= Solution
{c}
Fix $d$ strictly between $f'(a)$ and $f'(b)$, and set $F(x)=f(x)-dx$. The <function> $F$ is <continuous> on $[a,b]$: interior <differentiability> gives interior <continuity>, and the finite one-sided <derivatives> give <continuity> at the endpoints. The <extreme value theorem> therefore gives a point $c$ where $F$ attains its <minimum>.
Since $F'_+(a)=f'(a)-d<0$, for sufficiently small $h>0$ we have $F(a+h)<F(a)$, so $a$ cannot be a minimizing point. Since $F'_-(b)=f'(b)-d>0$, the quotient $(F(b-h)-F(b))/(-h)$ is positive for sufficiently small $h>0$, giving $F(b-h)<F(b)$. Thus $b$ cannot be a minimizing point either, and $c\in(a,b)$.
<Fermat's theorem on stationary points> states that a <differentiable function> at an interior <local minimum> or <local maximum> has <derivative> zero there. Applying it to $F$ gives
$$
\boxed{0=F'(c)=f'(c)-d,\qquad f'(c)=d.}
$$
This proves the relevant case of <Darboux's theorem>; the argument uses <continuity> of $f$, with no assumption of <continuity> of $f'$.
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