题目

求函数f(x)=(x-1)2(x+1)3的极值与单调区间.

提问时间：2020-12-30

答案

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