Scribe Post Friday November 19, 2010

We started class with working on a problem that Mr. O'Brien had on his AP exam all those years ago. Back in that day they didn't use graphing calculators, so the problems were a little different. After working on this question for a few minutes Mr. O'Brien told us some about how our exam will be in May. Then, he continued by going over the problem on the board.

Steps to solve related rates word problems:
1) Find our what it is about, and draw a picture
2) Identifying variables and constants
3) Write equation(s) relating the variables
4) Differentiate LHS and RHS with respect to t
5) Solve by adding in variables

This struck the end of related rates and lead us into our next set of word problems.

The next set of word problems are for extreme values of functions. So, we started learning the Calculus we will need to do to solve this problems by doing a few example problems...

f(x)=-x^3+6x^2
on [-1,5]

f'(x) = -3x^2+12x
Where f'(x) = 0

0=-3x^2-12x

=-3x(x-4)

x = 0,4

Now we can plug these numbers back into the f(x) function and we get
f(0) = 0

f(4) = 32

So, The minimum is 0 because it is the lower of the two and the maximum is 32 because it is the higher.

We graphed this as well... however my computer is having problems posting images. I'll work on it and try to get these up for everyone soon.

Next, we did the same problem however with a limit of [-1,7].
What was interesting about this was that when we graphed this we could see that the graph starts to have a steady negative slope between the x values of 5 and 7, and this made it so the the absolute minimum changed quite a lot. This is easy to get a visual on geogebra, but again it won't let me upload the images onto the blog and therefore I will have to add this images later.

However, we can solve the minimum without a graph by plugging in 7 as x in the function. And.... f(7) = -49. Since this value is the lesser now it is apparent that -49 is the minimum.

When looking for extreme values of a function:
1) Locate places where the derivative of the function is 0 or undefined, and these are the critical points
2) Evaluate the function at these critical points
3) Evaluate the function at the end points
4) Pick out biggest and smallest and call these the extreme values.

An important note is that if derivatives change from (+) to (-) it is a max. And, if the derivative changes from a (-) to a (+) then it is a min. Remember this can be seen by the slope of the graph. This is called the first derivative test... however there is a problem with it that you must always have in the back of your mind. It is only true with continuous functions as Robin "Beans" Wilder pointed out to us.

Next we went on with more examples...

Find the extreme values of f(theta) = tan(theta)
for [-pi/3, pi/4]

F'(theta) = sec^2 = 1/cos^2

There is no points where its zero or undefined so, there are no critical points. Therefore plug is -pi/3 and pi/4.

tan(-pi/3) = -sqrt(3)<------ This is the minimum

tan(pi/4 = 1 <----------- This is the maximum

Here are some links:

http://tutorial.math.lamar.edu/Classes/CalcI/MinMaxValues.aspx

http://www.youtube.com/watch?v=r56mq7iBH4o

http://www.themathpage.com/acalc/max.htm ----- UPDATE

http://library.thinkquest.org/3616/Calc/S2/FMME.html ----- UPDATE