Find the interval in which the function f(x) = x^x, x > 0 is strictly…

The correct answer is Option (2) → $[\frac{1}{e},∞)$

Given $f(x) = x^x, x > 0$.

Taking logarithms of both sides, we get

$\log f(x) = x \log x$, diff. w.r.t. $x$,

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

$⇒ f'(x) = f(x) (1 + \log x) ⇒ f'(x) = x^x(1+ \log x)$.

Now $f'(x) > 0$ iff $x^x(1+ \log x) > 0$ but $x^x > 0$ for all $x > 0$

$⇒1 + \log x > 0⇒ \log x > -1$

$⇒ x>e^{-1}$ i.e. $x> \frac{1}{e}$.

Thus, $f'(x) > 0$ for all $x ∈ \left(\frac{1}{e},∞\right)$, therefore, $f$ is strictly increasing in $[\frac{1}{e},∞)$.