Question

求y=x³–6x²+9x–5的单调区间和极值

Answer 1

解求导f'(x)=3x^2-12x+9=3(x^2-4x+3)=3(x-1)(x-3)
令f'(x)=0
解得x=1或x=3
令f'(x)＞0，解得x属于(3,正无穷大)和(负无穷大，1)
令f'(x)＜0，解得x属于(1,3)
故函数的增区间为(3,正无穷大)和(负无穷大，1)
减区间为(1,3)
极大值f(1)=1-6+9-5=-1
极小值f(3)=27-54+27-5=-5

追问

按照下面的题写谢谢

Answer 2

y=x^3-6x^2+9x-5
y'=3x^2-12x+9
y'=0
3x^2-12x+9 =0
x^2-4x+3=0
(x-1)(x-3)=0
x=1 or 3
y''=6x-12
y''(1) =6-12<0 (max)
max y= y(1)=1-6+9-5=-1
y''(3) =18-12 >0 (min)
min y = y(3) = 27-54+27-5 = -8
单调
增加 =(-∞, 1]U [3,+∞)
减小= [1,3]