How to find the "dip point" in the graph of $y=x^{x}$ graph?
I was exploring various graphs on Desmos, and I stumbled upon the graph of $x^x$.
Now I tried putting various values to examine the nature of this graph between $0$ and $1$ but how do I find that turning point of this graph where it starts rising up? Is there any elementary method to find it? Thank you
On
Since $y=x^x=e^{x\ln x}$, it is enough to find the minumum of $y=x\ln x$, $0<x<1$. Now, let $x=e^{-z}$, $0<z<\infty$. Then, the problem is finding the minumum of $y=-ze^{-z}$ and thus finding the maximum of $y=\frac{z}{e^z}$.
We claim that maximum of $y=\frac{z}{e^z}$ occurs at $z=1$ and it is $\frac{1}{e}$. If we increase $z=1$ to $z=1+\epsilon$, $\epsilon>0$, the the inequality $\frac{1+\epsilon}{e^{1+\epsilon}}<\frac{1}{e}$ gives $1+\epsilon<e^\epsilon$ which is true. Similarly for the decrease from $z=1$ to $z=1-\epsilon$. Since, this also gives a similar thing.
Hence, the maximum of $y=x^x$ occurs at $x=e^{-z}=e^{-1}=\frac{1}{e}$ and it is $\frac{1}{\sqrt[e]{e}}$.
Derivative of $x^x$ is $x^x(1+\ln(x))$. Solving $1+\ln(x) = 0$ gives us $x = e^{-1}$, which is the point you are looking for. Of course having derivative equal to $0$ isn't a sufficient condition, but you can easily check that in fact there is a local minimum there.