Is it possible to find the local maximum of $\sqrt[x]{x}$ without using derivative?

Question

Let $f(x)=\sqrt[x]{x}$, where $x\in\mathbb{R^{+}}$

Using derivative,

$$\frac{d}{dx}(x^{\frac 1x}) = -x^{\frac 1x - 2} (\log(x) - 1)$$

$$f'(x)=0 \longrightarrow x=e$$

$$\text{max}\left\{\sqrt[x]{x}\right\}=\sqrt[e]{e}$$

But,I'm looking for a way, which is doesn't include derivative.

Answer 1

This answer fleshes out the hint of Edward H. in the comments under the question.

Hint Since $\log$ and $\exp$ are strictly increasing (and since $\exp$ is onto the domain of $f$), if we can find the maximum value $m$ of $$\log f(e^t) = t e^{-t} ,$$ then the maximum value of $f(x)$ is $e^m$. The binomial expansion gives $$e^{t - 1} = \lim_{n \to \infty} \left(1 + \frac{t - 1}{n}\right)^n \geq 1 + n\cdot\frac{t - 1}{n} = t .$$

Rearranging gives $\frac{1}{e} \geq t e^{-t}$, but equality holds for $t = 1$, so $m = \frac{1}{e}$ and the (in fact, global) maximum of $f$ is $e^\frac{1}{e} = \sqrt[e]{e}$.