## Section 2.13 The Mean Value Theorem

¶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 }Unfortunately there are many obvious things that are decidedly false — for example “There are more rational numbers than integers.” or “Viking helmets had horns on them”. 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 }Recall that speed and velocity are not the same. Velocity specifies the direction of motion as well as the rate of change. Objects moving along a straight line have velocities that are positive or negative numbers indicating which direction the object is moving along the line. Speed, on the other hand, is the distance travelled per unit time and is always a non-negative number — it is the absolute value of velocity. .

Let us turn this around a little bit. Consider the premise of a 90s action film ^{ 3 }The sequel won a Raspberry award for “Worst remake or sequel”. — 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 let's 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 }There must be at least one such point — there could be more than one — but there cannot be zero. (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 \(f(a)=f(b)\text{.}\)