The Mean Value Theorem

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Here we see a key theorem of calculus.

Here are some interesting questions involving derivatives:

(a): Suppose you toss a ball upward into the air and then catch it. Must the ball’s velocity have been zero at some point?
(b): Suppose you drive a car on a straight stretch of highway from toll booth to another toll booth $30$ miles away in half of an hour. Must you have been driving at $60$ miles per hour at some point?
(c): Suppose two different functions have the same derivative. What can you say about the relationship between the two functions?

While these problems sound very different, it turns out that the problems are very closely related. We’ll start simply:

Rolle’s Theorem Suppose that $f$ is differentiable on the interval $(a,b)$ , continuous on the interval $[a,b]$ , and that $f(a)=f(b)$ .

Then $f'(c)=0$ for some $c$ in the open interval $(a,b)$ .

If the function $f$ happens to be a constant, then $f'(c)=0$ for all points $c$ in the open interval $(a,b)$ .

If $f$ is not a constant, then it must attain both global extrema in the interval $[a,b]$ , by the Extreme Value Theorem. One of these two different extrema is attained in the open interval $(a,b)$ at some point $c$ , and the function $f$ has a critical point there. Since $f$ is differentiable on $(a,b)$ , it follows that $f'(c)=0$ .

We can now answer our first question above.

Suppose you toss a ball upward into the air and then catch it at the same point where you tossed it from. Must the ball’s velocity have been zero at some point?

Let $p(t)$ be the position of the ball (on the vertical line) at time $t$ . Our time interval in question will be $[t_\mathrm {start},t_\mathrm {finish}].$ We may assume that $p$ is continuous on $[t_\mathrm {start},t_\mathrm {finish}]$ and differentiable on $(t_\mathrm {start},t_\mathrm {finish})$ . Since $p(t_\mathrm {start})=p(t_\mathrm {finish})$ , we may now apply Rolle’s Theorem and conclude that at some time $c$ in the given time interval, $p'(c)=\answer [given]{0}$ . Hence the velocity must be zero at some point.

Rolle’s Theorem is a special case of a more general theorem.

Mean Value Theorem Suppose that $f$ has a derivative on the interval $(a,b)$ and is continuous on the interval $[a,b]$ .

Then $f'(c)=\frac {f(b)-f(a)}{b-a}$ for some $c$ in $(a,b)$ .

We can now answer our second question above.

Suppose you drive a car on a straight stretch of highway from toll booth to another toll booth $30$ miles away in half of an hour. Must you have been driving at $60$ miles per hour at some point?

Let $p(t)$ is the position of the car at time $t$ , and $0$ hours is the starting time with $1/2$ hours being the final time, where $p(0)=0$ , and $p(1/2)=30$ . We can assume that $p$ is continuous on $[0,1/2]$ and differentiable on $(0,1/2)$ . Now the Mean Value Theorem states that there is a time $c$ such that $p'(c) = \frac {30-0}{1/2-0} = \answer [given]{60}\qquad \text {where $0<c<1/2$.}$ Since the derivative of position is velocity, this says that the car must have been driving at $60$ miles per hour at some point.

Now we will address the unthinkable: could there be a continuous function $f$ on $[a,b]$ whose derivative is zero on $(a,b)$ that is not constant? As we will see, the answer is “no.”

If $f'(x)=0$ for all $x$ in an interval $I$ , then $f(x)$ is constant on $I$ .

Let $a< b$ be two points in $I$ . Since $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , by the Mean Value Theorem we know $\frac {f(b)-f(a)}{b-a} = f'(c)$ for some $c$ in the interval $(a,b)$ . Since $f'(c)=0$ we see that $f(b)=f(a)$ . Moreover, since $a$ and $b$ were arbitrarily chosen, $f(x)$ must be the constant function.

Now let’s answer our third question.

Suppose two different functions have the same derivative on some interval $I$ . What can you say about the relationship between the two functions?

Set $h(x) = f(x) - g(x)$ , so $h'(x) = f'(x) -g'(x)$ . Now $h'(x) = 0$ on the interval $I$ . This means that $h(x) = k$ , for all $x$ in $I$ , where $k$ is some constant. Hence, for all $x$ in $I$ $f(x) = g(x) + k.$

Describe all functions whose derivative is $\sin (x)$ .

One such function is $-\cos (x)$ , so all such functions have the form $-\cos (x)+k$ ,

Finally, let us investigate two young mathematicians who run to class.

Two students Devyn and Riley raced to class down a straight hall. They start at the same time and finish in a tie. Was there a point during the race that Devyn and Riley were running at exactly the same velocity?

Let $P_\mathrm {Devyn}$ represent Devyn’s position with respect to time, and let $P_\mathrm {Riley}$ represent Riley’s position with respect to time. Let $t_\mathrm {start}$ be the starting time of the race, and $t_\mathrm {finish}$ be the end of the race. Set $f(t) =P_\mathrm {Devyn}(t)-P_\mathrm {Riley}(t).$ Note, we may assume that $P_\mathrm {Devyn}$ and $P_\mathrm {Riley}(t)$ are continuous on $[t_\mathrm {start},t_\mathrm {finish}]$ and that they are differentiable on $(t_\mathrm {start},t_\mathrm {finish})$ . Hence the same is true for $f$ . Since both runners start and finish at the same place, $\begin{align*} f(t_\mathrm{start}) = P_\mathrm{Devyn}(t_\mathrm{start})-P_\mathrm{Riley}(t_\mathrm{start}) &= \answer[given]{0}\qquad\text{and}\\ f(t_\mathrm{finish})=P_\mathrm{Devyn}(t_\mathrm{finish})-P_\mathrm{Riley}(t_\mathrm{finish}) &= \answer[given]{0}.\\ \end{align*}$ In fact, this shows us that the average rate of change of $f(t) = P_\mathrm {Devyn}(t)-P_\mathrm {Riley}(t) \qquad \text {on}\qquad [t_\mathrm {start},t_\mathrm {finish}]$ is $\answer [given]{0}$ . Hence by the mean value theorem, there is a point $c$ $t_\mathrm {start}\le c \le t_{\mathrm {finish}}$ with $f'(c) = \answer [given]{0}$ . However, $\answer [given]{0} = f'(c) = P_\mathrm {Devyn}'(c)-P_\mathrm {Riley}'(c).$ Hence at $c$ , $P_\mathrm {Devyn}'(c)=P_\mathrm {Riley}'(c),$ this means that there was a time when they were running at exactly the same velocity.

In conclusion, the Mean Value Theorem relates the function $f$ and its derivative, $f'$ . Since the derivative has many interpretations, e.g. instantaneous rate of change, slope of the tangent line, velocity, it is no surprise that we can use the MVT in different contexts.

Therefore, if the the function $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$ , then the MVT guarantees that

(a): there is a point $c$ in $(a,b)$ where the instantaneous rate of change of $f$ is equal to the average rate of change of $f$ over the interval $[a,b]$ ;
or
(b): there is a point $c$ in $(a,b)$ where the slope of the tangent line to the curve $y=f(x)$ is equal to the slope of the secant line through the points $(a,f(a))$ and $(b,f(b))$ ;
or
(c): there is a point $c$ in the time interval $(a,b)$ where the instantaneous velocity is equal to the average velocity over the time interval $[a,b]$ . Here, we assume that $f$ is the position function for the object moving along a straight line.