Question

设函数f(x)=2x^3-3(a+1)x^2+6ax+8,其中a属于R。若f（x)在x=3处取得极值，求常数a的值

Answer 1

f(x)=2x^3-3(a+1)x^2+6ax+8
f`(x)=6x^2-6(a+1)x+6a
f`(3)=54-18(a+1)+6a=0
54-18a-18+6a=0
a=3

Answer 2

f(x)=2x^3-3(a+1)x^2+6ax+8, f'(x)=6x²-6（a+1)x+6a
f（x)在x=3处取得极值 f'(3)=6*3²-6（a+1)* 3+6a =0 a=3