2. [3] Let $\ell$ be the line through $(0,0)$ and tangent to the curve $y=x^{3}+x+16$. Find the slope of $\ell$.
Answer: 13 Let the point of tangency be $\left(t, t^{3}+t+16\right)$, then the slope of $\ell$ is $\left(t^{3}+t+16\right) / t$. On the other hand, since $d y / d x=3 x^{2}+1$, the slope of $\ell$ is $3 t^{2}+1$. Therefore,
$$
\frac{t^{3}+t+16}{t}=3 t^{2}+1
$$

Simplifying, we get $t^{3}=8$, so $t=2$. It follows that the slope is $3(2)^{2}+1=13$.