求函数f(x)=2x^3-9x^2+12x-3的单调区间与极值
f(x)=2x^3-9x^2+12x-3
f’(x)=6x^2-18x+12
令f’(x)=0
6x^2-18x+12=0
x=1 x=2
f’’(x)=12x-18
f’’(1)=-6<0
f(1)=2为极大值
f’’(2)=6>0
f(2)=1为极小值
(-∞,1)增
(1，2)减
(2，∞)增
f(x)=2x^3-9x^2+12x-3
f'(x) = 6x^2 -18x+12
f'(x)=0
6x^2 -18x+12 =0
x^2 -3x+2=0
(x-1)(x-2)=0
x=1 or 2
f''(x) = 12x-18
f''(1) <0 (max)
f"(2) >0 (min)
max f(x) = f(1)= 2-9+12-3 =2
min f(x) = f(2) =16-36+24-3 =1
单调
增加 = (-∞, 1]U[2,+∞)
减小: [1,2]
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