The interval in which the function $f(x)=x^x, x0$, is strictly increasing is

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The interval in which the function $f(x)=x^x, x>0$, is strictly increasing is

Aicharya · Accepted Answer

$\begin{aligned} & \mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{x}} ; \mathrm{x}>0 \ & \ell \operatorname{nn}=\mathrm{x} \ell \mathrm{n} \mathrm{x} \ & \frac{1}{\mathrm{y}} \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{x}}{\mathrm{x}}+\ln \mathrm{n} \ & \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{x}^{\mathrm{x}}(1+\ell \mathrm{n} x) \end{aligned}$ for strictly increasing $\begin{aligned} & \frac{d y}{d x} \geq 0 \Rightarrow x^x(1+\ell n x) \geq 0 \ & \Rightarrow \ell n x \geq-1 \end{aligned}$ $\begin{aligned} & x \geq e^{-1} \ & x \geq \frac{1}{e} \ & x \in\left[\frac{1}{e}, \inftyight)\end{aligned}$

The interval in which the function $f(x)=x^x, x>0$, is strictly increasing is

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