Explanation

(c)
Given, $f(x) = x^{- x}$
$\Rightarrow \ \ \ \log{f(x)} = - x\log x$
On differentiating w.r.t. $x$, we get
$\frac{1}{f(x)}.f^{'}(x) = - \log x - 1$
$\Rightarrow \ \ f^{'}(x) = - f(x)(1 + \log x)$
Put $f^{'}(x) = )0$
$\Rightarrow \ \ \log x = - 1\ \ \ \Rightarrow \ \ \ x = e^{- 1}$
$\therefore\ \ \ f^{''}(x) = - f^{'}(x)\left( 1 + \log x \right) - f(x)\frac{1}{x}$
$= f(x)\left( 1 + \log x \right)^{2} - \frac{f(x)}{x}$
At $x = \frac{1}{e}$,
$f^{''}(x) = - ef\left( \frac{1}{e} \right) < 0$, maxima
Hence, at $x = \frac{1}{e},\ \ f(x)$ is maximum.