3. Given the function $f(x)=x^{2}-2 t x+t$. When $x \in$ $[-1,1]$, let the minimum value of $f(x)$ be $m$. Then the maximum value of $m$ is $(\quad)$.
(A) -2
(B) 0
(C) $\frac{1}{4}$
(D) 1
3. C.

When $t \leqslant -1$, $m=f(-1)=3t+1$, the maximum value of $m$ is -2;
When $-1<t<1$,
$$
m=f(t)=t-t^{2}=-\left(t-\frac{1}{2}\right)^{2}+\frac{1}{4} \text{, }
$$
the maximum value of $m$ is $\frac{1}{4}$;
When $t \geqslant 1$, $m=f(1)=1-t$, the maximum value of $m$ is 0.
In summary, the maximum value of $m$ is $\frac{1}{4}$.