
Problem 8.1. Let $f(x)$ be a linear function such that $f(0)=-5$ and $f(f(0))=-15$. Find all values of $m$, for which the set of the solutions of the inequality $f(x) \cdot f(m-x)>0$ is an interval of length 2.


Solution: Let $f(x)=a x+b$. It follows from $f(0)=-5$ that $b=-5$, and from $f(f(0))=-15$ that $a=2$. Therefore the function is $f(x)=$ $2 x-5$. Consider the inequality $(2 x-5)(2(m-x)-5)>0 \Longleftrightarrow$ $(2 x-5)(2 m-5-2 x)>0$. The solution of the last inequality is an interval with end points $\frac{5}{2}$ and $\frac{2 m-5}{2}$. Therefore $\left|\frac{5}{2}-\frac{2 m-5}{2}\right|=2$, so $|5-m|=2$. Finally, we obtain that $m=3$ and $m=7$.
