= Solution
Each coordinate lies in $[0,1]$. Since $0<c<1$, $x_i^c\geq x_i$, with equality exactly when $x_i=0$ or $1$. Summing gives $f(\mathbf x)\geq1$. Equality under the simplex constraint forces exactly one coordinate to be one. Thus \b[the minimum is $1$, attained exactly at the vertices] of the <probability simplex>.
For the maximum, $t\mapsto t^c$ is strictly concave on $[0,\infty)$. The <Jensen inequality> gives
$$
\frac1n\sum_{i=1}^nx_i^c\leq\left(\frac1n\sum_{i=1}^nx_i\right)^c=n^{-c}.
$$
Strict concavity makes equality possible exactly when all coordinates agree. Hence the <extrema of a concave power sum on a simplex> are
$$
\boxed{\min f=1,\qquad \max f=n^{1-c},\quad x_1=\cdots=x_n=1/n\ \text{at the maximum}.}
$$
For $n=1$ the simplex is a single point, and both descriptions coincide.
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