#### Explain solution RD Sharma class 12 chapter 17 Maxima and Minima exercise Very short answer type question  8 maths

Answer: $(e)^{-\frac{1}{c}}$

Given: $f(x)=x^{x}$

Solution:

$f(x)=x^{x}$

Taking logarithm on both sides ,we get

$\log f(x)=\mathrm{x} \log \mathrm{x}$

Differentiating with respect to x ,we get

\begin{aligned} &\frac{1}{f(x)} f^{\prime}(x)=\log x+1 \\\\ &f^{\prime}(x)=f(x)(\log x+1) \\\\ &f^{\prime}(x)=x^{x}(\log x+1) \ldots \ldots . .(i) \end{aligned}

For a local maxima or local minima we must have

\begin{aligned} &f^{\prime}(x)=0 \\\\ &x^{x}(\log x+1)=0 \\\\ &\log x=-1 \\\\ &x=\frac{1}{e} \end{aligned}

Now

\begin{aligned} &f^{\prime \prime}(x)=x^{x}(\log x+1)^{2}+\frac{x^{x}}{x} \\\\ &=x^{x}(\log x+1)^{2}+x^{x-1} \\\\ &\text { At } x=\frac{1}{e} \end{aligned}

\begin{aligned} &f^{\prime \prime}\left(\frac{1}{e}\right)=\left(\frac{1}{e}\right)^{\frac{1}{e}}\left(\log \frac{1}{e}+1\right)^{2}+\left(\frac{1}{e}\right)^{\frac{1}{e}-1} \\\\ &=\left(\frac{1}{e}\right)^{\frac{1}{e}-1}>0 \end{aligned}

So,$x=\frac{1}{e}$   is a point local minimum.

Thus the minimum value is given by

\begin{aligned} &f\left(\frac{1}{e}\right)=\left(\frac{1}{e}\right)^{\frac{1}{e}} \\\\ &=e^{-\frac{1}{e}} \\\\ &f\left(\frac{1}{\theta}\right)=e^{-\frac{1}{e}} \end{aligned}