Finding global extrema

[fred2028]

So I know that, with the first and 2nd derivatives, one can find local extrema. To find global, I've been told to find all locals, and see which one has the greatest/least y-value. However, if you have a graph like

f(x) = x^3 + 2x^2

then you get 2 critical points, and you find 2 local extrema.

f'(x) = 3x^2 + 2x
f'(x) = x(3x+2)
Critical at x = 0, x = -2/3

Comparing the 2, you would see that 1 on the left is higher, so you'd think that it's a global max, since it's higher than the other critical point. However, the graph extends to infinity to the far right, meaning no global max exists.
If I didn't know the general shape of a cubic graph, how can I determine that the global max of the above function does not exist (without graphing)? The method taught in class, comparing local maxs, does not work here since you wouldn't expect a local max at infinity right?

 

[skeeter]

in the interval x < -2/3, f'(x) > 0 ... f(x) is increasing

in the interval -2/3 < x < 0, f'(x) < 0 ... f(x) is decreasing

in the interval x > 0, f'(x) > 0 ... f(x) is increasing

note the end behavior of the function ... on the left, it increases from -infinity
and on the right, it increases w/o bound to infinity.

the sign of the derivative gives all the info you need to determine if the function has any absolute extrema.