abs max and min ? PLEASE HELP !?

Question

Whats the abs max and min ofthis problem : f(x) = x^2 + (432/x)

Answer 1

Just take the derivative:
f'(x) = 2x - (432/x^2)
Set this equal to 0.
2x - (432/x^2) = 0.
x = 6
Now, to figure out if it's a max or min, test regions.
x < 6, so test -7 in f'(x), and you get a negative answer.
x > 6, so test 7 in f'(x) and you get a positive answer. Because the derivative is going from a - to a +, 6 is a rel. min.

Edit: I did relative min... no clue why, sincerest apologies. I need to learn to read.

Answer 2

"Absolute max" is ∞ at x = -∞, 0+, and +∞
Absolute min is -∞ at x = 0-