The Extreme Value theorem simply states that there must contain both a max and min on a closed interval of a continuous function. That then brings you to Rolle's:
- 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 one number "c" in (a, b) such that f'(c)=0.
- Remember, if you are not given an interval assume the x intercepts.
An example problem of Rolle's is f(x)= x^4 - 2x^2 on [-2, 2] find all values c for which f'(c)=0.
First you must check if the function is continuous and differentiable: yes and yes
Now check if f(-2) = f(2)
f(-2)=(-2)^4 - 2(-2)^2=8
f(2)= (2)^4 - 2(2)^2=8
f(-2) and f(2) both equal 8, so you can move on to the next step.
Take the derivative and set it equal to zero, then solve for x.
f'=4x^3 - 4x=0
4x(x^2 - 1) = 0
x=0, 1, -1
All of these numbers are in the interval [-2,2], so your answer is c=-1, 0, 1.
I caught on to Rolle's theorem and Mean Value Theorem pretty quickly and was excited that i finally understood something. Until of course we learned optimization. I understood optimization in class when we were working the problems together but i can't seem to do them by myself. I don't know what to do first to the problem and how to find the primary and secondary equation. If anyone can help me with optimization i'd really appreciate it, or once again i'll receive another grade to not be spoken about on the quiz Tuesday.
Let's hope for a good seventh week of calculus :)
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