This week in calculus started off pretty easy. We learned Rolle’s Theorem and Mean Value Theorem. To use Rolle’s Theorem, you the equation giving has to be continuous and differentiable on the interval giving and f(a)=f(b). If it meets these requirements, you then have to take the derivative and set it equal to zero. Then solve for x and that is your c (which is your maxs and mins on that interval).
An example of Rolle’s Theorem is x^2-2x, [0,2]
Since the equation is a parabola, it is both continuous and differentiable.
F(0) = 0^2-2(0) = 0
F(2)= 2^2-2(2) = 0
Since f(0)=f(2)=0 , Rolle’s theorem can be used.
Derivative is 2x-2
2x-2 = 0
X= 1 so c=1
For Mean Value Theorem, you have to check to see if it is continuous and differentiable on the interval given such that f’(c)= f(b)-f(a)/b-a. Once you plug into that formula, you take the derivative and set it equal to the number you got then solve for x.
Example: x^2 , [-2,1]
It is both continuous and differential therefore mean value theorem can be used.
F(-2) = -2^2= 4
F(1) = 1^2 = 1
Plug into f(b)-f(a)/b-a = 4-1/ 4-1 = 1
Take the derivative: 2x
Set the derivative = to 1
2x=1
X= ½
C=1/2
The thing I did not understand this week was optimization. I have always been horrible with word problems. I get confused with the words and also for optimization, I never know what formula to plug into or how to get the primary formula ( if it’s not the one given) in the first place.
Subscribe to:
Post Comments (Atom)
For optimization, you have to mimimize or maximize a certain equation. A way to figure out which equation to set up is just to google geometric equations and learn them.
ReplyDeleteFirst you have to find your primary and secondary equations. I've noticed that the secondary equation is usually always the one set equal to a number. The primary will always have either of the words minimize or maximize near it. After finding these equations 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 your zeros into secondary equation
Once you take the derivative of your primary function, set it equal to zero, and get a zero, that zero will be your first answer. You plug this number into your secondary equation to get your second answer.
when the word problems come up your basicallly looking for key words to use. there will be a couple words your familiar with like perimeter ( p=2l+2w) or area (A=lw) onnce you know you need one of these formulas you use it and use the five steps mamie just pointed out
ReplyDelete