Find the absolute maximum value and the absolute minimum value, if any, of the function. (If an answer does not exist, enter DNE.)
f(x)=x^3+3x^2-1 on [-3,1]
f(-3) = -1
f(1) = 3
f'(x) = 3x^2 + 6x = 3x(x+2)
so, there are relative max/min at x= -2,0
f(-2) = 3
f(0) = -1
So, on [-3,1] we have
min = -1
max = 3