= Solution
{c}
Write the symmetric <difference quotient> as the average of two ordinary <difference quotients>:
$$
\frac{f(x+h)-f(x-h)}{2h}
=\frac12\left(\frac{f(x+h)-f(x)}h+\frac{f(x-h)-f(x)}{-h}\right).
$$
If $f$ is <differentiable> at $x$, both terms tend to the <derivative> $f'(x)$. Hence the <symmetric derivative> satisfies $\boxed{f^s(x)=f'(x)}$.
The <absolute value function> $f(x)=|x|$ is <continuous>, and $f^s(0)=0$ because $|h|=|-h|$. Its right <one-sided derivative> at zero is $1$ and its left <one-sided derivative> is $-1$, so it is not <differentiable> at zero. Away from zero it is <differentiable>, hence has a <symmetric derivative> there as well. This supplies an example meeting the hypotheses on, for example, $(-1,1)$.
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