Question

设函数fx=px-2lnx,(1)若p＞0，求函数fx的最小值

设函数fx=px-2lnx,(1)若p＞0，求函数fx的最小值（2）若函数gx=fx-p/x在其定义域内为单调函数，求P的取值范围

Answer 1

f(x) = px -2lnx
f'(x) = p -2/x
f'(x) =0
=> x = 2/p
f''(x) = 2/x^2 > 0 ( min )
min f(x) = f(2/p) = 2- 2ln(2/p)
g(x) = f(x) - p/x
= px -2lnx - p/x
g'(x) = p - 2/x +p/x^2
= (px^2-2x+p)/x^2
g(x) 在其定义域内为单调函数
=> g'(x) >0
=> px^2-2x+p >0
p> [1+√(1+p^2)]/ p or p< [1-√(1+p^2)]/ p