Application of Derivatives — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSApplication of Derivatives3 Marks
Question
Show that the function given by f(x) = $\frac{\log x}{x}$ has maximum at x = e.
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Answer
It is given that f(x) = $\frac{\log x}{x}$ Then, f'(x) = $\frac{\mathrm{x}\left(\frac{1}{\mathrm{x}}\right)-\log \mathrm{x}}{\mathrm{x}^{2}}=\frac{1-\log \mathrm{x}}{\mathrm{x}^{2}}$ Now, f'(x) = 0 $\Rightarrow$ 1 - log x = 0 $\Rightarrow$ log x = 1 $\Rightarrow$ log x = log e $\Rightarrow$ x = e Further, $f^{\prime \prime}(x)=\frac{x^{2}\left(-\frac{1}{x}\right)-(1-\log x)(2 x)}{x^{4}}$ = $\frac{-x-2 x(1-\log x)}{x^{4}}$ = $\frac{-3+2 \log x}{x^{3}}$ Now, $f^\prime{^\prime}$(e) = $\frac{-3+2 \log e}{e^{3}}=\frac{-3+2}{e^{3}}=\frac{-1}{e^{3}}<0$ Therefore, by second derivative test, f is the maximum at x = e.
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