Behavior of $y = x^x$ when $x$ approaches zero

Question

I am trying to understand the behavior of $y = x^x$ when $x$ approaches zero. When $x \le 1$, $y$ initally decreases but becomes larger at some point (around $x = 0.35$) and starts approaching one as $x\to0$ (see graph below).

This might likely be a simple/obvious answer but I don't have a huge math background and also couldn't find any help otherwise.

So my question is why does $y$ get larger at $x \approx 0.35$?

Answer 1

You need calculus to understand the behaviour of the function fully.

$y = f(x) = x^x$

$f'(x) = x^x(1+\ln x)$

For stationary point(s), set $f'(x) = 0$ to get $x^x(1+\ln x)$. Since the first term is always non-zero for real $x$, that leaves us with $(1+\ln x) = 0 \implies \ln x = -1 \implies x = e^{-1}$. At this point, $y = e^{-\frac 1e}$. You can show this is a minimum using the second derivative test.

As to the limiting behaviour of the function as $x$ approaches zero from the right hand side, that is also best addressed through calculus. Noting that $f(x) = e^{x\ln x}$, we set $g(x) = x\ln x$. This can be transformed into a form suitable for analysis with L'Hopital's Rule (LHR) for the limit $x \rightarrow 0^+$ by rewriting it $g(x) = \frac{\ln x}{\frac 1x}$. By LHR, that limit is equivalent to that of $\frac{\frac 1x}{-\frac 1{x^2}} = -x$, which is $0$. So the limit $\lim_{x \to 0^+}f(x) = e^0 = 1$.

Answer 2

Derivative of $x^x$ can be calculated by using log functions. $$ y=x^x \\ \ln{y}=x\ln{x}\\ \frac{y'}{y}=\ln{x}+1\\ y'=(\ln{x}+1)\times x^x $$ The derivative of the formula becomes $0$ when $\ln{x}+1=0$, which means the graph will have the minimum value when $x=\frac{1}{e}=0.367879...$.