= Solution
The <intermediate value theorem> states that if $f:[a,b]\to\mathbb R$ is <continuous function>[continuous] and $y$ lies between $f(a)$ and $f(b)$, then some $c\in[a,b]$ satisfies $f(c)=y$.
Assume $f(a)<y<f(b)$ and let
$$
S=\{x\in[a,b]:f(x)\leq y\},
\qquad c=\sup S.
$$
The <least-upper-bound property> makes $c$ well defined. If $f(c)<y$, continuity gives points immediately to the right of $c$ in $S$, contradicting that $c$ is an upper bound. If $f(c)>y$, continuity gives a left neighborhood of $c$ disjoint from $S$, contradicting the definition of the <supremum>. Thus $f(c)=y$. The endpoint and reversed-order cases follow directly or by replacing $f$ with $-f$.
The <mean value theorem> states that if $g$ is continuous on $[\alpha,\beta]$ and differentiable on $(\alpha,\beta)$, then some $c\in(\alpha,\beta)$ satisfies
$$
g'(c)=\frac{g(\beta)-g(\alpha)}{\beta-\alpha}.
$$
For the functions in the question, the definition of the <derivative> makes $h$ continuous at $a$ and $f$ continuous at $b$, with
$$
h(a)=g'(a)<k,
\qquad
h(b)=\frac{g(b)-g(a)}{b-a}=f(a),
\qquad
f(b)=g'(b)>k.
$$
If the middle secant slope equals $k$, take $[\alpha,\beta]=[a,b]$. If it exceeds $k$, the <intermediate value theorem> applied to $h$ gives $\beta\in(a,b)$ with $h(\beta)=k$, and we take $\alpha=a$. If it is below $k$, apply the theorem to $f$ to obtain $\alpha\in(a,b)$ with $f(\alpha)=k$, and take $\beta=b$. In every case
$$
\frac{g(\beta)-g(\alpha)}{\beta-\alpha}=k.
$$
The <mean value theorem> on this subinterval then produces $c\in(\alpha,\beta)\subset(a,b)$ with
$$
\boxed{g'(c)=k}.
$$
This is the <Darboux theorem for derivatives> for the stated pair of derivative values.
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