Question

Write the minimum value of f(x) = x^x .

Answer 1

$$\begin{gathered} \mathrm{Given}:\hspace{0.17em}f\left(x\right)=x^x \\ \mathrm{Taking}\log \mathrm{on}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ \log f\left(x\right)=x\log x \\ \mathrm{Differentiating}\mathrm{w}.\mathrm{r}.\mathrm{t}.x,\mathrm{we}\mathrm{get} \\ \frac{1}{f\left(x\right)}f\text{'}\left(x\right)=\log x+1 \\ \Rightarrow f\text{'}\left(x\right)=f\left(x\right)\left(\log x+1\right) \\ \Rightarrow f\text{'}\left(x\right)=x^x\left(\log x+1\right)....\left(1\right) \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow x^x\left(\log x+1\right)=0 \\ \Rightarrow \log x=-1 \\ \Rightarrow x=\frac{1}{e} \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=x^x{\left(\log x+1\right)}^2+x^x\times \frac{1}{x}=x^x{\left(\log x+1\right)}^2+x^{x-1} \\ \mathrm{At}x=\frac{1}{e}: \\ f\text{'}\text{'}\left(\frac{1}{e}\right)={\frac{1}{e}}^{\frac{1}{e}}{\left(\log \frac{1}{e}+1\right)}^2+{\frac{1}{e}}^{\frac{1}{e}-1}={\frac{1}{e}}^{\frac{1}{e}-1}>0 \\ \mathrm{So},x=\frac{1}{e}\mathrm{is}\mathrm{a}\mathrm{point}\mathrm{of}\mathrm{local}\mathrm{minimum}. \\ \mathrm{Thus},\mathrm{the}\mathrm{minimum}\mathrm{value}\mathrm{is}\mathrm{given}\mathrm{by} \\ f\left(\frac{1}{e}\right)={\frac{1}{e}}^{\frac{1}{e}}=e^{\frac{-1}{e}} \end{gathered}$$