The theorem in question (Darboux's theorem) basically states that the conclusion of the intermediate value theorem holds for the derivative of an everywhere differentiable function, even if the derivative is discontinuous.
For a simple discontinuity of either kind, it is true that either $f'(x-) \ne f'(x)$ or $f'(x+) \ne f'(x)$.
To be specific, let us treat the case where $f'(x)<f'(x+)$.
Let $\lambda\in(f'(x),f'(x+))$, and pick $y>x$ with $f'(z)>\lambda$ whenever $z\in(x,y]$. Thus $f'(x)<\lambda<f'(x+)$, and $f'(z)\ne\lambda$ for all $z\in[x,y]$, contradicting Darboux's theorem.
The case $f'(x)>f'(x+)$ is treated similarly (or replace $f$ by $-f$ and use the case already treated). The case $f'(x-) \ne f'(x)$ is also treated the same way (or replace $f(x)$ by $f(-x)$).
Edited to fix a flaw pointed out in the comments, no less than seven years later!
One keeps all definitions as in the question.
Lemma. Let $I$ be an interval either of the form $(\alpha ,0]$ or $[0,\beta)$. Let $J:=I\setminus \{0\}$. The $f'(J)$ contains the interval $(-1,1)$.
Proof. It is clear that $$\limsup_{x\rightarrow 0^+}f'(x)=1,\liminf_{x\rightarrow 0^+}f'(x)=-1,$$ and similarly
$$\limsup_{x\rightarrow 0^-}f'(x)=1,\liminf_{x\rightarrow 0^-}f'(x)=-1.$$
One prove the case when $J=(0,\beta)$ (the other case being similar). If $y_0\in (-1,1)$, then $y_0\in (-1+\epsilon,1-\epsilon)$ for som positive $\epsilon$. From the above observations, there exists $x_2\in J$ such that $f'(x_2)>1-\epsilon$. And then there exists $0<x_1<x_2$ such that $f'(x_1)<-1+\epsilon$. Now since $(x_1,x_2)\subseteq J$ and $f'$ is continuous there, there exists by IVP for $f'$ that $\exists x_0\in (x_1,x_2)$ with $f'(x_0)=y_0.$ This completes the proof.
Proposition. $f'$ satisfies the IVP, namely for $I=[a,b]$, $f'$ assumes any value $y$ between $f'(a)$ and $f'(b)$.
Proof. As the OP remarked, it suffices to prove the case when $0\in I$ (so $a\leq 0,b\geq 0$). Also the case when $f'(a)=f'(b)$ being void, one assumes that $f'(a)\neq f'(b).$ Let $y$ be any value between $f'(a)$ and $f'(b)$. Now one may divide into two cases.
Case 1. $|y|<1.$
This follows from the lemma above.
Case 2. There are $4$ subcases (possibly overlapping) as follows (where one applies the lemma above to choose $x_0$):
Subcase 1. $1\leq y<f'(b)$: Take $0<x_0<b$ such that $f'(x_0)<1$ and apply IVP to $f'$ on $[x_0,b]$.
Subcase 2. $1\leq y<f'(a)$: Take $a<x_0<0$ such that $f'(x_0)<1$ and apply IVP to $f'$ on $[a,x_0]$.
Subcase 3. $f'(a) < y\leq -1$: Take $a<x_0<0$ such that $f'(x_0)>-1$ and apply IVP to $f'$ on $[a,x_0]$.
Subcase 4. $f'(b)<y\leq -1$: Take $0<x_0<b$ such that $f'(x_0)>-1$ and apply IVP to $f'$ on $[x_0,b]$.
Combining all cases, the proposition is proven.
Best Answer
A real valued function has the “intermediate value property” if
With that definition, your function in Figure 2 does not have the intermediate property, it fails e.g. for $(a, b) = (4.5, 5.5)$.
But the following is true:
If $f$ is a real-valued differentiable function on an interval then $f'$ has the intermediate value property. That is Darboux's theorem.
A function with the intermediate value property does not have simple discontinuities.
For the latter, assume that $f$ is discontinuous at $x=a$, and has both a right limit $r$ and a left limit $l$ at $x=a$. Then the three numbers $f(a), l, r$ are not all identical (otherwise $f$ were continuous at that point). In a sufficiently small interval $J = (a-\epsilon, a+\epsilon)$ takes $f$ only values close to $l$, $r$, and $f(a)$, so that $f(J)$ is not an interval. It follows that $f$ does not have the intermediate value property.