Post #4

Sooooo, I have many questions.

First, I don't remember even talking about a zero of derivative, is that used for anything? It is one of our words, but I don't know anything about it.

So what I do understand is the rules for the first derivative:
1. take the derivative
2. set it equal to 0
3. solve for x for max, mins, horizonal tangents, your critical points
4. set up intervals using step 3
5. plug in first derivative
6. (if needed) to find an absoulute max/ min plug values from #5 into original function, check endpoints

but I'm having troubles within the problem
Example: Find the critical points of f(x)=(x^2-4)^2/3
=2/3(x^2-4)^-1/3 * 2x=0
4x/3(x^2-4)^1/3 =0
4x=0 x=0, + or - 2 --> critical points Now something I don't get is what did she say about the 0 and how do you check for points of discontinuity?

I do get what's next you set up intervals and plug into your first derivative:
(-infinity,-2) f(-3)= -ve so decreasing
(-2,0) f(-1)= +ve so increasing
(0,2) f(1)= -ve so decreasing
(2,infinity) f(3)= +ve so increasing
So your mins are x=-2, x=2, but can your max be x=0?

Now I know for the second derivative test the main thing that is different is you plug into the second derivative and you deal with concave up or down and points of inflection instead of increasing or decreasing. Example: 6/x^2+3
(x^2+3)(0)-[6(2x)]/(x^2+3)^2 = -12x/(x^2+3)^2
then take the derivative of that which you should get = 36(x+1)(x-1)/(x^2+3)^3
Do I then set equal to 0 to get x= + or - 1? I do fully understand the interval part which I showed earlier in a different problem expect you would use concave up or down.

I know no one can help me on this, but I'm still having trouble simplifying derivatives. I guess it will still take more and more practice.

I'm out.