Show that the function given by `f(x)=(logx)/x`has maximum at `x = e`.

Explanation

Let ` f(x) = (log x )/x , x gt 0 `
`f'(x)=(x*1/x-log x)/x^(2) = (1-log x)/x^(2)`
` and f''(x) = (x^(2)(-1/x)-(1-log x)* 2x)/x^(4)`
` = (-3+2 log x)/x^(3)`
For maxima/minima
f'(x) = 0
` rArr 1- log x = 0 `
` rArr x = e" "(.:' log e = 1)`
at ` x = e f''(x) = (-3+2 log e)/e^(3) = - 1/e^(3) lt 0 `
`:. ` Function is maximum at x = e,
Thus, the maximum value of function = `(log e)/e = 1/e` .