415. At what point does the curve $y=e^{x}$ have the greatest curvature?
Solution. We find the derivatives $y^{\prime}=y^{\prime \prime}=e^{x}$ and the curvature of the given curve at any point:

$$
K(x)=\frac{e^{x}}{\left(1+e^{2 x}\right)^{\frac{3}{3}}}
$$

Next, we find the maximum value of the function $K(x)$, which is defined and continuous over the entire number line:

$$
K^{\prime}(x)=\frac{e^{x}\left(1-2 e^{2 x}\right)}{\left(1+e^{2 x}\right)^{\frac{5}{2}}} ; \quad K^{\prime}(x)=0 \text { when } 1-2 e^{2 x}=0
$$

i.e., at the unique point $x_{0}=-\frac{\ln 2}{2}$. Determining the signs of $K^{\prime}(x)$ to the left and right of this critical point: $K^{\prime}(-10)>0, K^{\prime}(0)<0$, we establish that it is a point of maximum of the function $K(x)$. Since $x_{0}$ is the only extremum point of the continuous function $K(x)$ over the entire interval $(-\infty,+\infty)$, it achieves its maximum value at this point. Therefore, the desired point is $\left(-\frac{\ln 2}{2} ; \frac{\sqrt{2}}{2}\right)$. (The ordinate of this point is calculated from the given equation of the curve using the known abscissa.)