## Task A-4.4.

Determine the set of all values that the function $f: \mathbb{R} \rightarrow \mathbb{R}$

$$
f(x)=\frac{2020 x}{x^{2}+x+1}
$$

achieves.
## Solution.

The set of all values that the function $f$ achieves is the set of all real numbers $a$ for which there exists a real number $x$ such that $f(x)=a$.

Therefore, we need to determine all real numbers $a$ for which the equation

$$
\frac{2020 x}{x^{2}+x+1}=a
$$

in the variable $x$ has at least one real solution.

Let's rewrite this equation as a quadratic equation

$$
a x^{2}+(a-2020) x+a=0
$$

Its discriminant is $D=(a-2020)^{2}-4 a^{2}$, and the equation has solutions if and only if $D \geqslant 0$.

The solutions to the quadratic equation $-3 a^{2}-4040 a+2020^{2}=(a-2020)^{2}-4 a^{2}=0$ are $a_{1}=-2020$ and $a_{2}=\frac{2020}{3}$.

Therefore, the set of all solutions to the inequality $(a-2020)^{2}-4 a^{2} \geqslant 0$, and thus the set of values that the function $f$ achieves, is

$$
\left[-2020, \frac{2020}{3}\right]
$$