Week #6

So this week in Calculus we learned three new concepts: Rolle's Theorem, Mean Value Theorem, and Optimization.

The first two are just theorems describing maxs and mins...

Rolle's Theorem: states that a differentiable and continuous function, which attains equal values at two points, must have a point somewhere between them where the slope of the tangent line to the graph of the function is zero.

So, to use this...you are given an interval...let's say [-4,4], and a function...let's say f(x)=x^2. To use this, you have to check that first it is differentiable...which it is. Second, it's continuous...which it is. Next, you do f(-4) and f(4). If those two values are equal, which for this example it is, then you can say that rolle's theorem applies. You then take the derivative, which is 2x in our case, and set it equal to 0. We find that @ x=0 there is a min or a max, which IS between -4 and 4. That's Rolle's Theorem.

The next theorem is Mean Value Theorem: states, that given a section of a curve, there is at least one point on that section at which the derivative of the curve is equal (parallel) to the "average" derivative of the section.

To do this, you have to check that it is differentiable and continous. Then you plug into (f(b)-f(a))/(b-a). Set the derivative of the function equal to whatever that comes out to be. Solve for x.

The other thing we learned is Optimization which...at first was very hard to grasp but now that I did a few problems, extremely easy.

Things to do for Optimization that might help you:

1) I like to circle the equation I'm trying to maximize or minimize...this just reminds me of what I'm working with plugging back into at the end of the problem.
2) Next, identify if you have more than one variable in the equation...if so, you need to isolate it to one.
3) If you need to, find a secondary equation and solve it in terms of the variable you want to isolate your primary equation with....this is a LOT like substitution...actually, that's all it is.
4) After getting the equation in terms of one variable, simplify and then take the derivative. Set the derivative = to 0 and solve. This is taking the max's and min's of the function...
5) Look back at your circled equation...did you want to minimize or maximize it? Plug in the values you found when setting the derivative = to 0 into this circled equation...if you are looking for minimums, take the smallest value. If you are looking for maximums, you take the largest value.

Optimization is pretty cool. I finally see the actual application of Calculus in the real world. :-) (Well, one of them.)


-John