Question

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

Answer 1

We have f (x) = `log x/x, x > 0`

Differentiating w.r.t. x , we get

⇒ `f' (x) = (x(1/x) - (log x)*1)/x^2`

`= (1 - log x)/x^2`

For maximum / minimum, f (x) = 0

⇒ `(1 - log x)/x^2 = 0`

⇒ log x = 1          .....(∵x² ≠ 0)

⇒ x = e

Again differentiating w.r.t x, we get

`f'' (x) = (x^2 (-1/x) - (1 - log x) 2x)/x^4`

`= (-x - 2x + 2x log x)/x^4`

`= (x (2 log x - 3))/x^4`

`= (2 log x - 3)/x^3`

Also, f'' (e) = `(2 log e - 3)/e^3`

`= (2.1 - 3)/e^3`              ....(∵ log_e e = 1)

`= -1/e^3 < 0`

⇒ f (x) has a maximum value at x = e.