Task 2. (10 points) It is known that the function $f(x)$ for each value of $x \in(-\infty ;+\infty)$ satisfies the equation $f(x)+(0.5+x) f(1-x)=1$. Find all such functions $f(x)$.
# Solution.

Substitute the argument $(1-x)$ into the equation and write the system of equations:

$\left\{\begin{array}{l}f(x)+(0.5+x) f(1-x)=1, \\ f(1-x)+(0.5+1-x) f(1-1+x)=1 ;\end{array}\left\{\begin{array}{l}f(x)+(0.5+x) f(1-x)=1, \\ f(1-x)+(1.5-x) f(x)=1 .\end{array}\right.\right.$

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

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

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

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

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

$f(0.5)=0.5$.

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