 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2011/paper-1/3/b/solution

For

x > 0

ln f (x) = \frac{l n x}{x}, \frac{f ^{'} ( x )}{f ( x )} = \frac{1 - l n x}{x ^{2}} .

Since

f (x) > 0

, the derivative vanishes precisely when

ln x = 1

. Hence

x_{0} = e, f (x_{0}) = e^{1/ e} .

The derivative is positive before

e

and negative after

e

, so this is a global maximum on the given positive domain.

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