Example 1 Given the function $f(x)=-x^{2}+2 t x-t$, $x \in[-1,1]$. Let the maximum value of $f(x)$ be $M$. Find the minimum value of $M$.
Solution: Since $f(x)=-x^{2}+2 t x-t=-(x-t)^{2}+t^{2}-t$, and $-1 \leqslant x \leqslant 1$, then
when $t \leqslant-1$, $M=f(-1)=-3 t-1$;
when $-1<t<1$, $M=f(t)=t^{2}-t$;
when $t \geqslant 1$, $M=f(1)=t-1$.
Therefore, $M=\left\{\begin{array}{ll}-3 t-1 & (t \leqslant-1), \\ t^{2}-t & (-1<t<1), \\ t-1 & (t \geqslant 1) .\end{array}\right.$
It is easy to see that $M_{\text {min }}=-\frac{1}{4}$.