题目
求函数y=(x-1)∧2(x+1)∧3的单调区间.
提问时间：2021-03-17
答案
y = (x-1)^[2(x+1)^3] 定义域 x>1.则
y = e^[2(x+1)^3ln(x-1)]
y' = [6(x+1)^2ln(x-1)+2(x+1)^3/(x-1)] e^[2(x+1)^3ln(x-1)]
= [2(x+1)^2/(x-1)] [3(x-1)ln(x-1)+(x+1)] e^[2(x+1)^3ln(x-1)]
x>1 时,2(x+1)^2/(x-1) > 0,e^[2(x+1)^3ln(x-1)] > 0,
假定 u = 3(x-1)ln(x-1)+x+1 = 0 ,则
u' = 3ln(x-1)+4=0,ln(x-1) = -4/3,x = 1+e^(-4/3) = 1+1/e^(4/3)
u'' = 3/(x-1) > 0,x = 1+1/e^(4/3) 是 u 的极小值点,也是最小值点,
u[1+1/e^(4/3)] = 2-3/e^(4/3) > 0,
故 y = (x-1)^[2(x+1)^3] 的单调增加区间是 (1,+∞) .
y = e^[2(x+1)^3ln(x-1)]
y' = [6(x+1)^2ln(x-1)+2(x+1)^3/(x-1)] e^[2(x+1)^3ln(x-1)]
= [2(x+1)^2/(x-1)] [3(x-1)ln(x-1)+(x+1)] e^[2(x+1)^3ln(x-1)]
x>1 时,2(x+1)^2/(x-1) > 0,e^[2(x+1)^3ln(x-1)] > 0,
假定 u = 3(x-1)ln(x-1)+x+1 = 0 ,则
u' = 3ln(x-1)+4=0,ln(x-1) = -4/3,x = 1+e^(-4/3) = 1+1/e^(4/3)
u'' = 3/(x-1) > 0,x = 1+1/e^(4/3) 是 u 的极小值点,也是最小值点,
u[1+1/e^(4/3)] = 2-3/e^(4/3) > 0,
故 y = (x-1)^[2(x+1)^3] 的单调增加区间是 (1,+∞) .
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