Question

The maximum value of x^1/x, x>0 is

(a) e^1/e

(b) ${\left(\frac{1}{e}\right)}^e$

(c) 1

(d) none of these

Answer 1

(a) $e^{\frac{1}{e}}$
$$\begin{gathered} \mathrm{Given}:f\left(x\right)=x^{\frac{1}{x}} \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \log f\left(x\right)=\frac{1}{x}\log x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x,\mathrm{we}\mathrm{get} \\ \frac{1}{f\left(x\right)}f\text{'}\left(x\right)=\frac{-1}{x^2}\log x+\frac{1}{x^2} \\ \Rightarrow f\text{'}\left(x\right)=f\left(x\right)\frac{1}{x^2}\left(1-\log x\right) \\ \Rightarrow f\text{'}\left(x\right)=x^{\frac{1}{x}}\left(\frac{1}{x^2}-\frac{1}{x^2}\log x\right)...\left(1\right) \\ \Rightarrow f\text{'}\left(x\right)=x^{\frac{1}{x}-2}\left(1-\log x\right) \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow x^{\frac{1}{x}-2}\left(1-\log x\right)=0 \\ \Rightarrow \log x=1 \\ \Rightarrow x=e \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=x^{\frac{1}{x}}{\left(\frac{1}{x^2}-\frac{1}{x^2}\log x\right)}^2+x^{\frac{1}{x}}\left(\frac{-2}{x^3}+\frac{2}{x^3}\log x-\frac{1}{x^3}\right)=x^{\frac{1}{x}}{\left(\frac{1}{x^2}-\frac{1}{x^2}\log x\right)}^2+x^{\frac{1}{x}}\left(-\frac{3}{x^3}+\frac{2}{x^3}\log x\right) \\ \mathrm{At}x=e: \\ f\text{'}\text{'}\left(e\right)=e^{\frac{1}{e}}{\left(\frac{1}{e^2}-\frac{1}{e^2}\log e\right)}^2+e^{\frac{1}{e}}\left(-\frac{3}{e^3}+\frac{2}{e^3}\log e\right)=-e^{\frac{1}{e}}\left(\frac{1}{e^3}\right)<0 \\ \mathrm{So},x=e\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{local}\mathrm{maxima}. \\ \therefore \mathrm{Maximum}\mathrm{value}=f\left(e\right)=e^{\frac{1}{e}} \end{gathered}$$

Disclaimer: The answer given in the book is incorrect. The solution provided here is according to the question.