Can you find the maximum

Let $y = x^\frac 1x$ . Then taking natural logarithm both sides,

$\begin{aligned} \ln y & = \frac {\ln x}x & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x. \\ \frac 1y \cdot \frac {dy}{dx} & = \frac {1-\ln x}{x^2} & \small \color{#3D99F6} \implies \frac {dy}{dx} = 0 \text{, when } \ln x = 1 \implies x = e \\ -\frac 1{y^2} \cdot \frac {dy}{dx} + \frac 1y \cdot \frac {d^2y}{dx^2} & = \frac {-1-2(1-\ln x)}{x^3} & \small \color{#3D99F6} \text{Differentiate both sides w.r.t. }x \text{ again.} \\ \implies 0 + \frac 1y \cdot \frac {d^2y}{dx^2}\bigg|_{x=e} & = - \frac 1{e^3} & \small \color{#3D99F6} \implies \frac {d^2y}{dx^2}\bigg|_{x=e} < 0 \text{, as }y = x^\frac 1x > 0 \end{aligned}$

Therefore, $\max \left(x^\frac 1x\right) = e^\frac 1e \approx \boxed{1.445}$ .