这题如何求极值

2025-06-29 12:47:14
推荐回答(1个)
回答1:

y= x^(1/x)
lny = (1/x)lnx
(1/y) dy/dx = ( 1/x^2)(1 - lnx)
dy/dx =(1-lnx). x^(1/x -2 )
y'=0
1-lnx =0
x=e
dy/dx | x=e+ <0
dy/dx | x=e- >0
x=e (max)
max y = y(e)= e^(1/e)