Linear Functions Tangent to Their Product

by

Stacy Musgrave

The problem at hand 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 first begin by eliminating a few cases:

- If f(x) and g(x) are both horizontal lines, their product will also be a horizontal line. The only possibilities are that h(x) is either f(x) or g(x) if one of these maps all x to 1, in which case h(x) never meets one and meets the other everywhere, or h(x) is disjoint from both orginial graphs and hence has no points of intersection. So our linear functions f(x) and g(x) cannot both have zero slope.
Picture: Let f(x) = 1 and g(x) = -2, so h(x) = g(x) = -2

- If f(x) is a horizontal line, say f(x) = b, and g(x) has non-zero slope, say g(x) = cx+d, then h(x) = (bc)x + bd. If b=1, h(x) = g(x) and has no point of tangency to either f(x) or g(x) and if not, then h(x) will pass through f(x) when cx + d = 1 (i.e. x = (1-d)/c). Hence this case is not applicable to the current task either, and both of our functions must have non-zero slope.
Picture: Let f(x) = 2 and g(x) = x - 1, so h(x) = 2(x - 1)

With these cases eliminated, we begin by letting f(x)=ax+b and g(x)=cx+d, with a and c non-zero real numbers and b and d any real numbers. Then define h(x) = f(x)g(x) so that h(x) has the same x-intercepts as f(x) and g(x). Namely, h(x) intersects the f(x) at x= -b/a and h(x) intersects g(x) at x= -d/c.

Since we want h(x) to be tangent to f(x) and g(x), these points of intersection prove to be ready candidates for the points of tangency.

We turn to our recollection of calculus and the derivative to find the slope-predictor formula for the slope of the tangent line to h(x) at x:

Expanding h(x), we get: h(x) = acx

^{2}+ (ad+bc)x + bdSo, h'(x) = 2acx + (ad+bc)

Now we want h(x) tangent to f(x) at their common x-intercept (x = -b/a), meaning the slope of the tangent line should be the same as the slope of f(x):

a = 2ac(-b/a) + ad + bc

Simplifying and solving this for a, we see a = (bc)/(d-1)

Likewise, we want h(x) tangent to g(x) at their common x-intercept (x= -d/c), meaning the slope of the tangent line should be the same as the slope of g(x):

c = 2ac(-d/c) + ad + bc

Simplifying and solving this for a, we see a = (cb-c)/d

Setting our two equations for a equal, we get the following equation: c(d + b - 1) = 0. Since c was taken to be nonzero, we conclude d + b - 1 = 0. Solving this for b and plugging into the second equation for a, we get

a = (c(1 - d) - c)/d = (c - cd - c)/d = (-cd)/d = -c.

So f(x) and g(x) must have opposite slopes with y-intercepts summing to 1.

Let's see an example of this:

Let f(x) = 3x - 2 and g(x) = -3x + 3. Then the graph of the f(x), g(x) and h(x) = f(x)g(x) is:

Further ExplorationNow, what if we want to choose our points of tangency? We still have the original restriction of the graphs intersecting at the x-axis, but let's say we choose these points and ask which lines give us the desired property.

Suppose we want the points of tangency to occur at P(p,0) and Q(q,0). Then if f(x) = ax+b, g(x) = cx+d and h(x) = f(x)g(x) as before with h(x) tangent to f(x) at A and to g(x) at B, we have:

Looking at x-intercepts: 0 = a(p) + b, so b = -ap

0 = c(q) + d, so d = -cq

So f(x) = ax - ap and g(x) = cx - cq.

Taking the derivative of h(x), we get h'(x) = 2acx - ac(p + q).

Setting this equal to a when x = p and c when x = q as before, we find a = 2acp - ac(p + q) implies 1 = c(p-q), giving c = 1/(p-q)

and c = 2acq - ac(p+q) implies 1 = a(q-p), giving a = 1/(q-p).

So f(x) = (1/(q-p))x - (p/(q-p)) and g(x) = (1/(p-q))x - (q/(p-q)).

Notice the slopes are opposites and the y-intercepts sum to 1 as we had established necessary in the above argument.

Example: Find two linear functions f(x) and g(x) such that their product is tangent to either at P(2,0) and Q(-3, 0).

As per the above calculations, f(x) = (-1/5)x + (2/5) and g(x) = (1/5)x + (3/5).

We graph these lines and their product:

There are a few more observations to make. The first is that we can discuss the tangency of the lines to the parabola without the use of calculus. By definition a line is tangent to a curve if they meet at exactly one point. So we may check, for example, that f(x) is greater than or equal to h(x) for all x.

Since the last inequality is true for all x, with equality holding at x=2, we conclude f and h are tangent at (2, f(2)).

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