Question

F(x) = lnx/x

(a): Find f'(x) and show that x = e is a maximum
(b): Deduce that e^x >= x^e, for x > 0

Answer 1

`f(x) = (lnx)/(x)`

(a)
`f'(x) = (d)/(dx)[(lnx)/(x)]`
Using the quotient rule:

`f'(x) = ((1)/(x) \cdot x - lnx)/(x^(2))`

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

For maximum, f'(x) = 0;

`(1 - lnx)/(x) = 0`

`lnx = 1, x = e`

(b) Deduce:

`e^(x) \geq x^(e), x > 0`

Soln: Since x = e is the greatest value, then f(e) ≥ f(x) > f(0)

`(ln(e))/(e) \geq (lnx)/(x)`

`(1)/(e) \geq (lnx)/(x)`, since ln(e) is simply equal to 1

Now, since x > 0, then we don't have to worry about flipping the signs when multiplying by x.

`(x)/(e) \geq lnx`

`x \geq elnx`

`x \geq ln(x^(e))`

Taking the exponential to both sides will cancel with the natural logarithmic function in the right hand side to produce:

`e^(x) \geq e^{ln(x^(e)})`

`e^(x) \geq x^(e)`, as required.