First we find the critical points by taking the first derivative and setting it equal to zero. So y'=27x^2-27=27(x^2-1)=27(x+1)(x-1)=0. The critical points are x=-1 and x=1. But we have a closed interval here and x=-1 is not in that interval. We must check the values of the endpoints and x=1, the critical point.
We substitute those back into the original and get
y(0)=9, y(1)=9-27+9=-9 and y(3)=9(3)^3-27*3+9=171.
When it asks for values, it is asking for y values.
The absolute minimum is -9 and the absolute maximum value is 171.
