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

Question

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

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Answer

We have $\text{f}(\text{x})=\text{x}^{\text{x}}$
$\therefore\ \text{f}'(\text{x})=\text{x}^{\text{x}}(\log\text{x}+1)$
For maxima and minima, f'(x) = 0
$\Rightarrow \text{x}^{\text{x}}(\log\text{x}+1)=0$
$\Rightarrow \text{x}=\text{e}^{-1}$
Now,
$\therefore\ \text{f}''(\text{x})=\text{x}^{\text{x}}(\log\text{x}+1)^{2}+\frac{\text{x}^{\text{x}}}{\text{x}}$
At $\text{x}=\frac{1}{\text{e}}$
$\therefore\ \text{f}''(\text{x})>0\ \text{as}\ \text{x}^{\text{x}}(\log\text{x}+1)^{2}+\frac{\text{x}^{\text{3}}}{\text{x}}>0$
$\therefore \text{x}=\frac{1}{\text{e}}$ is the point of local minima.
Hence, minimum value $=\text{f}\Big(\frac{1}{\text{e}}\Big)=\Big(\frac{1}{\text{e}}\Big)^\frac{1}{\text{e}}=\text{e}^\frac{-1}{\text{e}}$ .

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