Task 4. (20 points) It is known that the function $f(x)$ for each value of $x \in(-\infty ;+\infty)$ satisfies the equation $f(x)-(x-0.5) f(-x-1)=1$. Find all such functions $f(x)$.
Solution.

Substitute the argument $(-x-1)$ into the equation and write the system of equations:
$\left\{\begin{array}{l}f(x)-(x-0.5) f(-x-1)=1, \\ f(-x-1)-(-x-1-0.5) f(x+1-1)=1 ;\end{array}\left\{\begin{array}{l}f(x)-(x-0.5) f(-x-1)=1, \\ f(-x-1)+(x+1.5) f(x)=1 .\end{array}\right.\right.$

Solve the system by the method of elimination, multiplying the second equation by $(x-0.5)$ and adding the results:

$f(x)(1+(x-0.5)(x+1.5))=1+(x-0.5)$,

$f(x)=\frac{0.5+x}{x^{2}+x+0.25}=\frac{0.5+x}{(0.5+x)^{2}}=\frac{1}{0.5+x}, x \neq-0.5$.

If $x=-0.5$, then from the original equation we get

$f(-0.5)-(-0.5-0.5) f(0.5-1)=1$,

$f(-0.5)=0.5$.

Finally,

$f(x)=\left\{\begin{array}{l}\frac{1}{0.5+x}, x \neq-0.5, \\ 0.5, x=-0.5 .\end{array}\right.$

Answer. $f(x)=\left\{\begin{array}{c}\frac{1}{0.5+x}, x \neq-0.5, \\ 0.5, x=-0.5 .\end{array}\right.$