The function $f(x)=x^x$ decreases on the interval

$(0, e)$ $(0, 1)$ $(0,\frac{1}{e})$ None of these

$(0,\frac{1}{e})$

Clearly, f(x) is defined for x > 0 Now, $f(x)=x^x⇒f'(x)=x^x(1+\log x)$ For to be decreasing, we must have $f'(x)<0⇒x^x(1+\log x)<0$ $⇒1+\log x<0⇒\log x< -1 ⇒ x<e^{-1}$ So, f(x) is decreasing on $(0,\frac{1}{e})$