Post #20

So the blog wouldn't let me sign in yesterday, but I finally got it working today.

Well, I think I have talked about the first derivative test way too much. So, I guess I will focus on the second derivative test this time.


PROBLEM: Finding the points of inflection for x^3-6x^2+12x.

1. take the derivative:
3x^2-12x+12
2. take the second derivative:
6x-12
3. set = to 0:
6x-12=0 x=2 --->gives possible pts. of inflection
4. set up intervals:
(-infinity,2) u (2,infinity)
5. plug in to SECOND DERIVATIVE:
6(0)-12= -ve --->concave down
6(3)-12= +ve --->concave up
So, x=2 is the point of inflection.


Also, I thought I might review implicit derivatives.


PROBLEM: y^3+y^2-5y-x^2=-4

1. take derivative of both sides: (implicit derivatives have an = sign)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
*remember every time you take the derivative of y, you have to note it by dy/dx or y^1
3. solve for dy/dx: (you are going to have to take out a dy/dx when solving)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)=2x
dy/dx(3y^2+2y-5)=2x
dy/dx=2x/3y^2+2y-5
*Also, if you want the slope you must plug in a x and y-value.


So, I would just like to point out that I understand most (definitely not all) things we have learned so far in Calculus. I just seem to forget a step, mess up a derivative, or the problem is just simply tough haha. I think I need to start reviewing everything again because now I'm getting worried about the ap test!