MathCalculus (check please)
posted by songyi .
FOr the function f(x)=x^2e^x
Find the critical points, increasing decreasing, first and second derivative and concavity
I am having trouble finding the concavity >_<
First derivative: xe^x(x+2)
Second : e^x(x^2+4x+2)
Critical points from first derivate:
x=0,2
Increasing: (infinity,0) (2, infinity)
Decreasing: (0,2)
I can't figure out how to do the concavity part of it.

The "concavity" of the function comes from the second derivative, where there is a critical point.
If the second derivative is positive at a point where the first derivative is zero, the function is "concave upward" there. If it is negative, it is concave downward. 
Increasing: (infinity,0) (2, infinity)
Decreasing: (0,2)
This does not make sense. Coming from oo you get to x = 2 before you get to x = 0
Now if I put in a large  number, say 100 for x
I get (100)^2 e^100= 10^4 *3.72*10^44
which sure looks like 0
If I put in x = 2 I get
(2)^2 e^2 = .54
so it increases from x = oo to x = 0
Now at x = 0, we know f(x) is zero, so it is decreasing from x = 2 to x = 0
We can be sure that it then heads up to the right so increasing from x = 0 to x > +oo
Well, I suppose concave means curving upward. That would happen where the second derivative is +
Now where is the second derivative zero?
x^2+4x + 2 = 0
x = [ 4 +/ sqrt(16 8) ] /2
= 2 +/ .5 sqrt 8
= 2 +/ sqrt 2
so the second derivative may change sign at 3.41 and at .586
in other words to the left and right of where the function is level at (2,.54)
To the left of 3.41 ,say x = 5, the second derivative is +, so curving up, concave from oo to 3.41
at x = 2, the second derivative is negative, so curves down from 3.41 to  .586
Then at x = 0 the second derivative is + again so it starts curving up at x = .586 and continues curving up from then on