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Subsection 2.13.1 Rolle's Theorem

Again, like the two scenarios above, this theorem says something intuitively obvious. Consider — if you throw a ball straight up into the air and then catch it, at some time in between the throw and the catch it must be stationary. Translating this into mathematical statements, let \(s(t)\) be the height of the ball above the ground in metres, and let \(t\) be time from the moment the ball is thrown in seconds. Then we have

\begin{align*} s(0) &= 1 & \text{we release the ball at about hip-height}\\ s(4) &= 1 & \text{we catch the ball $4s$ later at hip-height} \end{align*}

Then we know there is some time in between — say at \(t=c\) — when the ball is stationary (in this case when the ball is at the top of its trajectory). I.e.

\begin{align*} v(c) = s'(c) &= 0. \end{align*}

Rolle's theorem guarantees that for any differentiable function that starts and ends at the same value, there will always be at least one point between the start and finish where the derivative is zero.

There can, of course, also be multiple points at which the derivative is zero — but there must always be at least one. Notice, however, the theorem  5 Notice this is very similar to the intermediate value theorem (see Theorem 1.6.12)

does not tell us the value of \(c\text{,}\) just that such a \(c\) must exist.

We can use Rolle's theorem to show that the function

\begin{align*} f(x) &= \sin(x)-\cos(x) \end{align*}

has a point \(c\) between \(0\) and \(\frac{3\pi}{2}\) so that \(f'(c)=0\text{.}\)

To apply Rolle's theorem we first have to show the function satisfies the conditions of the theorem on the interval \([0,\frac{3\pi}{2}]\text{.}\)

  • Since \(f\) is the sum of sine and cosine it is continuous on the interval and also differentiable on the interval.
  • Further, since
    \begin{align*} f(0) &= \sin 0 - \cos 0 = 0-1 = -1\\ f\left(\frac{3\pi}{2}\right) &= \sin\frac{3\pi}{2} - \cos\frac{3\pi}{2} = -1-0 = -1 \end{align*}
    we can now apply Rolle's theorem.
  • Rolle's theorem implies that there must be a point \(c \in (0,3\pi/2)\) so that \(f'(c) =0\text{.}\)

While Rolle's theorem doesn't tell us the value of \(c\text{,}\) this example is sufficiently simple that we can find it directly.

\begin{align*} f'(x) &= \cos x + \sin x\\ f'(c) &= \cos c + \sin c = 0 & \text{rearrange}\\ \sin c&= - \cos c & \text{and divide by $\cos c$}\\ \tan c &= -1 \end{align*}

Hence \(c = \frac{3\pi}{4}\text{.}\) We have sketched the function and the relevant points below.

A more substantial application of Rolle's theorem (in conjunction with the intermediate value theorem — Theorem 1.6.12) is to show that a function does not have multiple zeros in an interval:

Show that the equation \(2x-1=\sin(x)\) has exactly 1 solution.

  • Start with a rough sketch of each side of the equation

    This seems like it should be true.

  • Notice that the problem we are trying to solve is equivalent to showing that the function
    \begin{align*} f(x) &= 2x-1-\sin(x) \end{align*}
    has only a single zero.
  • Since \(f(x)\) is the sum of a polynomial and a sine function, it is continuous and differentiable everywhere. Thus we can apply both the IVT and Rolle's theorem.
  • Notice that \(f(0)=-1\) and \(f(2) = 4-1-\sin(2) = 3-\sin(2) \geq 2\text{,}\) since \(-1\leq \sin(2) \leq 1\text{.}\) Thus by the IVT we know there is at least one number \(c\) between \(0\) and \(2\) so that \(f(c)=0\text{.}\)
  • But our job is only half done — this shows that there is at least one zero, but it does not tell us there is no more than one. We have more work to do, and Rolle's theorem is the tool we need.

  • Consider what would happen if \(f(x)\) is zero in 2 places — that is, there are numbers \(a,b\) so that \(f(a)=f(b)=0\text{.}\)

    • Since \(f(x)\) is differentiable everywhere and \(f(a)=f(b)=0\text{,}\) we can apply Rolle's theorem.
    • Hence we know there is a point \(c\) between \(a\) and \(b\) so that \(f'(c)=0\text{.}\)
    • But let us examine \(f'(x)\text{:}\)
      \begin{align*} f'(x) &= 2- \cos x \end{align*}
      Since \(-1\leq \cos x \leq 1\text{,}\) we must have that \(f'(x) \geq 1\text{.}\)
    • But this contradicts Rolle's theorem which tells us there must be a point at which the derivative is zero.

    Thus the function cannot be zero at two different places — otherwise we'd have a contradiction.

We can actually nail down the value of \(c\) using the bisection approach we used in example 1.6.15. If we do this carefully we find that \(c \approx 0.887862\dots\)

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