637. Investigate the convexity, concavity, and points of inflection of the curve $y=x^{3}-6 x^{2}+4$.
Solution. $1^{0} . y^{\prime}=3 x^{2}-12 x ; y^{\prime \prime}=6 x-12$.

$2^{0}$. From the equation $6 x-12=0$ we find $x=2$.

$3^{0}$. If $x<2$, then $y^{\prime \prime}<0$; hence, in the interval $(-\infty, 2)$ the curve is convex. If $x>2$, then $y^{\prime \prime}>0$; hence, in the interval $(2, \infty)$ the curve is concave.

Thus, as we pass through $x=2$, the second derivative $y^{\prime \prime}(x)$ changes sign from negative to positive. Therefore, the curve has an inflection point, the abscissa of which is 2.

$4^{\circ}$. Substituting $x=2$ into the equation of the curve, we find the ordinate of the inflection point: $y=2^{3}-6 \cdot 2^{2}+4=-12$. Thus, $(2 ;-12)$ is the inflection point.