Sunday, September 27, 2009

Week Six

This week was very hectic and we learned a lot. We had a test on monday, which didn't really go well for me. On Tuesday we learned about the Extreme Value Theorem, the Rolle's Theorem, and the Mean Value Theorem. On Wednesday, we learned about Optimization. On Thursday and Friday, we did example problems with Optimization.

Things I now understand:

1.) Everything I got wrong on my test, thanks to doing corrections.

2.) Extreme Value Theorem

3.) Rolle's Theorem
F(x) must be continuous on a closed intervale. It must be differentiable on the open interval. And f(a) must equal f(b). If all this is true, then there is at least one number 'c' within (a,b) such that f'(c) = 0.

Example:

You are given: f(x) = x^4 -2x^2 on [-2,2]. Find all values c such that f '(c) = 0.

Since it is continous (polynomial) and differentiable (no corners/discontinuities), you now check that f(a) = f(b).

f(-2) = f(2) = 8

Since all this works, you take the derivative of f(x).

F'(x) = 4x^3 -4x

You then set f '(x) = 0

You get that x = 0, 1, -1
Since all these are within the interval:

c= -1, 0, 1

4.) Mean Value Theorem

Same as Rolle's such that: must be continous and differentiable.

Different because: f '(c) = [f(b) - f(a)] / (b-a).


Something I do not understand is Optimization.

I understand most of the steps, but usually get stuck when trying to find the primary and secondary equations and which is one is which.

And example would be Example 5 on the handout Ms. Robinson gave us.


I am still kicking myself in the butt over not studying hard enough for the test on Monday, but I guess I know now.

1 comment:

  1. If you have an area, you can use that as my secondary. the primary function is what you are solving for. and if the give you a perimeter and ask for area, the perimeter would be your secondary equation and the area would be your primary equation.
    AFter you find that you solve for a variable using your secondary eqation and after that you plug that into the primary equation.
    Simplify waht you get take the derivative then set it equal to 0 and solve for x after that you find your otherr unkown vaible and you then plug both your varibales into the orginal to get your final awnser

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