Sarah Hofmann

EMAT 6680 Summer 2006

Assignment 1

Exploration 3: Our assignment is to 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.  We are to discuss and illustrat our method and results

To just get started, we try two random linear equations.  Who knows, we may get lucky!  Let f(x) = 3x + 1 and g(x) = 5x - 4.  The the graph of f(x), g(x), and h(x) in red, green and yellow respectively are

Well this wasn't very productive.  There are too many different numbers flying around in these equations to make much sense of what is happening.  Let's try making our equations simpler.  What if we make the slope the same in both equations.  What would happen?  Lets try a slope of 1. Now if we make the y-intercepts the same, we will have two of the same line.  That won't help.  So lets try to make one y-intercept +1 and one -1.  So we are trying the equations f(x) = 1x + 1 and g(x) = 1x - 1.  Graphing f,g, and h in red, green, and yellow respectively we find

This is definately better.  The green line is getting close, it just needs to move down a little.  But all we have to do with the red line is move it down too, but then both lines would be tangent at the same point.  So we nee to change the red line.  What if we switched it so that it had negative slope instead of positive slope?  Then we would just need to move the red line down on the other side of the parabola.  So lets try f(x) = -1x + 1 and g(x) = 1x -1.  Then we get the following graph

Wow! This is a lot better.  Maybe we can make this even easier if we made the center of the parabola (and the point where the lines cross) are centered on the y-axis.  So we need to make sure that f(0) = g(0).  That is that -0+b = 0+c or b=c.  So we need to have an equation like f(x) = -x + 1 and g(x) = x + 1.

Now it looks like all we have to do is slide the graphs up and down until we find the right y-intercepts.  Let's look at an animation that has different values for the y-intercepts and see if that helps.   Click on the graph above to find the animation .

If we look at this animation it looks like the correct answer happens when the y-intercept is between 0 and 1, so lets try a y-intercept of 1/2.  So we will graph the equations for f(x) = -x + 0.5 and g(x) = x + 0.5 below.

This looks right!  Lets double check.  We can see that f(x) = h(x) = f(x)g(x) implies -x+0.5 = (-x+0.5)(x+0.5) implies 0 = x + 0.5 implies x = -0.5.  The equation of the tangent line of h(x) = (-x +0.5)(x + 0.5) at the point x = -0.5 is y = -x + 0.5 = f(x).  So f(x) is tangent to h(x) at x = -0.5.  Similarly g(x) is tangent to h(x) at x = 0.5.  Therefore, we have found the two needed points!

Just for fun, lets see if we can find more sets of lines based on our first set.  Lets try an animation so that both the y - intercepts are 0.5 and the slopes of f(x) and g(x) are n and -n respectively.   Click on the above graph for the animation .

We can see from the animation that in fact, the lines f(x) = nx + 0.5 and g(x) = -nx + 0.5 give the desired result for any nonzero value of n.