Question

Show that $\frac{\log x}{x}$ has a maximum value at x = e.

Answer 1

$$\begin{gathered} \mathrm{Here}, \\ f\left(x\right)=\frac{\log x}{x} \\ \Rightarrow f\text{'}\left(x\right)=\frac{1-\log x}{x^2} \\ \mathrm{For}\mathrm{the}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow \frac{1-\log x}{x^2}=0 \\ \Rightarrow 1=\log x \\ \Rightarrow \log e=\log x \\ \Rightarrow x=e \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=\frac{x^2\left({\displaystyle \frac{-1}{x}}\right)-2x\left(1-\log x\right)}{x^4}=\frac{-3+2\log x}{x^3} \\ \Rightarrow f\text{'}\text{'}\left(e\right)=\frac{-3+2\log e}{e^3}=\frac{-1}{e^3}<0 \\ \mathrm{So},x=e\mathrm{is}\mathrm{the}\mathrm{point}\mathrm{of}\mathrm{local}\mathrm{maximum}. \end{gathered}$$