2. Show that the function given by f ( x ) = \frac{\log x}{x} has maximum at x = e.

Answers (1)

Given function is 
f ( x ) = \frac{\log x}{x}
f^{'}(x) = \frac{1}{x}.\frac{1}{x} + log x\frac{-1}{x^2} = \frac{1}{x^2}(1-\log x)
f^{'}(x) =0 \\ \frac{1}{x^2}(1-\log x) = 0\\ \frac{1}{x^2} \neq 0 \ So \ log x = 1\Rightarrow x = e
Hence, x =e is the critical point 
Now,
f^{''}(x) = \frac{-2x}{x^3}(1-\log x)+\frac{1}{x^2}(-\frac{1}{x}) = \frac{1}{x^3}(-2x+2xlog x-1)\\ f^{''(e)} = \frac{-1}{e^3} < 0
Hence, x = e is the point of maxima

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