10/31/10

Soooo, this week was NOT a good week for me at all..Basically from what I remember, we didn't learn anything new. We just practiced with the concepts that we recently learned in Chapter 3, and of course some things from Chapters 1 and 2 were thrown in there too. So here are a few examples:

Ex. 1) Find all critical numbers of the function g(x)=x^4-26x^2
*key words in this problem are "critical numbers"..that tells you that all you have to do is take the derivative of the function and set it equal to zero. Then solve for x
*So for the derivative you get 4x^3-52x
4x^3-52x=0
4x(x^2-13)=0 *take out a 4x
4x=0 x^2=13 *divide
x=0, +/- sqrt of 13
^^So those are your critical numbers

Ex. 2) Locate the absolute extrema of the function f(x)=x^2+6x+4 on the closed interval [-6,6]
*Key words in this problem are "absolute extrema"..which tells you you're going to have to take the derivative of the function and set it equal to zero to find your x-values first. Then, you're going to plug your endpoints and the points of your interval into your original equation to find the max/mins.
*So for the derivative you're going to get 2x+6
2x+6=0
2x=-6
x=-3
*Plug in:
f(-3)=(-3)^2+6(-3)+4=-5
f(-6)=(-6)^2+6(-6)+4=4
f(6)=(6)^2+6(6)+4=76
**absolute max is (6,76); absolute min is (-3,-5)

Ex. 3) Determine whether Rolle's Theorem can be applied to f(x)=(x^2-16)/(x) on the closed interval [-16,16]. If Rolle's Theorem can be applied, find all values of c in the open interval (-16,16) such that f'(c)=0.
*First of all let's look at the function..Since they're asking us to use Rolle's Theorem we must first check if the function is BOTH continuous and differentiable before we can even use the theorem.
*The function is differentiable, butttttttttttt it is NOT continuous. It isn't continuous because there is a vertical asymptote at x=0. Therefore, the theorem CANNOT be applied.

Ex. 4) Determine whether the Mean Value Theorem can be applied to the function f(x)=x^2 on the closed interval [-8,2]. If the Mean Value Theorem can be applied, find all numbers c in the open interval (-8,2) such that f'(c)=f(2)-f(-8)/(2-(-8)).
*First check for differentiability and continuity.....and the function is both continuous and differentiable so you can continue using the theorem
*Next take the slope of the function and you should get this:
m=(4-64)/10
m=-6
*Now take the derivative of f(x)=x^2 and you should get 2x
*Now set your derivative equal to your slope (-6) and solve for x.
2x=-6
x=-3 *this is your c value

**Well for the most part I pretty much get the hang of all that we learned in Chapter 3. I could use some more practice though, but that's about it. Hope I do good on my test tomorrow!