Analytic Properties of $f(x) = x^x$ The function $f(x) = x^x, x > 0$ can be plotted on graphing software and inspected to see a local minimum around .367. The function is convex, decreasing from 0 to its minimum, and increasing thereafter. The derivative of the function can be found by implicit differentiation to be $f'(x) = x^x(\ln(x)+1)$.


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*Can the exact value of the local minimum be found?

*Can someone explain intuitively why the function first decreases and then increases?

*Is there anything interesting about the class of functions $\{ \, f(x) \, | \, f'(x) = f(x) \cdot g(x) \, \}$

Edit: I forgot, I'll be subject to a firing squad if I don't explain what I have tried!

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*Setting the derivative equal to 0 doesn't do much for me. If this can be solved exactly without optimization algorithms, I suspect something really clever will need to happen.


*I tried thinking about what is happing for different types of inputs (irrational $x$, rational $x$, natural $x$). Still didn't get anywhere.


*Nothing to try really, does this class of functions come up anywhere in math?
 A: The minimum is attained at $x=\frac1e$ as can be seen by setting the derivative equal to $0$ and solving for $x$.
Your statement in a comment that it is not always true that if $x>y>0$ then $x^x>y^x$ is false.  We have $$x>y\implies \log x > \log x\implies x\log x > x\log y \implies x^x>y^x  $$
Perhaps you meant to say it is not always true that $x>y>0$ implies $x^x>y^y$.  This is true.  Taking logarithms, there's no particular reason to believe that $x\log x > y\log y$ and indeed, it isn't always true.
To try to explain it intuitively, note first that we're concerned with numbers $0<x<y<1$.  When you raise such a number to a power, the larger the power, the smaller the result.  When we compare $x^x$ and $y^y$, for some choices of $x$ and $y$, this effect dominates, and $y^y$ is the smaller of the two numbers.  For other choices of $x$ and $y$, the fact that $x$ is smaller dominates.
A: It's good habit to rewrite $a^b$ as $\exp(b\ln(a))$. In our case this is $f(x)=x^x = \exp(x\ln(x))$, which is easier to study. The expression makes sense for $x>0$, doesn't make sense for $x<0$, and leaves some room for debate at $x=0$ although it's sensible to take $0^0=1$. The function is then continuous on $\{x\colon x \ge 0\}$.
The derivative, like you said, is $f'(x)=x^x(1+\ln(x))$. For this to be $0$, either $x^x$ should be $0$ (never happens, as $\exp$ never vanishes) or $1+\ln(x)=0$ which means $\ln(x)=-1$ or $x = \frac{1}{e}$. If there is a minimum – and indeed there is – it must be there. You can verify this in any number of ways, such as taking a second derivative or just checking nearby values.
The function decreases when $0 \le x \le \frac{1}{e}$ and increases afterwards.
I can't imagine why you wanted to understand the behavior for rational and irrational values of the argument.
Exponential functions arise all over the place in mathematics. This particular function isn't super common, but it pops up here and there.
A: For your other question, we can rewrite the equation as an ODE
$$\frac{dy}{dx} = g(x)\cdot y$$
which we can separate variables and integrate to solve. The final answer is always of the form
$$y = C\exp\left(G(x)\right)$$
where $C$ is an arbitrary constant and $G$ is an antiderivative of $g$.
A: for question 3:
$f'(x) = f(x) g(x) \implies \dfrac{f'(x)}{f(x)} = g(x) \implies ln|f(x)| = \int g(x) = G(x) \implies f(x) = e^{G(x)}$
there are 2 cases... if $G(x) = ln(h(x))$, then $f(x) = h(x)$ otherwise it is an exponent. Either ways $f(x)>0$ is the only property I can see... Can we say that means f(x) is exponential?
A: Among the properties of the function $x^x$, don't forget

*

*The antiderivative $\quad \text{Sphd}(x)=\int_0^x t^{\,t}\, dt\quad$ namely the Sophomore's dream function (Note below).


*The inverse function of $\quad x^x=y(x)\quad$ which is
$$x(y)=e^{W\big(\ln(y)\big)}$$
$W$ is the Lambert W function (graph below)

The graph is a copy of page 13 in https://fr.scribd.com/doc/34977341/Sophomore-s-Dream-Function .
Note : In this paper one can find many properties of the antiderivative of $x^x$ wich is a special function (not standardized).
