact the latter case that we are looking for!

Furthermore:

A rather obvious but no less important statement is the fact that h(x) is determined by f(x) and g(x), in other words f(x) and g(x) are our variables in this problem.

Stab 1:

Consider (1):

We expect h(x) to open up in (i) and to open down in (ii):



fig. 1: The graphs of f(x) , g(x) and h(x).
The left hand graph depicts case (i) while the right hand graph depicts case (ii).

As we look at (i) it seems obvious that the product of two linear functions will never be tangent to BOTH linear functions while they have the "same gradient" (ie both positive or both negative). It is left to the reader to explore this statement.

Looking at (ii) we have a better starting point. We need to "move" the parabola down, in fact the turning point of the parabola needs to "move" below the intersection of the two straight lines. Notice also that the y-intercept of the parabola is equal to the product of the y-intercepts of the straight lines.

Stab 2:

Let us focus on getting the y-intercept of the parabola to be below the respective y-intercepts of the straight lines. Think about this for a moment (be sure that you check that you are happy with the statements I make before going on):

(i) if the two straight line graphs both have y-intercepts greater than or equal to 1 then the parabola will have a y-intercept that is greater than that of either of the straight line graphs and we are no closer to a solution,

(ii) if the two straight line graphs have y-intercepts less than or equal to -1 we have the same scenario as above,

(iii) if the two straight line graphs have intercepts that satisfy: -1 < c < 1 then we may expect the y-intercept of the parabola to be less than either of the y-intercepts of the straight line graphs (compare with the experience of multiplying fractions which leads to smaller numbers......), and

(iv) if the two straight line graphs have y-intercepts so that one is greater than 1 and the other less than -1 then the y-intercept of the parabola will be below the y-intercepts of either linear function.
Based on (iii) lets try: (2)


fig. 2. the graphs f (x), g(x) and h(x), with f(x) and g(x) as in (2)

This is a LOT better. h(x) seems to be "closer" to being a tangent to g(x) than to f(x). Notice that g(x) it is also steeper than f(x) . Some further study of the sketch would suggest that this problem (h(x) being "more" tangent to one of the linear functions than to the other) will continue while the linear functions have different (absolute) gradients (stop for a moment and see if you agree with this statement).

Stab 3:

Lets adjust the gradients of the linear functions: (3)


fig. 3. the graphs f (x), g(x) and h(x), with f(x) and g(x) as in (3)

We are getting closer! This time h(x) seems "more tangent" to f(x) than g(x). The only significant difference between the linear functions remains their y-intercepts (g(x)'s y-intercept is greater than f(x)'s).

Stab 4:


This time let's work with the graphs so that they are symmetrical about the y axis to make things easier for the moment - we can see if this is important later. That is let f(x) and g(x) have the same y intercept and the same (absolute) gradient.
Lets try: (4)


fig. 4. the graphs f (x), g(x) and h(x), with f(x) and g(x) as in (4)

This looks good - lets check by zooming in on the intersections:


fig. 5. detail of the intersection/tangency of f(x), g(x) and h(x)

It would seem as if we have found a case which satisfies our objective. Let us have Algebra Xpresser confirm that h(x) touches each of the linear functions:

Well we have definitely found a case where h(x) is tangent to the two linear functions f(x) and g(x). We now need to generalise our findings - that is we need to explore this special case to see if we can make general predictions about the properties of f(x) and g(x) that will ensure that h(x) is tangent to the two linear functions.

Let's back track and pose some questions:

1. Were we lucky in stab 4 with the choice of 1/2 as the y-intercepts of the linear functions or would 4/5 also have worked?

2. Was the gradient of 2 (and -2) critical or will other gradients also work?

3. Returning to the opening remarks in stab 2: can we satisfy the objective of the problem with linear functions that have y-intercepts of opposite signs and not worry about their magnitude?

As we explore these questions let us keep in mind our objective - to list requirements of f(x) and g(x) so that h(x) will be tangent to both linear functions.

Stab 5:

In response to the first question lets vary the y-intercepts:

* (5)

* (6)


fig. 6: the left hand graph describes (5) while the right hand graph (6)

It would seem as if the y-intercepts of the linear functions are critical.

Stab 6:


Lets explore the question of gradients (keeping the y-intercepts constant).

* (7)

* (8)


fig. 7: the left hand graph describes (7) while the right hand graph (8)

It would seem as if h(x) is again tangent to both f(x) and g(x) in each case. To see a zoomed version of case (7) click here and for a zoomed version of case (8) click here. It is left to the reader to confirm that indeed h(x) touches f(x) and g(x) in each case (as per calculations previously).

Stab 7:

One question from earlier remains (can we satisfy the objective of the problem with linear functions that have y-intercepts of opposite signs and not worry about their magnitude?). Let us answer this question in a different manner. Let us shift the axes (or the graphs) left and right so that we both answer this question, but also address the matter of using the y-axis as an axis of symmetry.......

* (9)

* (10)


fig. 8: the left hand graph describes (9) while the right hand graph (10)

Again we seem to be successful - it is left to the reader to confirm that indeed h(x) is tangent to both f(x) and g(x).

Well where does that leave us......

It would seem as if we can make the following observations from this exploration:

1. We need f(x) and g(x) to have gradients that are symmetrical about the vertical - phrased differently one must have a positive gradient while the other has a negative gradient and the absolute value of the coefficient of x is the same in both cases (though their seems no evidence that value matters). ie:

(11)

2. It would seem as if the only other critical matter is that the y-value of the point of intersection of f(x) and g(x) is 1/2.

This suggests that: and are the general linear equations that will generate h(x) = f(x) . g(x) which is tangent to both f(x) and g(x).

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