d
dx
 ef(x) = f¢(x)ef(x)
and, as e is never zero, the turning points of ef(x) occur for the same values of x as the turning points of f(x).

Now x1/x = e1/xlogx and
d
dx
( 1
x
 logx) = 1-logx
x2
 .
The derivative is positive for logx < 1, that is x < e, it is is zero when logx = 1, x = e and it is negative for x > e. Hence the function is increasing for x < e and decreasing for x > e.
d2
dx2
( 1
x
 logx) = -x-2x(1-logx)
x4
= 2logx -3
x3
 .
At e this is -1/e3 so (e, e1/e) is a maximum point. As the graph is decreasing as x® ¥ for each value of x there is a value of n for which n1/n < x1/x and vice versa so both tend to the same limit which is 1.

Writing t=1/x,
x1/x = æ
ç
è 		1
t
 ö
÷
ø 		t

 
 = 1
tt
 .
As x® 0 we have t® ¥ and tt ® ¥ so
x1/x = 1
tt
 ® 0
.

Hence the graph of y=x1/x is always above the line y=0 although the function is undefined at x=0.

Graph of y = x^(1/x)