> >Is the function y = x^(1/3) considered to be locally linear at x = 0? > > > >Mary Thomas > > > I would come down on the side of saying that while y=x(1/3) has a vertical > tangent at x=0, it is not locally linear there since there is no > approximating linear function at x=0. Maybe we should just say that the > function is locally vertical at the origin and locally affine elsewhere.
Thoughtful response to an interesting question. How about this--the function whose graph is described by y = x^3 has the x-axis as its ÃÂbestÃÂ local linear approximator at the origin. I would propose that ÃÂbest linear approximatorÃÂ characterizes tangent lines. Then, consider mapping the graph described by y = x^3 and its local linear approximator at (0,0) with a reflection in the y = x line. Would this not lead to the conclusion that the y-axis is the best local linear approximator (that is, the tangent line) for the graph described by y = x^(1/3) at (0,0). In this case we would simple ackowledge that the best linear approximator does not happen to be the graph of a function.
I have mentioned in a post some time ago that Donald Kreider at Dartmouth created an interesting investigation using a graphing calculator of how the tangent line to a particular funciton at a particular point compares to a line with a slightly different slope through the same point. Each year I take the time with my AB students to run a variation of his investigation. The students see for themselves that the tangent line dominates all competitors in a neat visual way. This has led me to appreciate in a new way the reasonableness of characterizing tangent lines as "BEST" linear approximators.
> > > Doug > > Doug Kuhlmann > Math Department > Phillips Academy > 180 Main Street > Andover, MA 01810 > firstname.lastname@example.org > > --- > You are currently subscribed to ap-calculus as: email@example.com > To unsubscribe send a blank email to leave-ap-calculus-5913831W@list.collegeboard.org
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