This week in math we didn’t learn anything new we just did our presentations and our study packets for our test that’s on Wednesday. Although we didn’t learn anything new ill recap you on the steps of
Optimization
1 Identify primary and secondary equations your primary is the one your or maximizing or minimizing and your secondary is the other equation
2. Solve for your secondary variable and plug into your primary equation if your primary only has on variable this isn’t necessary
3. Plug into secondary equation to find the other value check your end points if necessary
Those are the steps of optimization hope that helped if you needed the recapped or you missed the notes.
What I need help on is Rolles theorem, mean value theorem and the intermediate value theorem. I don’t understand the first two but the intermediate value I can grasp I just would like a little help in remembering it.
But that’s it for me this week
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To use Rolle's Theorem, you first have to check to make sure the function is conintuous and differentiable on the interval giving. f(a) has to equal f(b) which has to equal the same number.
ReplyDeleteEXAMPLE: f(x)= x^2-3x+2 on [1,2]
The function is both continuous and differentiable.
1: (1)^2 - 3(1)+2= 0
2: (2)^2 - 3(2)+2= 0
Because f(1=f(2)=0 then Rolle's theorem can be used.
Take the derivative: 2x-3
and solve for x: x= 3/2 or c=3/2
For mean value theorem, you check to make sure the function is continuous and differentiable on the interval given. Then you plug into the formula f(b)-f(a)/b-a. Once you get an answer, you take the derivative of the function and set it equal to the answer you got by plugging into the formula then solve of x.
When using Rolle's theorum you are given an equation and a point. First you figure out if the function is continuous and differentiable. If it is not continuous or differentiable on the point give, then the problem cannot be worked. If it is continuous and differentiable, then you can move on to the next steps.
ReplyDeleteLet's say they give you x^2-2x and (0,2) you would first plug in
f(0)=0^2-2(0)=0
f(2)=(2)^2-2(2)=0
The y values have to match, if they do not the problem cannot be worked further.
Now you take the derivative of the function and set it equal to the y value 0
2x-2=0
which gives you x=1. Therefor c=1
To do Rolles theorem you first have to check if the function is continuous or differentiable on the interval given. If not the problem cannot be worked. The next step is to take the derivative and then solve for x.
ReplyDeleteFor the Intermediate Value Theorem, I just remember plug in answer choices to original equation. The problem will give what f(c) needs to qual. Whichever number is plugged in and equals the same as f(c) is your answer.
ReplyDeleteExample:
f(x)=x^2-2x-3,[4,8], f(c)=12. And the choices are A) 3 B) 2 C) 5 D) 7 E) 6
f(x)=x^2-2x-3
=5^2-2(5)-3 =12 So, the answer is C) 5.
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).
ReplyDelete------------------
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.