6.29. Find the angles at which the curves $f(x)=-x^{2}+4 x+4$ and $f(x)=x^{2}-2 x+4$ intersect.
Solution. Let's find the abscissa of the intersection point of the curves: $-x^{2}+4 x+4=x^{2}-2 x+4, x_{1}=0, x_{2}=3$. The tangent of the angle between the tangents at each point is determined by the slopes of the tangents:

$$
\begin{aligned}
\operatorname{tg} \alpha_{i}= & \frac{f_{1}^{\prime}\left(x_{i}\right)-f_{2}^{\prime}\left(x_{i}\right)}{1+f_{1}^{\prime}\left(x_{i}\right) \cdot f_{2}^{\prime}\left(x_{i}\right)}=\frac{\left(2 x_{i}-2\right)-\left(-2 x_{i}+4\right)}{1+\left(2 x_{i}-2\right)\left(-2 x_{i}+4\right)}= \\
& =\frac{4 x_{i}-6}{1+\left(2 x_{i}-2\right)\left(-2 x_{i}+4\right)}(i=1,2)
\end{aligned}
$$

We have $\operatorname{tg} \alpha_{1}=6 / 7, \alpha_{1}=\operatorname{arctg}(6 / 7), \operatorname{tg} \alpha_{2}=-6 / 7, \alpha_{2}=$ $=-\operatorname{arctg} 6 / 7$.

## CALCULATION OF THE FIRST DIFFERENTIAL