Question

已知函数f(x)=lnx+k/x,k属于R (1) 若k=1,求函数f(x)的单调区间;(2)若f(x)>=2+(1-e) / x衡成立,求实数k...

已知函数f(x)=lnx+k/x,k属于R (1) 若k=1,求函数f(x)的单调区间;(2)若f(x)>=2+(1-e) / x衡成立,求实数k的取值范围

Answer 1

(1)
k = 1, f(x) = lnx + 1/x, f'(x) = 1/x - 1/x² = (x - 1)/x² = 0, x = 1
0 < x < 1: f'(x) < 0, 递减增
x > 1: f'(x) > 0, 递增

(2)
令g(x) = f(x) - [2 + (1 - e)/x] = lnx + (k + e - 1)/x - 2
只须g(x)值域为g(x) ≥ 0即可
g'(x) = 1/x - (k + e -1)/x² = [x - (k + e - 1)]/x² = 0
x = k + e - 1
k + e - 1 ≤ 0时, g'(x) ≥ 0, g(x)无最小值，舍去。
k + e - 1 > 0时, g'(x) = 0, x = k + e - 1时, g(x)有最小值
g(k + e - 1) = ln(k + e - 1) + (k + e - 1)/(k + e - 1) - 2 = ln(k + e - 1) - 1 ≥ 0
k + e - 1 ≥ e
k ≥ 1

Answer 2

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