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

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

Let \(y = x^x\). By logarithmic differentiation:

    \(\log y = x\log x\)

    \( \displaystyle \frac{1}{y}\frac{dy}{dx} = \log x + 1 \)

    \(f'(x) = x^x(\log x + 1)\)

Since \(x^x \gt 0\), the sign of f′(x) is the sign of \(\log x + 1\):

    \(f'(x) \lt 0\) when \(\log x \lt -1\)

    \( \displaystyle \text{that is, when } 0 \lt x \lt \frac{1}{e} \)

So f decreases on \(\left(0, \frac{1}{e}\right)\) and increases on \(\left(\frac{1}{e}, \infty\right)\).