0
$\begingroup$

Question: Where does the function $f(x) = x^{1-x}$ attain its local extrema?

Now analytically one can say that for $x > 0$, we have $x^{1-x} = e^{(1-x)\ln(x)}$, and then we can find the derivative and so forth to find a local extremum at x = 1. But there are more local extrema for $x < 0$ as can be seen from the graph here.

My question is: is there an analytic way to find the other extrema?

1 Answers 1

0

The function $f(x)=x^{1-x}$, as a real-valued function, is only defined for $x\geq0$. For $x<0$ you can define it only for some values. for example, for all $x=\frac{p}{q}$ where $q$ is even and $p<0$, the function is undefined. The graph you see is the graph of the real part and the imaginary part, when thinking of $f(x)$ as a complex valued function.