## Task A-2.3.

Two quadratic functions $f_{1}(x)$ and $f_{2}(x)$ are given.

Function $f_{1}(x)$ achieves its minimum value at $x=-1$, and one of its roots is $x=3$. Function $f_{2}(x)$ achieves its maximum value at $x=3$, and one of its roots is $x=-1$.

Determine all values of $x$ for which the product $f_{1}(x) f_{2}(x)$ achieves its maximum value.
## Solution.

Since the zeros of a quadratic function are symmetric with respect to the vertex, the second zero of the function $f_{1}(x)$ is equal to $-1-(3-(-1))=-5$. Similarly, the second zero of the function $f_{2}(x)$ is equal to $3+(3-(-1))=7$.

Therefore, the functions $f_{1}(x)$ and $f_{2}(x)$ have the form

$$
f_{1}(x)=a_{1}(x-3)(x+5), \quad f_{2}(x)=a_{2}(x+1)(x-7)
$$

for some real numbers $a_{1}$ and $a_{2}$ different from zero.

Since the function $f_{1}(x)$ achieves a minimum, and $f_{2}(x)$ a maximum, it must be that $a_{1}>0$ and $a_{2}<0$.

Now we have:

$$
\begin{aligned}
f_{1}(x) f_{2}(x) & =a_{1}(x-3)(x+5) \cdot a_{2}(x+1)(x-7) & & 1 \text { point } \\
& =a_{1} a_{2}((x-3)(x+1))((x+5)(x-7)) & & \\
& =a_{1} a_{2}\left(x^{2}-2 x-3\right)\left(x^{2}-2 x-35\right) . & & 1 \text { point }
\end{aligned}
$$

Let $t=x^{2}-2 x-3$. Then the product of the functions $f_{1}(x)$ and $f_{2}(x)$ is equal to $a_{1} a_{2} t(t-32)$. The vertex is located at $t=16$. Also, since $a_{1} a_{2}<0$, the function $g(t)$ achieves a maximum at the vertex.

Therefore, the desired maximum value is achieved for $t=16$, i.e., when

$$
x^{2}-2 x-3=16
$$

Thus, the points at which the maximum of the product of the functions $f_{1}(x)$ and $f_{2}(x)$ is achieved are the solutions of the quadratic equation

$$
x^{2}-2 x-19=0
$$

i.e., $x_{1}=1+2 \sqrt{5}$ and $x_{2}=1-2 \sqrt{5}$.

Note: The problem can also be solved by introducing similar substitutions for $t$ (e.g., $t_{1}=$ $x^{2}-2 x, t_{2}=x^{2}-2 x-19, t_{3}=x^{2}-2 x-35$). Such solutions should be graded in an analogous manner to the official solution.