I mean, the highest point of the $f(x)=\sqrt[x]{x}$ is when $x=e$.
I'm trying to calculate how can I prove that or how can it be calculated.
I mean, the highest point of the $f(x)=\sqrt[x]{x}$ is when $x=e$.
I'm trying to calculate how can I prove that or how can it be calculated.
The typical proof is to use derivatives and find critical points etc which is a general method and you ought to have it in your toolbox. You already have multiple answers for that.
Here is a different approach.
A simple trick works for this problem:
Use the inequality
$e^t \ge 1 + t$
which is valid for all real $t$.
Let $x$ be any real $\gt 0$.
Then
$ e^{(x/e) - 1} \ge 1 + (x/e) - 1 = x/e$
Thus
$e^{x/e} \ge x$
and so
$e^{1/e} \ge x^{1/x}$
Write the function as $y = x^{1/x}$. To take the derivative, use logarithmic differentiation. That is, let $\ln y = \ln x^{1/x} = \frac{1}{x} \ln x.$ The derivative of this is \frac{y'}{y} = -x^{-2} \ln x + x^{-2} = \frac{1}{x^2} (1 - \ln x) Since we know $y$, we can multiply both sides by $y$ to have a solution of y': y' = x^{1/x} \frac{1}{x^2} (1 - \ln x) Now, the max will occur when this is 0. The only solution is when $1 - \ln x = 0$, or when $x = e$. Now, check to make sure this actually gives you the absolute max by seeing that the derivative is positive to the left of $e$ and negative to the right of $e$. That is easy, since the only term that will ever change sign is the $1 - \ln x$. Done.
Well, write $f(x) = e^{\frac{1}{x}\ln(x)}$ and differentiate and set equal to 0 to get: $\dfrac{d}{dx}(e^{\frac{1}{x}\ln(x)})=\bigg(-\frac{1}{x^2}\ln(x)+\frac{1}{x^2}\bigg)e^{\frac{1}{x}\ln(x)}=0$ Which implies (after dividing by the exponential term) that $\frac{1}{x^2}(1-\ln(x))=0$ Whence $1=\ln(x)$ or $x=e$.
Now you just need to check whether this gives you a local minimum or maximum via the second derivative.
It really only makes sense for $x\gt 0$, at least if you stick to real numbers.
On $(0,\infty)$ can rewrite the function as $f(x) = x^{1/x} = e^{(\ln x)/x}.$ Note that as $x\to\infty$, $\lim_{x\to\infty}\frac{\ln x}{x} = 0,$ so $\lim\limits_{x\to\infty}f(x) = e^0 = 1$ and as $x\to 0^+$, we have $\lim_{x\to 0^+}\frac{\ln x}{x} = -\infty$ so $\lim\limits_{x\to 0^+}f(x) = \lim\limits_{t\to-\infty}e^t = 0$.
So that means that the function is bounded. We find its critical points by taking the derivative: $\begin{align*} \frac{d}{dx}f(x) &= \frac{d}{dx} e^{(\ln x)/x}\\ &= e^{(\ln x)/x}\left(\frac{d}{dx}\frac{\ln x}{x}\right)\\ &= e^{(\ln x)/x}\left(\frac{x\frac{1}{x} - \ln x}{x^2}\right)\\ &= e^{(\ln x)/x}\left(\frac{1-\ln x}{x^2}\right). \end{align*}$ This is zero if and only if $1-\ln x=0$, if and only if $\ln x = 1$, if and only if $x=e$. So the only critical point is at $x=e$.
If $0\lt x \lt e$, then f'(x)\gt 0 (since $\ln x \lt 1$), so the function is increasing on $(0,e)$, and if $e\lt x$, then f'(x)\lt 0, so the function is decreasing on $(e,\infty)$. Thus, $f$ has a local maximum at $x=e$, and since it is the only local extreme of the function, which is continuous, $f(x)$ has a global extreme at $x=e$.