Without loss of generality, assume $f(x_0) > g(x_0)$. Then, in some neighborhood of $x_0$, $f(x) > g(x)$, and thus $h(x) = f(x)$. Since $f(x)$ is differentiable at $x_0$, so is $h(x)$. This confirms option (c).
Now, consider the case when $f(x_0) = g(x_0)$. In
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