Consider the following situation. Two towns are separated by a 120km long stretch of road. The police in town $A$ observe a car leaving at 1pm. Their colleagues in town $B$ see the car arriving at 2pm. After a quick phone call between the two police stations, the driver is issued a fine for going $120km/h$ at some time between 1pm and 2pm. It is intuitively obvious  1  that, because his average velocity was $120km/h\text{,}$ the driver must have been going at least $120km/h$ at some point. From a knowledge of the average velocity of the car, we are able to deduce something about an instantaneous velocity  2  .
Let us turn this around a little bit. Consider the premise of a 90s action film  3  — a bus must travel at a velocity of no less than $80km/h\text{.}$ Being a bus, it is unable to go faster than, say, $120km/h\text{.}$ The film runs for about 2 hours, and lets assume that there is about thirty minutes of non-action — so the bus' velocity is constrained between $80$ and $120km/h$ for a total of $1.5$ hours.
It is again obvious that the bus must have travelled between $80 \times 1.5 = 120$ and $120\times 1.5 = 180km$ during the film. This time, from a knowledge of the instantaneous rate of change of position — the derivative — throughout a 90 minute time interval, we are able to say something about the net change of position during the 90 minutes.
In both of these scenarios we are making use of a piece of mathematics called the Mean Value Theorem. It says that, under appropriate hypotheses, the average rate of change $\frac{f(b)-f(a)}{b-a}$ of a function over an interval is achieved exactly by the instantaneous rate of change $f'(c)$ of the function at some  4
(unknown) point $a\le c\le b\text{.}$ We shall get to a precise statement in Theorem 2.13.5. We start working up to it by first considering the special case in which $f(a)=f(b)\text{.}$