Sunday, September 27, 2009

week 6 post

This week we learned a lot of stuff, like, the Extreme Value Theorem, Rolles Theorem, and Mean Value Theorem.

EVT: the EVT states that a continuous function on a close interval [a,b], must have both a minimum and a maximum on the interval. However, the max and min can occur at the endpoints.

Rolles theorem gives the conditions that guarantee the existance of an extrema in the interior of a closed inteval.
Rolles: Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f(a)= f(b) then there is at least one number, “c” in (a,b) such the f '(c)=0

MVT: If f is continuous on the closed interval [a,b] and differentiable on the interval (a,b) then there exists a number c in (a,b) such that f '(c)= (f(b)-f(a))/(b-a).

Example: f(x)-x^(2)-3x+2, show that f ' (x)=0 on some interval. (Hint: if they don't give an interval, assume it's x-intercepts)

so first you factor and find that x=1 and x=2

so then, you check and see that f(1)=f(2)=0 so it's continuous and it's differentiable.

So after you've found that, you take the derivative and set it equal to zero.

F '(x)=2x-3
2x-3=0
2x=3
x=3/2
c=3/2. And that's it. You've found c.
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Another example

f(x)=x^(4)-2x^(2) on [-2,2] find all values of c for which f '(c)=0
continuous: yes
differentiable: yes. So let's get started.

First, you have to plug in to see if f(-2) is equal to f(2)
so, f(-2)=(-2)^(4) -2 (-2)^(2)= 8
f(2)= (2)^(4)-2(2)^(2)=8
so that's good. They equal each other. By the way, i'm not sure if i'm typing up all my parentheses correctly, so sorry if I mess that up.

So then you take the derivative and set = to zero
f '(x)=4x^(3)-4x=0
4x(x^(2)-1)=0
x= 0, 1, -1
c= -1, 0, 1. and you're finished again.


one thing i don't understand is optimization. one reason for this is because i have like no notes on it... i'm not exactly sure why though. i have a bunch of examples in my notebook, but i can't find any notes. so i pretty much don't understand it at all. thanks.

3 comments:

  1. I second this and just about everyone else's post...since it's all about optimization

    ReplyDelete
  2. Ok, at first I didn't understand optimization either, but once I really just sat down and looked at it for a while, I learned it was really easy.

    If you're given a word problem asking you to find the demensions of a cube and the mamimum area of a cube, you first pick out what would be your primary and secondary functions.

    What I've noticed is when there's a number connected to an equation i.e. if the word problem says the perimiter is equal to 24, that equation would most likely be your secondary. The primary is always the one where they ask you to maximize or minimize.

    Once these equations are found, the steps are as follows:

    1. Solve secondary for one variable
    2. Plug into primary
    3. Take the derivative
    4. Set equal to zero
    5. Plug zeros back into secondary

    From here you'll get your answer

    ReplyDelete
  3. Solve secondary for one variable
    Plug into primary
    Take the derivative
    Set equal to zero
    Plug zeros back into secondary

    then you get ya answer

    ReplyDelete