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Question
Write the minimum value of f(x) = xx .
Solution
\[\text { Given: } \hspace{0.167em} f\left( x \right) = x^x \]
\[\text { Taking log on both sides, we get }\]
\[\log f\left( x \right) = x \log x\]
\[\text { Differentiating w . r . t . x, we get }\]
\[\frac{1}{f\left( x \right)} f'\left( x \right) = \log x + 1\]
\[ \Rightarrow f'\left( x \right) = f\left( x \right) \left( \log x + 1 \right)\]
\[ \Rightarrow f'\left( x \right) = x^x \left( \log x + 1 \right) .............. \left( 1 \right)\]
\[\text { For a local maxima or a local minima, we must have } \]
\[f'\left( x \right) = 0\]
\[ \Rightarrow x^x \left( \log x + 1 \right) = 0\]
\[ \Rightarrow \log x = - 1\]
\[ \Rightarrow x = \frac{1}{e}\]
\[\text { Now }, \]
\[f''\left( x \right) = x^x \left( \log x + 1 \right)^2 + x^x \times \frac{1}{x} = x^x \left( \log x + 1 \right)^2 + x^{x - 1} \]
\[\text { At }x = \frac{1}{e}: \]
\[f''\left( \frac{1}{e} \right) = \frac{1}{e}^\frac{1}{e} \left( \log\frac{1}{e} + 1 \right)^2 + \frac{1}{e}^\frac{1}{e} - 1 = \frac{1}{e}^\frac{1}{e} - 1 > 0\]
\[\text { So,} x = \frac{1}{e}\text { is a point of local minimum .} \]
\[\text { Thus, the minimum value is given by }\]
\[f\left( \frac{1}{e} \right) = \frac{1}{e}^\frac{1}{e} = e^\frac{- 1}{e} \]
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