Some days ago, I did a puzzle using exp(x) and ln(x) and it occurred to me that these functions can be used to start off a different type of puzzle.
Let x = 1. Then you have the points (1,e) and (1,0). Now draw tangents to the curves at those points. The tangent lines (most likely - I haven't formally figured it out this far) would intersect at a point. Figure out what this point is. Now let x = 2 to get the points (2, exp(2)) and (2, ln(2)). Now draw tangents to those two points and figure out what the point of intersection is for the next tangent lines. Let x take on various different values and derive more points of intersection.
Can you produce a graph showing the locus of intersection points? As an extra bonus question, can you derive an equation for these locus points? (to take it a step further, you may want to explore what happens letting y = x +1 or using parabolas for variety sakes).
October 23rd 2010, 12:29 PM
This was a fun problem. Here's my answer. For the function we write the equation of the tangent line to the point as Similarly, for the function we write the equation of the tangent line as Now, given that and we can immediately write down the equations of the tangent lines:
We want to set these equal. Obviously, this will have to happen when so re-define and set
Solving for yields
Plugging this back into either or (it shouldn't matter, although I did it in both as a check, and they both come out the same), you get