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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.

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This is not really number theory; it's calculus. – Arturo Magidin Apr 4 '12 at 20:32
When I see things like $x^{1/x}$ I usually say something like "That is an exponential, re-write it as such first"... – AD. Apr 4 '12 at 20:49
up vote 4 down vote accepted

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 $\gt 0$.


$$ e^{(x/e) - 1} \ge 1 + (x/e) - 1 = x/e$$


$$e^{x/e} \ge x$$

and so

$$e^{1/e} \ge x^{1/x}$$

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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.

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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.

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Why $\frac{1}{2x}\ln(x)$? That would be $f(x) = \sqrt[2x]{x}$. – Arturo Magidin Apr 4 '12 at 20:36
Oops, I will edit it. – Keaton Apr 4 '12 at 20:44

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 $x=e$.

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