Function f(x) = x^x,x>0 decreases on the interval

The correct answer is Option (1) → $(0,\frac{1}{e})$

Given: $f(x) = x^{x},\; x > 0$

Take $\log$ on both sides: $\ln f = x\ln x$

Differentiate w.r.t. $x$:

$\frac{f'(x)}{f(x)} = \ln x + 1$

$\Rightarrow f'(x) = x^{x}(\ln x + 1)$

$f'(x) < 0 \Rightarrow \ln x + 1 < 0 \Rightarrow \ln x < -1 \Rightarrow x < e^{-1}$

Hence, $f(x)$ decreases on $(0,\frac{1}{e})$