15. (12 points) Given vectors
$$
\boldsymbol{a}=\left(x^{2}, x+1\right), \boldsymbol{b}=(1-x, t) \text {. }
$$

If the function $f(x)=\boldsymbol{a} \cdot \boldsymbol{b}$ is a monotonically increasing function on the interval $(-1,1)$, find the range of values for $t$.
Three, 15. By definition, we have
$$
\begin{array}{l}
f(x)=x^{2}(1-x)+t(x+1) \\
=-x^{3}+x^{2}+t x+t . \\
\text { Then } f^{\prime}(x)=-3 x^{2}+2 x+t .
\end{array}
$$

If $f(x)$ is a monotonically increasing function on the interval $(-1,1)$, then $f^{\prime}(x) \geqslant 0$ on $(-1,1)$.

Thus, $-3 x^{2}+2 x+t \geqslant 0$ must hold for all $x \in(-1,1)$, which means $t \geqslant 3 x^{2}-2 x$ must hold for all $x \in(-1,1)$.
Consider the function $g(x)=3 x^{2}-2 x$.
Since the graph of $g(x)$ is a parabola that opens upwards with its axis of symmetry at $x=\frac{1}{3}$, to ensure that $t \geqslant 3 x^{2}-2 x$ holds for all $x \in(-1,1)$, we need
$$
t \geqslant g(-1)=5 \text {. }
$$

When $t \geqslant 5$, $f^{\prime}(x) \geqslant 0$ holds on $(-1,1)$, meaning $f(x)$ is a monotonically increasing function on the interval $(-1,1)$.
Therefore, the range of values for $t$ is $t \geqslant 5$.