4. [4] Find the real number $\alpha$ such that the curve $f(x)=e^{x}$ is tangent to the curve $g(x)=\alpha x^{2}$.
Answer: $\mathbf{e}^{2} / \mathbf{4}$. Suppose tangency occurs at $x=x_{0}$. Then $e^{x_{0}}=\alpha x_{0}^{2}$ and $f^{\prime}\left(x_{0}\right)=2 \alpha x_{0}$. On the other hand, $f^{\prime}(x)=f(x)$, so $\alpha x_{0}^{2}=2 \alpha x_{0}$. Clearly, $\alpha=0$ and $x_{0}=0$ are impossible, so it must be that $x_{0}=2$. Then $\alpha=e^{x_{0}} /\left(x_{0}^{2}\right)=e^{2} / 4$