= Solution
{c}
We prove the <symmetric derivative criterion for monotonicity> using an <antiderivative>. Fix $a<u<v<b$ and define
$$
F(t)=\int_u^t f(s)\,ds.
$$
The <fundamental theorem of calculus> gives $F'=f$ because $f$ is <continuous>. For each interior $x$,
$$
F(x+h)-2F(x)+F(x-h)=\int_0^h\bigl(f(x+t)-f(x-t)\bigr)\,dt.
$$
Since $f^s(x)>0$, the definition of <symmetric derivative> gives $\delta>0$ such that $f(x+t)-f(x-t)>t f^s(x)$ for $0<t<\delta$. Therefore, for $0<h<\delta$,
$$
F(x+h)-2F(x)+F(x-h)>\frac12 f^s(x)h^2>0.
$$
We next show that $F$ is a <convex function>. On any closed <interval> $[r,s]\subset(a,b)$ let $\ell$ be the <affine function> joining $(r,F(r))$ to $(s,F(s))$. If $F-\ell$ were positive anywhere, the <extreme value theorem> would give a positive <maximum> at an interior point $x$. For every sufficiently small $h>0$, that <maximum> would imply
$$
(F-\ell)(x+h)-2(F-\ell)(x)+(F-\ell)(x-h)\leq0.
$$
The <affine function> contributes zero to this symmetric difference, contradicting the strict positive <inequality> above. Thus $F\leq\ell$ on every such <interval>, exactly <convexity>.
For a <differentiable> <convex function>, secant slopes bound endpoint <derivatives>. Explicitly, the chord <inequality> at $u+h$ and at $v-h$, followed by $h\downarrow0$, gives
$$
f(u)=F'(u)\leq\frac{F(v)-F(u)}{v-u}\leq F'(v)=f(v).
$$
So $f$ is <nondecreasing>. If $f(u)=f(v)$ for some $u<v$, <monotonicity> would make $f$ constant on $[u,v]$, forcing $f^s(x)=0$ for $u<x<v$, a contradiction. Consequently $\boxed{f(u)<f(v)\text{ for every }u<v}$: $f$ is <strictly increasing>. This proof requires no <continuity> of $f^s$ and never applies the ordinary <mean value theorem> to a merely symmetrically <differentiable> <function>.
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