so this was another week in calculus, and we had more ap test to take. For some stuff that i understand a little bit:
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.
I need some serious help with intergration. I suck at it
so integration is.. simple..i guess
ReplyDeleteif you have to integrate 2x^1/2
then add one to the exponent: 2x^3/2
multiply the coefficent by the reciprical of the new exponent: 3x^3-2
so that's your integral
as for anything else, these are simple terms...
you can look at like the integral of [sin(x)]^2
it would be 1/3[sin(x)]^3 TIMES -cos(x)
you have to times it by -cos because the derivative of -cos is sin..therefore the integral of sin is -cos
HOPE THIS HELPS!