One, (20 points) Given that when $-1 \leqslant x \leqslant 0$, the function $y=\left|x^{2}+(m-1) x+\left(m^{2}-3 m+1\right)\right|$ increases as $x$ increases. Find the range of real numbers for $m$.

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One, (20 points) Given that when $-1 \leqslant x \leqslant 0$, the function $y=\left|x^{2}+(m-1) x+\left(m^{2}-3 m+1\right)\right|$ increases as $x$ increases. Find the range of real numbers for $m$.
Let $f(x)=x^{2}+(m-1) x+\left(m^{2}-3 m+1\right)$.
Then $\Delta=(m-1)^{2}-4\left(m^{2}-3 m+1\right)$
$$
=-(3 m-1)(m-3) \text {, }
$$
The axis of symmetry of $f(x)$ is $x=\frac{1-m}{2}$.
When $\Delta \leqslant 0$, i.e., $m \leqslant \frac{1}{3}$ or $m \geqslant 3$,
$$
\frac{1-m}{2} \leqslant-1 \Rightarrow m \geqslant 3 \text {. }
$$

When $\Delta>0$, i.e., $\frac{1}{3}<m<3$,
$$
\begin{array}{l}
-1<\frac{1-m}{2}<\frac{1}{3} \\
\Rightarrow\left\{\begin{array}{l}
\frac{1-m}{2} \geqslant 0, \\
f(-1)=m^{2}-4 m+3 \leqslant 0
\end{array}\right. \\
\Rightarrow m=1 .
\end{array}
$$

In summary, the range of values for $m$ is $m=1$ or $m \geqslant 3$.