OCTOBER 8: INFLECTION POINTS, CRITICAL POINTS CONCAVITY AND DERIVS REVIEW

Hello blog. It's me Jamie again with today's scribe post or yesterday's really. I'm starting it up again, after it abruptly stopped for that day. At the moment the class is basically at the peak of chapter 2. It's still a bit vague for me but having class with Mrs. Karras and her famous notations like "fcn" helps.

First we spent the first seven minutes or so settling in and having a few warm up discussions, relating to math of course. Most of it. Time is golden, right? We were deciding amongst ourselves whether we were audio/visual learners who absolutely adore diagrams [and possibly lectures] and those theoretic learners who love...well, theorems and of course, there are the extraordinary who are both. But let's get down to business.

We started the class reviewing what we should know... and I guess emphasizing that knowledge on derivatives. We began with the FIRST DERIVATIVE TEST, although it was not called that in the text. Of course, this is one way to find the relative maxima or minima of a function, without being too tedious.

Like on the slides, there are the notes describing the function itself. But to carve this inside your head, I guess, I'll repeat. If this derivative test is to be used, it follows these conditions:

• f is a continuous function located on a closed interval [a, b]
• f is differentiable on an open interval (a, b)

Along with those two conditions, are indicators of which direction the intervals would be, which are:

1. If f '(x) is pos^ve for the x values in the open interval, then f '(x) is INCREASING on the closed interval, [a, b]
2. If f '(x) is neg^ve for the x values in the open interval, then f '(x) is DECREASING on the closed interval.
   

Secondly, to make sure we were all on the same train of thought, we defined EXTREMA. This the term describing either the local minimum or maximum of a differentiable fcn. 

Extrema occur in a fcn where f is on an interval (I) and c (a critical point; more info on critical points here) is a number in I. They also follow these conditions:

• f (c) is the max value of the function on the interval given that f '(x) is less than or equal to f (c) or vice versa, which is shown in the next pt. [LOCAL MAX]
• f (c) is the min value of the fcn on the interval given that f '(x) is greater than or equal to f (c). [LOCAL MIN]
  

But if f (c) is constant, then the function is BOTH a max and min value for EVERY real number c, of course considering that not every function has both a minimum and/or a maximum.

[more on...] CRITICAL NUMBERS 

This next section was a further explanation of critical numbers. But basically, if a function has a local extrema then either f '(c) equals zero or is undefined.

CONCAVITY and the 2nd DERIV TEST

This was one moment in class where I was quite intrigued. It was one of those moments where you learn a concept through a different perspective, in this case, I learned a new way to describe what concavity is. Concave UP was when the function was curving upwards above the tangent line and concave DOWN was if it was curving downwards below the tangent line.

**NOTE VOCABULARY: INFLECTION POINT [click on the word for a link]

THE SECOND DERIVATIVE TEST

This occurs when a function is differentiable on an open interval containing "c" and f '(c) = 0. Once again, this indicates where there is a local extrema. If f "(c) is negative, then f has a local MAX at "c". If f "(c) is positive, then f has a local MINIMUM at "c".

With taking notes aside, we were given a warm up question that applied some of the things we've learned so far:

If f (x) = 12 + 2x² - x⁴ , use the 2nd derivative test to find the local max and min for f . Discuss concavity, find points of inflection and sketch the graph.

THE SOLUTION:

Use the power law to differentiate the first and second derivative. The power rule is also known as: f ' (x) = nx^{n - 1} 

f ' (x) = 4x - 4x³  first deriv

f " (x) = 4 - 12x^{2 =} 4x (1 - x)(1 + x) second deriv

To find critical numbers, we must know that critical numbers exist when f '(x) [or the 2nd deriv] = 0, meaning that the critical numbers were {-1, 0, 1}.

Then to continue this problem, finding the inflected points makes finding the concavity easier since we can find whether the function increases or decreases by separating the intervals accordingly by using the critical numbers and plugging numbers that were in b/w intervals into the first derivative. But then the bell rang and we are to conclude this problem on Tuesday. Also, Mrs. Karras will be back to teach us about limits. 

Later that week though, just a reminder, when Dr. Eviatar comes back, we will have our pre-test for this unit and the test the following day. My post today wasn't quite exciting, but at least there was no major advertizing... [or is there?] But anyhow... I'm famished-- so are my fingers typing, so I'm done for tonight. The next scribe derived from randomosity is Richard.