Generally speaking, yes, your intuition is in fact the standard way in which, for instance, I explain the Mean Value Theorem: if you think of $f(t)$ as the position function, and the slope of the line from $(a,f(a))$ to $(b,f(b))$ is the average velocity.
If the average velocity is $m$ units of distance per units of time, then we expect there to be some instant at which we were going at exactly $m$ units of distance per units of time: intuitively, we can't always be going slower than the average, nor always faster than the average, so "sometime between an instant where we are going slower than the average and an instant where we are going faster than the average, we should be going at exactly the average velocity."
Of course, getting all the details nailed down is somewhat complicated, and you have to take into account functions whose derivatives may not be continuous (for continuous derivatives, you can invoke the Intermediate Value Theorem). So, after a first flush of being sure this will definitely be true, it is suddenly not entirely so obvious that this will work for all functions that are differentiable everywhere on $(a,b)$ (and continuous on $[a,b]$), so there actually is something to the Mean Value Theorem.
Also related is Darboux's Theorem, which is the one that tells you that the property we want always holds, even if the derivative is not continuous: if $f(x)$ is differentiable on $(a,b)$ and continuous on $[a,b]$, then $f'(x)$ has the intermediate value property: if $f'_{+}(a)$ is the right derivative at $a$ ($\lim\limits_{h\to 0^+}\frac{f(a+h)-f(a)}{h}$) and $f'_{-}(b)$ is the left derivative at $b$ ($\lim\limits_{h\to 0^-}\frac{f(b+h)-f(b)}{h}$), and $k$ is a number between $f'_{+}(a)$ and $f'_{-}(b)$, then there is a point $t$, $a\lt t\lt b$, such that $f'(t)=k$.
Note: Except some technicality issues the following example gives a good intuition behind the mean value theorems.
In the 2012 Olympics Usain Bolt won the 100 metres gold medal with a time of 9.63 seconds.
His average speed was total distance, $d(t_2)-d(t_1)$, over total time, $t_2-t_1$:
$$V_a=\dfrac{d(t_2)-d(t_1)}{t_2-t_1}=\dfrac{100}{9.63}=10.384 \ \text{m/s}=37.38 \ \text{km/h}.$$ Mean value theorem
$$
f'(c)=\dfrac{f(b)-f(a)}{b-a}
$$
says that at some point (which is $c$ seconds) Bolt was actually running at the average speed of $37.38$ km/h.
Powell Asafa was participating in that race also, with a time $11.99=1.245\times9.63$ seconds, so Bolt's average speed was $1.245$ times the average speed of Powell. Generalized mean value theorem: $$\dfrac{f'(c)}{g'(c)}=\dfrac{f(b)-f(a)}{g(b)-g(a)}=\dfrac{\frac{f(b)-f(a)}{b-a}}{\frac{g(b)-g(a)}{b-a}},$$ says that at some point Bolt was actually running at a speed exactly $1.245$ times of Powell's speed!
Best Answer
You could write that as
$$\exists c \in (a,b)\, :\, f'(c) = \frac{f(b)-f(a)}{b-a}$$
Some people might prefer to use other conventions, such as $$\exists c.\ c \in (a,b) \wedge f'(c) = \frac{f(b)-f(a)}{b-a}$$
Frankly, though, there's nothing wrong with words, especially in something like analysis. It doesn't make it any less rigorous as long as the language used isn't ambiguous.
Edit:
Amused by one of the comments, I thought I'd write out in full the statement of the mean value theorem using only symbols.
$(\forall a,b \in \mathbb{R})(\forall f \in \mathbb{R}^{[a,b]})(((\forall x \in [a,b])(\forall \epsilon > 0)(\exists \delta > 0)$ $(\left|x-y\right|<\delta \Rightarrow \left|f(x)-f(y)\right|<\varepsilon)) \wedge ((\forall x \in (a,b))(\exists f'(x) \in \mathbb{R})$ $(\forall \varepsilon > 0)(\exists \delta > 0)(\forall h>0)((h<\delta) \Rightarrow \left| \frac{f(x+h)-f(x)}{h} - f'(x)\right| < \varepsilon))$ $\Rightarrow (\exists c \in (a,b))(f'(c) = \frac{f(b)-f(a)}{b-a}))$