This week I actually understood something on my own....and explained it to people! My gosh I felt accomplished. Also, I almost hyperventilated when I saw my quiz grade. I was excited, I'm sorry! =P Anyway, Rolle's Theorem.
Let's look at the definition
Rolle's Theorem gives the conditions that guarantee the existence of an extrema in the interior of a closed interval.
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 1 number c in (a,b) such that f'(c)=0
Okay, what that basically says is
that if the function is continuous on the interval and it's differentiable on the interval, when you plug in the endpoints of your interval (or x intercepts) and they equal each other, when solved for x in the first derivative, there will be numbers inside your interval.
I have no idea if that made sense to you or not...here's an example problem:
f(x)=x^4-2x^2 on [-2,2] find all values c for which f'(x)=0
1) Is it continuous?
Yes
2) Is it differentiable?
Yes
3) Plug in your first end point.
f(-2)=(-2)^4-2(-2)^2 = 8
4) Plug in your second end point
f(2)=(2)^4-2(2)^2 = 8
5) Are they equal?
f(2)=f(-2)=8
Yes
6) Take the first derivative
f'(x)=4x^3-4x
7) Solve for x
x= -1, 0, 1
8) Which points are on your interval?
Yes so c=-1, 0, 1
9) Justify
Rolle's Theorem is applied to this function because it is continuous and differentiable and f(-2)=f(2)=8. When Rolle's Theorem is applied, the first derivative is taken and c is found to equal -1, 0, and 1.
Now, can someone please help me with optimization? I have NO clue what I'm doing at all for this. I don't even know how to start it! =0
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Well when i first start optimization, i look at what i'm given. if i'm given area, i use that as my secondary.. the primary function is what you are solving for. So if they give you perimeter and ask for area, the perimeter would be your secondary equation and the area would be your primary.
ReplyDeleteWhen you figure out the equations, you solve for one variable in your secondary equation, when that is done, you plug it back into your primary equation for the variable you just solved for.
Then you will simplify that and then take the derivative of the function. Then set the derivative equal to zero and solve for x. Once you have your x, you plug the x back into the secondary equation and find the other unknown variable. Then plug your numbers you found into your primary equation to get your final answer.
i hope this helps :)
Oh My Gosh!!
ReplyDeleteThat helps SO much!
Thank you!!!
ok so if you look on the first page of your optimization packet, you can find five steps for it. mrs. robinson wrote down the steps on the board today (9/29) in "lamen's terms" haha. but i forgot to copy it down :( idk if you did or not though. but here are the steps, it's kind of complicated to understand.
ReplyDelete1. Identify all given quantities and quantities to be determined. if possible, give a sketch.
2. write a primary equation for the quantity that is to be maximized or minimized.
3. reduce the primary equation to one having a single independent variable. (may involve use of secondary equations)
4. determine the feasible domain of the primary equation. (determine values for which the stated problem makes sense)
5. determine the desire max/min value by using calculus techniques.
hope i helped a little :)