# Section2.13The Mean Value Theorem¶ permalink

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 \(f(a)=f(b)\text{.}\)