How to find the minimum of the \begin{align} f(x)=c|1+x|^n+|1-x|^n \end{align} for $n \ge 1$ and $c > 0$.
If we take the derivative of $f(x)$ we get \begin{align} f'(x)=-c {\rm sign}(1+x) |1+x|^{p-1} - {\rm sign}(1-x) |1-x|^{p-1} \end{align} but how do solve for $f'(x)=0$?
I suggest splitting your equation into multiple parts. Indeed we have $$ f(x) = \begin{cases} c(1 + x)^n + (x - 1)^n, \ x \in [1,\infty) \\ c(1 + x)^n + (1 - x)^n, \ x \in [-1,1) \\ c(-1)^n (1 + x)^n + (1 - x)^n, \ x \in (- \infty, -1) \end{cases}$$ Now you may compute the derivative for each of these intervals separately and consider in which of the intervals $f$ attains its smallest value compared to the others!