The Department of Mathematics and Science Education
Find two linear functions f(x) and g(x) such that the product
h(x) = f(x).g(x) is tangent to each.
This problem was posed by a group of teachers during a workshop
in which the use of function graphers was being explored. Our
analysis is presented as a sort of stream of consciousness account
of how one might explore the problem with the tools at hand. In
fact, we came up with two different streams of consciousness and
so we have two senarios that are parallel in that they cover alternative
approaches to the problem. The senarios represent a composite
of several discussions of the problem with teachers, students,
and colleagues. Note, the goal here is using this problem context,
not only to solve the problem posed, but to understand the concepts
and procedures underlying the problem.
Function graphers are available for almost any computing platform
or graphic calculator. Such tools make it possible to look at
new topics in the mathematics curriculum or to look at current
topics in different ways. This problem has some elements of each.
In general it would not be included in the school curriculum,
but there is no reason it should not. Further, the use of technological
tools to examine visualizations of the functions makes for a different
approach to the problem.
Is it possible to find two linear functions, f(x) = mx + b and
g(x) = nx + c, such that the function h(x) = f(x).g(x) is tangent
to each. A traditional approach would begin with algebraic manipulation.
This is useful because it keeps the students occupied, but what
do they learn from it? Clearly, h(x) = (mx + b)(nx + c) is a polynomial
of degree 2 and h(x) has two roots. The respective roots are when
f(x) = 0 and g(x) = 0. This means the graph of h(x) crosses the
x-axis at the same two points as f(x) and g(x). Thus, if there
are points of tangency then they must occur at these common points
on the x-axis. Experienced students, very bright students, and
good problem solvers could whittle this information and a lot
more information out of this algebraic analysis. Novice students
would be more tentative.
Lets open up the function grapher and explore
with some specific f(x), g(x), and the resulting h(x). Lets try
The graphs on the same set of axes are
For some novices, seeing the graph of the product
h(x) = (3x + 2)(2x+1) and the graphs of the two straight lines
from the factors on the same coordinate axes provides a new experience.
This particular graph has one of the two lines "close"
to being tangent to the product curve but the other one is not
close. How could the picture be changed?
One idea is to spread the two lines so that one has negative slope.
Try
The graphs are
This is better? What can be observed? How can
the graph of h(x) be "moved down?" What if the graphs
of f(x) and g(x) had smaller y-intercepts? Try
The graphs are
Still not too good but at least the graph of
h(x) was "moved down."
Try smaller y-intercepts, such as
and result in graphs of
These graphs seem close, but clearly the line
with negative slope is not tangent to the graph of h(x). Looking
back over the sequence of graphs (and perhaps generating some
others) the graph of h(x) always has a line of symmetry parallel
to the y-axis. It seems that the pair of tangent lines will have
to have this same symmetry. How? Try making the slopes 3 and -3.
The functions are
and the graphs are
A zoom to the right hand side of the graph
with give
showing tangency has not been achieved. A zoom
to the left hand side shows a similar problem.
One could try adjusting the y-intercepts. In
fact, if the y-intercepts were equal, the y-axis would be the
line of symmetry. Try
The resulting graphs are
Worse. Try
The graphs are
A zoom to the left shows
and to the right shows
The result seems on target. It remains to confirm
f(x) and g(x) each share exactly one point in common with h(x).
Again the tradition is to do so algebraically, but it might be
instructive to look at some graphs of h(x) - f(x) and h(x) - g(x),
such as the following:
Can we immediately generate graphs of other
f(x), g(x), and h(x) satisfying the conditions of the problem?
Reviewing the graphs and the strategy, its seems that the slopes
of the lines can vary, the only condition being that they are
m and -m. So, a simpler case might be to let the slopes be 1 and
-1, giving
and
It is also of interest to see both of the solutions
on the same graph:
Other solutions could be generated by making
some other vertical line the axis of symmetry. Indeed substituting
gives the graphs
for which the equations simplify to
Try
First consider the graphs of f(x) and g(x)
and try to sketch in h(x). The graph is
What do these lines tell you about the parabola?
What points do you know the curve will go through? Why? What causes
it to open the way it does? Now let's add the graph of the parabola
and compare it with our sketch.
How is it like our sketch? How is it different?
Is there anything we should notice or consider?
It appears that all three graphs seem to intersect at 1 on the
y axis. Lets zoom for a closer look.
Maybe changing one of the functions will help
with the explanation.
Consider
The three functions no longer intersect at
1 on the y-axis. However, the changed function, f(x), does intersect
the curve at its y-intercept.
When g(x) = 1 the parabola intersects f(x). Is the opposite true?
Lets graph and see.
It seems that if f(x) = 1 then h(x) = g(x)
and if g(x) = 1 then h(x) = f(x). Lets test this by trying to
generate h(x) from a new f(x) and g(x). Let
Then add the sketch h(x) and compare with .
. .
.In this process we seem to have also noticed
that the lines and the parabola intersect at the points when the
lines cross the x-axis. Why would this be true?
Now the goal is to get one line tangent to the parabola. The function
g(x) is close to being tangent. If we could just get the two points
to slid together then they would become one point -- the point
of tangency. (If a line intersects a parabola in exactly one point,
what is true about the line?)
Since f(x) takes on a value of 1 when x = 1, then lets try to
change g(x) so that g(1) = 0. Lets see we could change the slope
or change g(x)'s position up and down.
Lets try them both.
To change slope, g(x) = -2x + 2 and test to see if g(1) = -2(1)
+ 2 = 0
Or change position, g(x) = -3x + 3 and test
g(1) = -3(1) + 3 = 0
This seems to imply that f(x) and h(x) are
tangent at the f(x) and h(x)'s common root, if the function g(x)
takes on the value 1 at this root. Or in other words if f(a) =
0 and g(a) = 1 then f(x) is tangent to h(x) at a.
What would we have to do to get both f and g tangent to h? That
would mean that when f(x) = 0, then g(x) = 1; and when g(x) =
0, then f(x) = 1. Lets first start with an easy function for f
and then try to generate a g(x) which satisfies what we want.
Lets begin with f(x) = x.
Now when f(x) = 0, then g(x) should take a
value of 1. In other words if f(0) = 0, then g(0) = 1. Likewise,
when g(x) = 0, f(x) should have a value of 1. Since f(1) = 1 then
g(1) = 0. We need a our linear function g(x) to go through (0,1)
and (1,0). So our g(x) = -x +1. Lets graph it to check.
That looks right! Now, add the graph of the
product and then test it to see if the curves are tangent.
That looks good. (What is the coordinates of
the vertex?) Lets zoom in at the roots.
and
.This seems to be a useful direction. Does it
work on the previous problem? When we left off, f(x) = x and g(x)
= -3x + 3. Can we use our technique to find a different f(x) that
works for g(x) to produce h(x) = f(x).g(x) with f(x) and g(x)
each tangent to h(x)? We have the following graph.
Since g(1) = 0, then f(1) = 1 and since g(2/3)
= 1 then f(2/3) = 0 will be necessary. So f(x) contains the points
(1,1) and (2/3, 0). Try g(x) = -3x - 2.
Zoom in for a closer look.
What are the coordinates of the upper vertex
of the triangle? What are the coordinates of the vertex of the
parabola? Are the lines really tangent to the parabola? How might
this be proved? Where else could the function f(x) possibly take
on the same value as h(x) if h(x) = f(x).g(x)? And how can we
interpret this on the graphs?
Summary
Many issues are hidden in these composite accounts of examination
of this problem. We still have the additional problem of writing
a concise argument of proof of the demonstration -- that the solution
will always have the two lines of slope m and -m crossing on y
= 1 and the vertex of the parabola on y = 1/2.
Each senario presents a somewhat different approach. Which would
be most helpful in finding two quadratic functions f(x) and g(x)
such that their product function h(x) = f(x).g(x) has each tangent?
The following graphs show such functions. How can they be generated?
What are some generalizations of the problem
(and the solutions)?