Write the minimum value of f(x)=x^1 x .

Question

Write the minimum value of $\text{f}(\text{x})=\text{x}^\frac{1}{\text{x}}.$

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Answer

We have $\text{f}(\text{x})=\text{x}^\frac{1}{\text{x}}$ Taking log on both side, we get $\log\text{f}(\text{x})=\frac{1}{\text{x}}\ \log\text{x}$ Differentitating W.r.t. x, we get $\frac{1}{\text{f}(\text{x})}\text{f}'(\text{x})=\frac{-1}{\text{x}^{2}}\ \log\text{x}+\frac{1}{\text{x}^{2}}$ $\Rightarrow\text{f}'(\text{x})=\text{f}'(\text{x})\frac{1}{\text{x}^{2}}\ (1-\log\text{x})$ $\Rightarrow\text{f}'(\text{x})=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big) ...(\text{i})$ $\Rightarrow\text{f}'(\text{x})=\text{x}^{\frac{1}{\text{x}}-2}(1-\log\text{x})$ For a local maximum or a local minima, we must have f'(x) = 0 $=\text{x}^{\frac{1}{\text{x}}-2}(1-\log\text{x}) =0$ $\Rightarrow \log\text{x}=1$ $\therefore \text{x}=\text{e}$ Now, $\text{f}''(\text{x})=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big)^{2}+\text{x}^\frac{1}{\text{x}}\Big(\frac{-2}{\text{x}^{3}}+\frac{2}{\text{x}^{3}}\log\text{x}-\frac{1}{\text{x}^{2}}\Big)$ $=\text{x}^\frac{1}{\text{x}}\Big(\frac{1}{\text{x}^{2}}-\frac{1}{\text{x}^{2}}\log\text{x}\Big)^{2}+\text{x}^\frac{1}{\text{x}}\Big(\frac{-2}{\text{x}^{3}}+\frac{2}{\text{x}^{3}}\log\text{x}-\frac{1}{\text{x}^{2}}\Big)$ At x = e $\text{f}''(\text{e})=\text{e}^\frac{1}{\text{e}}\Big(\frac{1}{\text{e}^{2}}-\frac{1}{\text{e}^{2}}\log\text{e}\Big)^{2}+\text{e}^\frac{1}{\text{e}}\Big(\frac{-3}{\text{e}^{3}}+\frac{2}{\text{e}^{3}}\log\text{e}\Big)$ $=-\text{e}^\frac{1}{\text{e}}(\frac{1}{\text{e}^{3}})<0$ So, x = e is a point of local maximum. Thus, the maximum value is given by $\text{f}(\text{e})=\text{e}^\frac{1}{\text{e}}$