Question

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

Answer 1

\[\text { Given: } \hspace{0.167em} f\left( x \right) = x^x \]

\[\text { Taking log on both sides, we get }\]

\[\log f\left( x \right) = x \log x\]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{1}{f\left( x \right)} f'\left( x \right) = \log x + 1\]

\[ \Rightarrow f'\left( x \right) = f\left( x \right) \left( \log x + 1 \right)\]

\[ \Rightarrow f'\left( x \right) = x^x \left( \log x + 1 \right) .............. \left( 1 \right)\]

\[\text { For a local maxima or a local minima, we must have } \]

\[f'\left( x \right) = 0\]

\[ \Rightarrow x^x \left( \log x + 1 \right) = 0\]

\[ \Rightarrow \log x = - 1\]

\[ \Rightarrow x = \frac{1}{e}\]

\[\text { Now }, \]

\[f''\left( x \right) = x^x \left( \log x + 1 \right)^2 + x^x \times \frac{1}{x} = x^x \left( \log x + 1 \right)^2 + x^{x - 1} \]

\[\text { At }x = \frac{1}{e}: \]

\[f''\left( \frac{1}{e} \right) = \frac{1}{e}^\frac{1}{e} \left( \log\frac{1}{e} + 1 \right)^2 + \frac{1}{e}^\frac{1}{e} - 1 = \frac{1}{e}^\frac{1}{e} - 1 > 0\]

\[\text { So,} x = \frac{1}{e}\text {  is a point of local minimum .} \]

\[\text { Thus, the minimum value is given by }\]

\[f\left( \frac{1}{e} \right) = \frac{1}{e}^\frac{1}{e} = e^\frac{- 1}{e} \]