Question

求函数f(x)=(x-1)²(x+1)³的单调区间与极ŀ

Answer 1

f(x)=(x-1)^2
*
(x+1)^3
f'(x)
=
(x-1)^2
*
3(x+1)^2
+
(x+1)^3
*
2(x-1)
=
(x+1)^2
*
(x-1)
*
{
3(x-1)+2(x+1)
}
=
(x+1)^2
*
(x-1)
*
(5x-1)
=
5(x+1)^2
*
(x-1/5)
*
(x-1)
x＜1/5或x＞1时，f'(x)＞0，f(x)单调增；1/5＜x＜1时，f'(x)＜0，f(x)单调减少
∴单调增区间：（-无穷大，1/5）u（1，+无穷大）；单调减区间（1/5，1)
x=1/5时有极大值f(1/5)
=
(1/5-1)^2
*
(1/5+1)^3
=
3456/3125
x=1时有极小值f(1)
=
0

Answer 2

方法如下，请作参考：

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