Warm up was to find the value of c guaranteed to exist for a continuous function by the IVT. Went over assessments some, then passed out AP Limits packet due Monday 9/17. Looked at a difficult limit problem where a secondary variable, h, changed instead of x. Started examining the problem of dividing by zero and how limits can help fix that, which motivates a central question in calculus: how to find the slope of a curve (or more specifically, the slope of a line tangent to a curve). Usually a line requires 2 pts to get slope, but a tangent by definition only has 1 point in common with the curve. The remedy is to take the slope of the line that crosses twice (a secant) and make the gap in between the two points (delta x) approach zero. This is done with a limit, and as delta x shrinks, the secant line becomes the tangent line. (Applet here). This expression is known in calculus as the "limit definition of derivative." It has 2 notations (actually more), owing to its history of having 2 inventors.
Notes from board
Limits Packet
Homework
p80 #87-90,95-98
limits packet due Monday
Resources
limit definition of derivative (summary of what we did in class today) LINK
showing that a function has a root in a given interval: link
finding the value of c guaranteed by the IVT: link