Find the absolute maximum and minimum values of f(x)=x^(2/3) on the interval [-2,3]

So i found f'(x) =(2/3)x^(-1/3)

Then when f'(x)=0 solved for x which i got x=0

Next i pluged in x=0,-2, 3 into f(x) and got

f(0)=0

f(-2)=-1.58

f(3)= 2.08

soo the absolute max is f(3)=2.08 and im having trouble finiding the absolute min.

I drew out f(x) on my graphing calculor and it came out looking like the absolute value of x so then f(0) would be the absolute min. But without looking at the calculator i would have assumed f(-2) is the absolute min.... So which is correct and why?