How to prove that derivatives have the Intermediate Value Property I'm reading a book which gives this theorem without proof:

If a and b are any two points in an interval on which ƒ is differentiable, then ƒ'
  takes on every value between ƒ'(a) and ƒ'(b).

As far as I can say, the theorem means that the fact ƒ' is the derivative of another function ƒ on [a, b] implies that ƒ' is continuous on [a, b].
Is my understanding correct? Is there a name of this theorem that I can use to find a proof of it?
 A: A proof without words:

The slope of the secant varies continuously from $f'(a)$ to $f'(b)$, so takes on every value in $[f'(a), f'(b)]$. By the mean value theorem, so does $f'$.
For the details, you can read the original proof by Lars Olsen that this animation is based on. Remarkably, this proof only seems to have been discovered in 2004. 
A: This is actually a nice exercise.  (In fact, if I recall correctly, it was given as a problem on the very first math exam I took in college.  Unfortunately all I was able to say was that it was true if $f'$ was assumed to be continuous, for which I received zero credit.)
Let me set it up a little bit and leave the rest to the interested readers: it is easy to reduce the general case to the following: suppose that $f'(a) > 0$ and $f'(b) < 0$.  Then there exists $c \in (a,b)$ with $f'(c) = 0$.
Here's the idea: an interior point with $f'(c) = 0$ is a stationary point of the curve (and conversely!).  In particular the derivative will be zero at any interior maximum or minimum of the curve.  Recall that since $f$ is differentiable, it is continuous and therefore assumes both a maximum and minimum value on $[a,b]$.  So we're set unless both the maximum and minimum are attained at the endpoints.  Perhaps the sign conditions of $f'$ at the endpoints have something to do with this...
A: The result is commonly known as Darboux’s theorem, and the Wikipedia article includes a proof.
A: Here's what I came up with off the top of my head: Let $c$ be between $f'(a)$ and $f'(b)$, and want to show there exists $x_0 \in (a,b)$ such that $f'(x_0) =c$.
Case 1:  $c \leq \frac{f(b)-f(a)}{b-a}$.
Then set 
$$
g(x) := \frac{f(x) - f(a)}{x-a}
$$
Then $g$ is clearly continuous for $x>a$, and $g(a) = f'(a)$, so $g$ is continuous at $a$ as well.  Since $g(a) = f'(a) <c$ and $g(b) \geq c$, by the intermediate value theorem there exists a point $x_1 \in (a,b]$ such that $c=g(x_1)$.  Then by the mean value theorem, there exists $x_0 \in (a,x_1]$ such that $f'(x_0) = c$.
Case 2: If $c > \frac{f(b)-f(a)}{b-a}$, then try a similar argument with
$$
h(x) := \frac{f(b) - f(x)}{b-x}
$$
