Calculus
posted by Jessica .
What values of x is there an absolute minimum for the function y= 1/3x^2 + 2x?
I found the derivative of the function which is y'=2/3x + 2
set it to 0=y'=2/3x + 2
x=3
What do I do now?

what is the second derivative of y?
y"=2/3, which means that the curve is downward, so that is a maximum...
The curve is a parabola, with a max at (x=3), and has no absolute min. 
Thank you!
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