11/14/10

Well this week went by super fast! Only one more week until the holidays!!!!!!! :D Anyway, this week all we really did was take time to review for our Chapter 5 test which we took Thursday I think? And we didn't learn anything new except for Friday. Friday we learned about concavity and the Second Derivative Test. Not too many steps changed for that so it's not too bad. It's pretty easy so far. Sooooo, here are some examples:

Ex. 1) Determine the open intervals on which the graph is concave upward or concave downward.
a.) f(x)=3x^2-x^3
*Okay when the directions ask you specifically for concavity, all you have to do is find the second derivative of the function, set it equal to zero, solve for x, check for differentiability, and set up intervals (intervals just like in the 1st derivative test..except this time, when you get a positive number, that means it is concave up on that interval and when you get a negative number that means it is concave down on that interval)
*So first let's take the derivative of this function. You should get 6x-3x^2
*Now let's take the 2nd derivative...You should end up with 6-6x
*Now set 6-6x equal to zero and solve for x
*You should get x=1; Now set up your intervals like this:
(-infinity,1)u(1,infinity)
*Now plug values in those intervals into the 2nd derivative to find out if it is negative or positive
*For the first interval you can plug in 0, and you should get a positive number
*For the second interval you can plug in 2, and you should get a negative number
**Soooo, your answer is:
(-infinity,1) concave up
(1,infinity) concave down
b.) f(x)=(x^2+1)/(x^2-1)
*We're doing the same thing as problem a. so first, take the derivative of the function using the quotient rule
*For the 1st step of the first derivative you should get this: (x^2-1)(2x)-[(x^2+1)(2x)/(x^2-1)^2
*f'(x)=(-4x)/(x^2-1)^2
*Now for the 2nd derivative...For the 1st step of that you should get this:
(x^2-1)^2(-4)-[(-4x)(2(x^2-1)(2x)]/(x^2-1)^4
*Simplifying all of that (which takes quite a while) you should get this:
f''(x)=4(3x^2+1)/(x^2-1)^3
*Now, take the top of that fraction^ and set it equal to zero to find your x-values
*4(3x^2+1)=0
3x^2=-1
...**You get an i, so this does NOT work..
*Now you have to check for differentiability [which I didn't really do in problem a., because it is a polynomial and they are differentiable everywhere :)]
-So to check for differentiability here you take the bottom of the fraction (where there is a vertical asymptote) and set it equal to zero to find the x-values
*So you get x^2-1=0
x^2=1
x=+/-1
*Now you set up your intervals like this:
(-infinity,-1)u(-1,1)u(1,infinity)
*Now plug values in the intervals into the 2nd derivative to find out if it's positve or negative
*For the first interval you can plug in -2, and you should get a positive number
*For the second interval you can plug in 0, and you should get a negative number
*For the third interval you can plug in 2, and you should get a positive number
***So your answer is:
(-infinity,-1)u(1,infinity) concave up
(-1,1) concave down

**I understood what we learned this week so I don't have any questions...yet.