Question

设函数f(x)=xlnx(x>0),求函数f(x)的最小值

Answer 1

x>0
f(x)=xlnx
f'(x) = x*1/x+lnx*1 = 1+lnx = lne+lnx = ln(ex)
当ex＞1时,f(x) 单调增；当ex＜1时,f(x) 单调减.
x=1/e时,最小值f(1/e) = 1/e ln（1/e) = 1/e*(-1) = -1/e