Question

y=(x+2)^2(x-1)^3 用导数求单调区间

Answer 1

y'=2(x+2)*(x-1)^3+(x+2)^2 *3(x-1)^2=(x+2)(x-1)^2*(2x-2+3x+6)=(x+2)(5x+4)(x-1)^2
零点为：-2,-4/5,1
在区间（负无穷，-2）及（-4/5，正无穷）上，f'(x)>=0
f(x)单调递增，
同理，在区间（-2，-4/5）上，f'(x)<=0,
f(x)单调减少

Answer 2

y=(x+2)^2(x-1)^3
=(x-1+3)^2(x-1)^3
=[(x-1)^2+6(x-1)+9](x-1)^3
=(x-1)^5+6(x-1)^4+9(x-1)^3
y'=5(x-1)^4+24(x-1)^3+27(x-1)^2
=(x-1)^2[(5(x-1)^2+24(x-1)+27]
=(x-1)^2*[5(x-1)+9][(x-1)+3]
=(x-1)^2 *(5x+4)(x+2)
x<-2时，或x>-4/5时，y'>0 单调递增
-2<x<-4/5时, y‘<0单调递减