The sixth week of calculus began with a test on chapter three. Unfortunately all of my hours of work didn't help me out so well. Tuesday we started learning something new.
First was Rolle's Theorem:
In order to be able to do Rolle's Theorem you have to have a continuous equation and it has to be differentual. When you plug in the points that are given to you, it has to equal to the same number that way f(a)=f(b)=0 [or any other number]
If all of this is passed, then you can start using the theorem:
find the first derivative, set it equal to zero and solve. The number[s] you get for x are the possible answers. If any of the numbers are not in range of the points given then those numbers cannot be your answer. The numbers that are in the range are your answer so you say c = ____ <--those numbers.
FOR EXAMPLE:
f(x)=x^2-2x [0,2]
it is continuous and differentiable.
plug in: f(0)=0
f(2)=0
take the first derivative: 2x-2=0
now solve: x= -1
since -1 isn't between 0 and 2 it is not an answer. Therefore there is no answers!
Then we learned Mean Value Theorem:
In order to be able to do Mean Value Theorem you have to have a continuous equation and it has to be differentual. When you have the points in the problem [a,b] you plug those to the equation. When you solve and find numbers you take [the "b" answer] - [the "a" answer] at the top of the equation, and from the original points given you take [b - a] for the bottom of the equation. When you solve and get a number that's the number that you set the first derivative to. [Take the derivative of the equation given and then set equal to that number] The number[s] you get for the x are the possible answers. If any of the numbers are not in range of the points given then those numbers cannot be you anwer. The numbers that are in the range are your answer so you say c = ____ <--those numbers.
FOR EXAMPLE:
f(x)=x^2 [-2,1]
it is continuous and differentiable.
plug in: f(-2)=4
f(1) =1
so you plug into the equation...1-4 = -3
----- ----
1+2 = 1
since that equals to -3 when you take the first derivative [2x] you will set it equal to -3.
giving you x= -3/2 [which is between [-2,1] so c = -3/2
Next we learned about Optomization. To be honest, i still don't get the concept of it so if someone would be so kind and explain that to me .. it would be greatly appreciated!
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If you look on the handout that mrs robinson gave us on the first page there are the steps to optimization. But also if you go read john's post from this week he explains optimization really good and it may benefit you to go look at it. Hope this helps.
ReplyDeletei think we just all need more practice...but follow these stepss...
ReplyDelete1. Identify all given quantities and quantities to be determined.
2. Write a primary equation for the quantity that you want to max or min
3. Reduce the primary equation so it only has one variable.
4. find the domain of the primary equation.
5. Then find max and mins by derivative set equal to zero.