求函数f(x)=(x-1)2(x+1)3的极值与单调区间.
题目
求函数f(x)=(x-1)2(x+1)3的极值与单调区间.
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
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