During the sixth week of calculus we learned a few new concepts such as: Rolle's theorem, the mean value theorem, and optimization.
Rolle's theorem starts off with a polynomial and a point. the first step is to plug both points into the f(x)= (equation). if both numbers come out to equal the same number then you take the first derivative. however, if both numbers do not equal the same number then you stop right there. on our last quiz mrs. robinson gave us one like that and asked us if it could be solved and then we had to justify our answer.
Another thing we learned about was the mean value theorem (MVT). the mean value theorem states that given a section of a smooth differentiable curve, there is at least one point on that section at which the derivative (slope) of the curve is equal to the "average" derivative of the section.Let f : [a, b] → R be a continuous function on the closed interval [a, b], and differentiable on the open interval (a, b), where a <>f1(c)=f(b)-f(a)/b-a. EXAMPLE:
Optimization refers to choosing the best element from some set of available alternatives.
In the simplest case, this means solving problems in which one seeks to minimize or maximize a real function by choosing the values of real or interger variables from within an allowed set. More generally, it means finding "best available" values of some objective function given a defined domain, including a variety of different types of objective functions and different types of domains.
In the simplest case, this means solving problems in which one seeks to minimize or maximize a real function by choosing the values of real or interger variables from within an allowed set. More generally, it means finding "best available" values of some objective function given a defined domain, including a variety of different types of objective functions and different types of domains.
I understand rolles theorem but im still having a little trouble with the mean value theorm so if anyone could explain it that would be helpful to me.
When you use the mean value theorum, your going to set the derivative of the function equal to the slope between a and b. Let's look at example 39 on the homework. it gives us x^2 and (-2,10) We know the function is continuous and differentiable. so now lets take f(-2)=(-2)^2=4 and f(1)=(1)^2=1. Now we set it up to find the slope between the two points.
ReplyDeletethe formula for that is: F(b)- F(a)/b-a
so it would be1-4/1--2)which would give us -1
so we will set the derivative of x^2 equal to -1. so that would be 2x=-1
that would give us x=-1/2 so c would equal -1/2. Hope this helps you :)
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.
ReplyDeleteFOR 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
Since you're still having trouble with Mean Value Theorem, i'll explain that.
ReplyDeleteFirst of all you have to determine if the function is continuous on the closed interval [a, b] and differentiable on the open interval (a, b). If it is, you may move on. You then need to find the number (c) in (a, b) such that the f'(c)=f(b) - f(a)/b - a. This just means the derivative must be equal to the slope between a and b. Remember if an interval is not given, assume x-intercepts. An example problem is f(x)=5 - 4/x find all values c on (1, 4) such that f'c = f(b) - f(a)/b - a.
Continuous-yes on the interval
Differentiable-yes on the interval
You then use the interval to plug in.
f(4)=5 -4/4 = 4
f(1) 5 - 4/1 = 1
This gives you that f(b) - f(a)=3
Also b-a=3 so you get 3/3, which is 1.
You then take the derivative of the original function and set it equal to 1.
f'(x)=4/x^2
4/x^2=1
x=+ or -2
so c=2 because you ignore -2 because it is not on the interval given.