This week in Calculus, the thing I understand most is Rolle's Theorem. Later on in the week we learned optimization. This is the thing I do not understand..while we are working it, I can follow along, but i cannot do it by myself, because I don't understand the steps..if there are any?
The first thing we learned this weed was Rolle's theorem. It states: "Let f be continuous on a closed interval [a,b] and differentiable on the open interval, (a,b), then there exists a number, c, in (a,b) such that f'(c) = ( f(b)-f(a) ) / ( b-a).
To put this in simpler terms it means there must be a point, c, where the instantaneous rate of change equals the average rate of change. Rolles Theorem gives conditions that promise the existance of an extrema in the interior of the closed interval.
An example of this would be:
f(x) = x^2 - 3x + 2 Show that f'(x) = 0 on some interval.
The first thing you should know, is if they do not give you an interval, you should ALWAYS assume the x-interval.
First, to find the interval, solve for x :
x^2 -3x + 2 = 0
(x-2)(x-1)
x-interval --> [1,2]
Then, begin to check and see if Rolle's theorem can be used. The standards are that the function must be continuous and differentiable (on the interval given or found), and the y-values MUST be equal.
Plug in the numbers in the x-interval to find out if the y-values are equal.
f(1)=f(2)=0
This part of the test is successful.
Next, you must check if the function is continuous and differentiable. It is, so then you can proceed by taking the derivative and setting it equal to zero.
f'(x) = 2x-3 = 0
2x = 3
x = 3/2
c = 3/2
After checking to make sure that the c-value is in between the values in the closed interval, you justify...of course.
Using Rolle's Theorem, because the function is continuous and differentiable, and the y-values are equal, when the derivative was taken and set equal to zero, i found that c is equal to 3/2.
Optimization is the only thing I do not understand..if anyone can explain to me the steps, it would probably save me from failure on the quiz tuesday..thankss.
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1. Identify all given quantities and quantities to be determined. (find all numbers and variables)
ReplyDelete2. Write a primary equation for the quantity that is to be maximized or minimized. (will be the formula before you plug anything in)
3. Reduce the primary equation so it only has a single independent variable. You will find secondary equations. (relates to the problem)
4. Determine the feasible domain of the primary equation. (where can it be?)
5. Determine the desired maximum or minimum value by the calculus techniques. (taking derivativative, etc.)
first find your variable, then find your primary equation by using the variable given, then reduce to find the secondary, then determine the domain and determine your max or min.
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