Question

设函数f(x)=2x^3+3ax^2+3bx+8c在x=1和x=2时取得极值,求a,b

Answer 1

(1)f'(x)=6x^2+6ax+3b.

因为函数f(x)在x=1及x=2时取得极值,故有

{f'(1)=0,f'(2)=0}

--->{6+6a+3b=0,24+12a+3b=0}

--->a=-3,b=4

(2)由以上知,f(x)=2x^3-9x^2+12x+8c,

f'(x)=6x^2-18x+12=6(x-1)(x-2)

当x属于[0,1)时,f'(x)>0

当x属于(1,2)时,f'(x)