Example 1. Find the intervals of concavity and convexity, as well as the inflection points of the Gaussian curve $y=e^{-x^{2}}$.
Solution. We find the first and second derivatives:

$$
y^{\prime}=-2 x e^{-x^{2}} ; y^{\prime \prime}=\left(4 x^{2}-2\right) e^{-x^{2}}
$$

Setting the second derivative to zero, we obtain the critical points of the second kind:

$$
x_{1}=-\frac{1}{\sqrt{2}} ; x_{2}=\frac{1}{\sqrt{2}}
$$

These points divide the number line into three intervals:

$$
\left(-\infty,-\frac{1}{\sqrt{2}}\right),\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right),\left(\frac{1}{\sqrt{2}}, \infty\right)
$$

Since the second derivative is positive in the first and third intervals, i.e.,

$$
y^{\prime \prime}>0 \text { when }-\infty0 \text { when } x>\frac{1}{\sqrt{2}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_22_7ef2114a321cf223ba99g-128.jpg?height=288&width=503&top_left_y=415&top_left_x=42)

Fig. 3.13 then the curve is concave up in these intervals. Since $y^{\prime \prime}<0$ when $-\frac{1}{\sqrt{2}}<x<\frac{1}{\sqrt{2}}, \quad$ the curve is concave down in the interval $\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$ (Fig. 3.13). The points $M_{1}\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{e}}\right)$ and $M_{2}\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{e}}\right)$ are points of inflection.