(15) (12 points) Given vectors $\vec{a}=\left(x^{2}, x+1\right), \vec{b}=(1-x, t)$. If the function $f(x)=\vec{a} \cdot \vec{b}$ is a monotonically increasing function on the interval $(-1,1)$, find the range of values for $t$.


15 According to the definition,
$$
f(x)=x^{2}(1-x)+t(x+1)=-x^{3}+x^{2}+t x+t,
$$

then $f^{\prime}(x)=-3 x^{2}+2 x+t$. If $f(x)$ is a monotonically increasing function on the interval $(-1,1)$, then $f^{\prime}(x) \geqslant 0$ on $(-1,1)$. That is, $-3 x^{2}+2 x+t \geqslant 0$ holds for all $x \in(-1,1)$, which means $t \geqslant 3 x^{2}-2 x$ holds 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 with the axis of symmetry at $x=\frac{1}{3}$ and opens upwards, to ensure that $t \geqslant 3 x^{2}-2 x$ holds for all $x \in(-1,1)$, it is necessary that $t \geqslant g(-1)=5$.

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