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Monday, December 1, 2008


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'.


Hi I'm Justus said...

Lol at the kids with the laptops

"What are these kids, like, ELEVEN?"

Haha, I lul'd

good one :]

zeph said...

Yes, I would also like to know what our situation is with our laptops.