638. Find the points of inflection of the curve $y=\frac{8}{x^{2}+4}$.
Solution. $1^{0} . y^{\prime}=-\frac{16 x}{\left(x^{2}+4\right)^{2}} ; y^{\prime \prime}=-\frac{16\left(x^{2}+4\right)^{2}-2\left(x^{2}+4\right) 2 x \cdot 16 x}{\left(x^{2}+4\right)^{4}}=$ $=\frac{16\left(x^{2}+4\right)-64 x^{2}}{\left(x^{2}+4\right)^{3}}=\frac{48 x^{2}-64}{\left(x^{2}+4\right)^{3}}$.

$2^{0} . \frac{48 x^{2}-64}{\left(x^{2}+4\right)^{3}}=0$, from this $48 x^{2}-64=0, x^{2}=\frac{4}{3}, x_{1,2}= \pm \frac{2}{\sqrt{3}}, x_{1}=$ $=\frac{2}{\sqrt{3}}, x_{2}=-\frac{2}{\sqrt{3}}$.

$3^{0}$. Investigate the value $x_{1}=\frac{2}{\sqrt{3}}$. If $x<\frac{2}{\sqrt{3}}$, for example $x=1$, then $y^{\prime \prime}=\frac{48-64}{5^{3}}<0$; if $x>\frac{2}{\sqrt{3}}$, for example $x=2$, then $y^{\prime \prime}=\frac{48.4-64}{8^{3}}>0$. Thus, at $x_{1}=\frac{2}{\sqrt{3}}$ we have a point of inflection.

Investigate the value $x_{2}=-\frac{2}{\sqrt{3}}$. If $x<-\frac{2}{\sqrt{3}}$, for example $x=-2$, then $y^{\prime \prime}=\frac{48.4-64}{8^{3}}>0$; if $x>-\frac{2}{\sqrt{3}}$, for example $x=0$, then $y^{\prime \prime}=-1<0$. Thus, at $x=-\frac{2}{\sqrt{3}}$ we also have a point of inflection.

$4^{0}$. Find

$$
y_{1}=\frac{8}{(2 / \sqrt{3})^{2}+4}=\frac{3}{2}, \quad y_{2}=\frac{8}{(-2 / \sqrt{3})^{2}+4}=\frac{3}{2}
$$

Therefore, $(2 / \sqrt{3} ; 3 / 2)$ and $(-2 / \sqrt{3} ; 3 / 2)$ are points of inflection.