630. Determine the intervals of convexity of the curve $y=x^{3}$.
Solution. $1^{0}$. Find the second derivative: $y^{\prime}=3 x^{2} ; y^{\prime \prime}=6 x$. It equals zero at the point $x=0$.

$2^{0}$. The point $x=0$ divides the domain of the function into intervals $(-\infty, 0)$ and $(0, \infty)$.

$3^{0}$. The condition for the convexity of the curve is $f^{\prime \prime}(x)<0$. Then $6 x<0$, $x<0$, i.e., the curve is convex in the interval $(-\infty, 0)$.