The Second Derivative Test

Hey guys, this is 'Not Paul' scribing for Dec 2nd 2008. Today (or rather, yesterday as of the time this is being written), we continued talking about the Second Derivative Test and how we can use it. Oh and by the way the next scribe is Francis.

Okay, first, what is the second derivative test?

The second derivative test is basically a little trick that lets you know the concavity of your function.

If the value of the second derivative at a critical point (of the original function) is positive, then the original function is concave up at that point and is a local maximum. If it is negative, it is a local minimum.

To help me explain why this is, I have drawn a little squiggly:

Okay, so on this squiggly, we have atleast 4 critical points. For now, we'll deal only with one of the critical points to demonstrate the second derivative test. The easiest point to use is the local maximum at approxmiately (-1,1). I have labelled it C here.

Well, how is this related to the second derivative test? I'll tell, if you'd stop being so impatient.

The second derivative is the derivative of the first derivative right? And what is the first derivative, but the slope of the tangent line at a point in an interval? So our first function here, its on the interval [-3,3]. And the first derivative of our point c (c') is the slope of the tangent line at c.

As you can see, at our critical point c, c' is a horizontal line, which means the slope of c' (and the derivative of c) is 0. This is what the graph of our first derivative would look like (roughly):

Note the roots of the first derivative function at point c.

On a side note, as Benchmen pointed out we can find inflection points by looking for local maximum and minimums in the first derivative. When the first derivative function has a maxmimum or mimimum, that is where the slope of the original function is largest or smallest locally. That is basically the definition of an inflection point.

We can see that in the graph as the local maximums and minimums of the first derivative graph occur at inflection points on the original graphs. 

So anyway, our second derivative is just a derivative of our first derivative. Easy.

Yes, messy, I know, but I promise, this has a point.

See, if we take a look at my rough graph of the second derivative, we can tell three things:

a) At c, c" is negative

b) At c, c' is zero

c) The roots of the second derivative function are where we have inflection points

I'll make another point, g (for grue), and label it on the graph.

So, as we can see, where g" is positive, g' is zero and g is concave up.

So by this, we know a few things. First, we know that you can easily find the concavity of the function two ways:

By inputting the critical value (value being the x coordinate) into our second derivative function. If the output of the second derivative function is positive at that critical value, we know the original function is concave up. If it's negative, we know it's concave down.

Or by looking at the next value of the first derivative. Since we have value c which is a critical value, and we know c' is zero at c, if the value of the derivative function is negative at c + 1 or c + h where h is any small value (maybe its epsilon!) then the function is concave down (because the slope is becoming negative therefore the original function is decreasing). Conversely, if it's positive then it must be concave up.

Second, we know where the inflections points are (usually, or where they SHOULD be):

They are where you have roots on the second derivative, or where the second derivative changes from negative to positive

Orrr, where the first derivative has a local max/min (makes sense, as we know the derivative has a root at a critical point in the original function and second derivative is the derivative of the first derivative!)

So there you have it, thats what the second derivative test is about. Explained in a really long roundabout way.

Simple right?

So then, this stuff we did today should be a breeze:

We have a function f(x) = 4e^(-x^2).

We find the derivative of it with the chain rule (yes the chain rule!). Think of -x^2 as a function, so we have f(g(x)) = f(x) where g(x) = -x^2. So f(g(x)) = 4e^(g(x)). Simple! The derivative of 4e^(g(x)) is, well, 4e^(g(x)), and the derivative of g(x) is -2x. Multiply! Volia, -8xe^(-x^2) is our first derivative.

Since the only number than will make f'(x) = 0 is x = 0, then at x = 0 we have our critical value.

Then we derive again to get the second derivative.  Input the critical value into the second derivative equation, and f''(0) < x =" 0">

We apply the same process to the second questions. Remember! The derivative of ln(x) is 1/x.

Because of the +1 in the numerator, our function will never be undefined. So we only have to worry about the numerator. At x = 0, the first derivative is 0 so at x = o is where our critical value is.

To find the second derivative, we have to use the quotient rule. LO-DEE HIGH MINUS HI-DEE LO ALL OVER LO LO [In other words, the derivative of the denominator function times the numerator function minus the derivative of the numerator function times the denominator, all divided by the denominator squared].

Input x = 0 into the second derivative equation. It is greater than zero, so it must be a local minimum and the graph is concave up.

We also did some stuff that involved Grade 11 stuff, namely a function that looks like this:

[(x-a)(x-b)]/[(x-c)(x-d)]

Which serves as a lead in to our next topic of INFINITE LIMITS! I'd go into detail but Mr.K didnt post the slides for this, and it's 6:55am. 

Dont forget, homework was exercise 5.2 all odd numbered questions!

And lastly, the next scribe is not Joseph because he'll be away! Nah, the next scribe will not be Lawrence either, because that would be cruel.

Oh, didn't I say who the next scribe is already? I swear I wrote it at the beginning of the scribe post...

Here's something I whipped up as well for the whole scribe post badge thing.

Teacher-less

So as others have also witnessed, the Smartboard has been reborn. Unfortunately everything comes with a price. Mr. K has disappeared again. The Scribal Council of Calculus called an emergency meeting. Mr. K was summoned to an emergency meeting and we were left to fend for ourselves during class.

Actually, I have no idea.

Another member of the Jabbamatheez was M.I.A. today after delivering a Level 5 scribe post and has since not been seen or heard from. Intel has it that codename 'Hi I'm Justus' is the missing Jabbamathee. No word from Secret-Op members Francis and Not Paul on the missing Jabba mathee.

Anyways [I had to fill it with something. I mean, come on, how do you follow up on those two posts with no teacher?]

Where's our laptops? These kids are like, 11. LOL.

To the point. So as you know, Mr. K was away for an emergency meeting. Therefore we were left to lead the class on our own. First, we went over the homework sheet that Mr.K gave us.

Courtesy of Kristina to do the graph.


So basically, for the first three boxes, the first derivative is greater than zero. That means that the slope of the tangent line is positive. The second derivative is less than zero which means that the slope is decreasing. Which is why the curve at the first three intervals is the way it is.

At -5/2, the first derivative is zero but the slope is still decreasing, which is why it's starting to decrease at that point.

At the interval of -2, slope is still negative, but there is no change in the slope, because the second derivative is 0.

At -1, both first and second derivatives are 0, meaning that at that point, there is an inflection point, or where the graph changes concavity.

At the interval of +2, both intervals are undefined, meaning that there is a discontinuity at that point. So at that point, it's either Chocolate (Jump), Vanilla(Infinite), or Ripple(Removable). In this case, it's Ripple flavour. Yum.

Next we had a two slides describing the first and second derivative test.

First Derivative Test

Suppose that 'c' is a critical point of the function and suppose that there is an interval (a,b) containing 'c'.

• If f'(x) > 0 for all x in (a, c) and f'(x) <>
• If f'(x) <> 0 for all x in (c, b) then 'c' is a local minimum of f.

Second Derivative Test

Suppose that 'c' is a critical point of the function f and suppose that there is an interval (a, b) containing 'c'.

• If f'(c) and f"(c) <>
• If f'(c) and f"(c) > 0 then 'c' is a local minimum of f.

Now of course, due to the absence of Mr. K, we didn't really elaborate on this and then... Benchoun threw an eraser at me, and we started working on the slides.



So everyone was there, except Justus, who seems to have a good grasp of derivatives judging by his scribe post, but for people like me, who haven't 100% grasped it yet, will try to explain what Ben did.

First ben took the derivative of the first function in which we had to use the product rule. We factored out x.

Now that I look at it, it seems that Mr. K did some edit to our work so I'll just explain that.

He factored out xe^x and found x = 0 and x = -2.

In all honesty, I lost myself, so I kinda have a brain cramp. I'm gonna post what Mr. K did utilizing the 2nd Derivative test.


So as we can see, x = 0 is a min and x = -2 is a max. If someone could help me out and explain these, that'd help. Sorry guys.

Next scribe is 'Not Paul'.

Continuing From Last Class.

Well first we started off on the wrong track because we all forgot where we left off we then realized how did we get those numbers. we came to the conclusion that we had skipped a few steps will here is the notes but in the right order. i think.

First you use the critical points that you found and plug the values into the second derivative test and you will find the local max and min.



Next you find all possible inflection points by making the second derivative equal to zero and solve for x.



After Figuring all that stuff out you solve for the original function and use the critical points that we have found for the x values.



Now you need to find out if it is concave up or concave down.
To find the that out you need to choose a value from the ranges and plug it in to the second derivative formula which is f ''(x) = 4 - 12x




Next you solve for f(x) but this time you use the infliction points that we found for the x value.



This next picture is a picture of the second derivative i think. it shows when the graph is concave up and concave down.




Well thats all for today tune in next time. Next scribe post will be featuring Shelly.

Rguy out.

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.

Slides October 9th

Here they are ....

AP Calculus October 9th, 2008

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Slides October 8th

AP Calculus October 8th, 2008

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Slides October 7th

FWIW. Sorry for the confusion. Together, we will defeat this book AND learn calculus (and have fun, I hope).

AP Calculus October 7th, 2008

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