求函数f(x)=ax^2+2ax+1,x∈[-3,2]的最值

f(x) = ax^2+ax +1
f'(x) = 2ax + a =0
x = -1/2
f''(x) = 2a
f(-3) = 9a- 3a + 1 = 6a +1
f(2) = 4a+2a + 1 = 6a+1
f(-1/2) = a/4 -a/2 +1 = -a/4 +1
if a 0 , f''(x) < a ( min )
minf(x) = f(-1/2) = -a/4 +1
maxf(x) = f(2) = 6a+1