= Solution
{c}
If there were no interior $x_2$ with $f^s(x_2)\leq0$, the <symmetric derivative> would be positive everywhere in $(a,b)$. Part <4/b/ii> would make $f$ <strictly increasing> there. Choose $a<u<v<b$. <Continuity> at the endpoints and <monotonicity> on the interior imply
$$
f(a)\leq f(u)<f(v)\leq f(b),
$$
contradicting $f(a)=f(b)$. Hence such an $x_2$ exists.
If there were no interior $x_1$ with $f^s(x_1)\geq0$, 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. Thus
$$
\boxed{\exists x_1,x_2\in(a,b):\quad f^s(x_1)\geq0,\quad f^s(x_2)\leq0.}
$$
The two points need not be distinct.
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