SUNY Geneseo Department of Mathematics
Monday, September 11
Math 221 05
Fall 2017
Prof. Doug Baldwin
Geneseo Reaching Out to Women and Under-Represented Groups in STEM Fields.
First event is this week:
Catherine Cerulli, URMC
“Creating a Career Based on Values”
Thursday (Sept. 14), 4:00 - 5:00
Newton 204
Beware that many questions have several subquestions not explicitly labeled as “parts.”
What is the number counting to? Maybe toward 10?
Is there a correlation between the pictures?
What are “yes” & “no” answers to? Which direction the bar moves?
Did the red line start at 0?
Is the red line ever going to overlap itself?
Is the red figure really a 2D view of a 3D shape?
Is the number the radius of the red figure?
Limits are a useful tool for answering some of these questions. At this point you know enough about limits to appreciate that, but unfortunately we don’t get to limits as x approaches infinity ’til chapter 4. We’ll come back to these questions then.
Section 2.4
I have this feeling that when I talk to students about continuous functions, some of them think there’s a horizontal line test to recognize continuous functions. There isn’t, but let’s think a bit about whether there could be.
Note: The horizontal line test is really a test to see if a function is 1-to-1.
A Supposed Discontinuity Test: If there is some horizontal line segment parallel to the interval (a,b) on the x axis such that the graph of f(x) lies below the segment at some point(s) and above it at other(s), but the segment never intersects the graph of f(x), then f is discontinuous on (a,b).
Show that this test is valid, i.e., any function that passes it really is discontinuous on (a,b).
Reading ideas:
Justification for the test: suppose there is a horizontal line segment as described in the test, but f is continuous between a and b. Then because the minimum value of f must be below the line, and the maximum above, the Intermediate Value Theorem says there must be some point where f(x) equals the y value of the line segment. But then the segment would intersect f’s graph, which we know it doesn’t. So the only way all the conditions of the test can hold is if f is discontinuous.
But It’s Not a Continuity Test. It’s possible to construct functions that have the property that every horizontal segment passing above some points on the graph and below others intersects the graph, yet the underlying function is not continuous. So failing to satisfy the test does not show that a function is continuous.
Formal definitions of limits
Read section 2.5