If G(x)=e^f(x), by the chain rule:

G¢(x)=ef(x)*f¢(x).

As e is never zero, the gradient function of G(x) is zero when the gradient function of f(x) is zero and it always has the same sign as f¢(x) so G(x) and f(x) have the same turning points.

To answer the question of turning points of G(x)=x^1/_x = e^1/xlogx let's first find the derivative of ¹/_xlog x.

d dx f(x)= 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 x=e the second derivative of f(x) is -1/e³ which is negative so f(x) has a maximum and so G(x) has a maximum. Therefore (e, e^1/e) is a maximum point of G(x)=x^1/_x.

To prove

lim

x1/x ® 1

as x® ¥ we may use the similar result for the discrete case n^1/_n. As the graph of f(x)=x^1/_x is decreasing as x® ¥ for each value of x there is a value of n for which n^1/n < x^1/x and vice versa so both tend to the same limit which is 1.

What if x® 0? Well, as suggested in the question, letting 1/x=t in x^1/_x, one can note that as x® 0, t® ¥. Writing t=1/x,

x1/x = æ ç è 1 t ö ÷ ø t   = 1 tt .

As x® 0 we have t® ¥ and t^t ® ¥ so

x1/x = 1 tt ® 0

.

Hence the graph of y=x^1/x is always above the line y=0 although the function is undefined at x=0.