 Solution

If there were no interior $x_2$ with $f^s(x_2)\le 0$, the symmetric derivative would be positive everywhere in $(a,b)$. Part ii would make $f$ strictly increasing there. Choose $a<u<v<b$. Continuity at the endpoints and monotonicity on the interior implycontradicting $f(a)=f(b)$. Hence such an $x_2$ exists.

If there were no interior $x_1$ with $f^s(x_1)\ge 0$, then $(-f{)}^s=-f^s>0$ everywhere. Apply the same argument to the continuous function $-f$, whose endpoint values are equal. This is again impossible. ThusThe two points need not be distinct.