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Monday, January 19, 2009

Pre-Testing Revised, David vs. Goliath Stories, and Area

Okay, so I'm the scribe today amirite? Sounds good to me. Before I begin I'd like to explain the David vs. Goliath story thinger.

Basically, today at my basketball game, we beat Oak Park by about 4 points. Now this might not mean anything to you, but it does to us (the basketball team.) On paper, Oak Park should destroy just about any team in manitoba. Hence the DvG story.

Anyways onward with the math. Today we had a revised version of the pre-test, which I liked much better. Coming out of that pre-test I felt much more prepared for the test tomorrow then I usually do, which is a good thing! :p Basically, what we changed was, instead of working by ourselves for a large portion of the pre-test, we worked on it independently for about 20 mins. We then went (read: Were placed) into small groups, and treated it somewhat like a workshop. After about 10 minutes of that, we spent the rest of the class going over the answers. It's those answers/ After pre-test discussions I'll be going over now ;p

Question 1.

What the writing means:

Essentially, how the problem is set-up means that F(x) = the stuff in the slide written in black.

Now we know that the derivative is the underlying function. It is from this notion that we end up with the next bit, (starting with F'(x))

What see you there in blue is the anti-derivative of F'(x) and therefore is F(x).

From there it's simply a matter of finding the values of those functions at (2) (the bit you see in the green there.) One thing to remember for this last step here, is that the integral of any function from x to x, is 0.

Answer: E

Question 2.

Alright, as you can probably guess by all the blue circles the main idea here is that the accumulation function is actually a COMPOSITE OF FUNCTIONS. This means, that you need to use chain rule when differentiating.

SO the procedure here is fairly standard, take the derivative of the outer at the inner, multiplied by the derivative of the inner. The result is the green stuff below. Then evaluate this function at -1 and you have your answer!

Answer: E

Question 3.

Okay, lets see if I remember this. Basically, the first thing to do, was sketch the graph (which is usually a good move.) Next you need to do some logistics work, by looking at the actual integral itself. You should notice that the inner function f(x) is all that there really is, and this is given in the question, so you don't need to do any guess work there. Next, start going through the options.

Option I : F(1) = 6.5

This is confirmed by counting up the squares. If you did it right you should end up with 6.5.

Option II : F'(1) = 3

The thing to remember here, is that the derivative of F(x) is f(x) which is shown in the graph. Therefore, all you need to do to confirm this, is go over to 1, and read the according value on the y axis.

Option III : F''(1) = 1

To check this one, you need to remember a continuation on the first idea from option 2. As said before, the derivative of F is f(x), so the next step would be to look at the second derivative of F. It just so happens that the second derivative of F is the first derivative of f(x). Looking at the function for f(x) and taking the derivative of it's two pieces you get f'(x) = 0 and f'(x) = 1. It just so happens, that f'(x) = 1 is just what we were looking for. Thus, myth confirmed!

Answer: E

Question 4.

This one was easy enough, a fact confirmed by the lack of needing an explanation.

Basically, use your calculator for this one. Plot the graphs, find the interval in which they overlap. 2nd calc > 7 select your lower and upper bounds. Repeat for the other graph, and subtract the values. If that all went according to plan you should have something like, 1.66666666666666666666666666666666666666666666666666667, or 4/3.

Answer : C

Question 5.

So, this problem is a simple definite integral.

Start by anti-differentiating, to end up with the first bit of blue seen in the slide.

After you anti-differentiate, substitute in the values seen on the definite integral (-2 and 2), and subtract. Its all simple algebra from there.

If it all went well, you should get k = 4, which happens to be answer d.

Answer : D

Question Free Response

Part a.

So, you need to find the value of the accumulation function, G, on the interval from - 4 to 4 at -4. This should be simple enough, since we know that if the interval of an integral is from x to x, then the integral is 0.

moving on.

Part. b

This is a bit trickier, although not much moreso. G'(-1) is the same thing as asking for the value of the inner function (f in this case) at that point. Since we have the graph of f right here, lets do that shall we? Simple look on the graph to discover that f(-1) = 2 so therefore, G'(-1) also is 2.

Part. c

on the graph you may notice that it on the intervals (-4 , -3) U (-1, 2) the graph of f is decreasing, and therefore the first derivative of G is decreasing on those intervals as well. So, by taking the derivative of f(x), you get the second derivative of G, and can go ahead and apply the second derivative test. (I think D:)

You also have to justify why this is true (as in write a short little ditty about why its decreasing on those intervals and things.)

Sorry if this doesnt make sense, I kinda need to clarify some things here D:

Part. d

This bit is also somewhat hazy for me, but what I have in my notes here, is that you need to use extreme value theorem, and examine both the endpoints, and where the derivative is zero.

I also recall hearing that if you dont examine the endpoints (read: forget them)

Okay guys, I think that about wraps it up for todays scribe post. Since it's a new cycle I think I will pick francis to be the next scribe mmkay? Hopefully you kinda understood what I was talking about here. I apoligize if you didn't. Any comments/critique would be appreciated!

~Justus out.

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