Rolle's Theorem

Rolle's Theorem is an important theorem that proves the existence of a stationary point between two points on a curve with the same value. Specifically, it states the following:

Rolle's Theorem: Suppose the function $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$. If $f(a)=f(b)$, then there exists a constant $c$ with $a<c<b$ such that $f'(c)=0$.

For example, let's look at the semicircle below, described by the equation $y=\sqrt{25-x^2}$:

Since the curve is continuous on $[-5,5]$ and differentiable on $(-5,5)$, Rolle's Theorem guarantees us a point $c$ with $a<c<b$ such that $f'(c)=0$. Indeed, we see from the picture that $c=0$ works. We can also figure this out algebraically, using Rolle's Theorem: since $$f'(x)=\frac{d}{dx}(\sqrt{25-x^2})=\frac{1}{2}(25-x^2)^{-\frac{1}{2}}\cdot(-2x)=\frac{-x}{\sqrt{25-x^2}},$$ solving the equation $f'(c)=0$ yields $$f'(c)=\frac{-c}{\sqrt{25-c^2}}=0\Longrightarrow c=0,$$ as desired.

Now you try!

Let's first graph the function to get a better idea:

Since $f'(x)=3x^2-1$, we can solve the equation $f'(c)=0$ as follows: $$f'(c)=3c^2-1=0\Longrightarrow c=\pm\frac{\sqrt{3}}{3}.$$ Since both values are in the interval $[-1,1]$, both values of $c$ satisfy Rolle's Theorem for this interval. This example shows you that there can be more than one value of $c$ that works.

Great, so now we know how to use this theorem for calculations. But can we use this theorem for anything else? Try to see if you can prove the following proposition:

Proposition: Suppose $f(x)$ is continuous and differentiable, and further suppose that it has two distinct zeroes. Prove that $f'(x)$ also has at least one zero.

To do so, let's call the two zeroes $a$ and $b$. By definition of zeroes, we have $$f(a)=0=f(b).$$ Since $f(x)$ is continuous on $[a,b]$ and differentiable in $(a,b)$, we know by Rolle's Theorem that there exists a $c$ with $a<c<b$ such that $f'(c)=0$. So, $c$ is a zero of $f'(x)$, which proves the proposition.

Now, one natural thing to ask would be whether we can generalize this idea: what if $f(a)\ne f(b)$? Can we find a similar case? You'll find out in the next section!

Suppose $f(x)$ is continuous on $[a,b]$ and differentiable on $(a,b)$, and further suppose $f(a)=f(b)$. By the Extreme Value Theorem, the function achieves both a minimum and a maximum on $[a,b)$. There are two possible cases:

1. Suppose both the maximum and minimum occur at endpoints. Since $f(a)=f(b)$, this means that the $f(x)\not < f(a)$ and $f(x) \not > f(a)$ for all $x$ in $(a,b)$. This means that $f(x)=f(a)$, meaning that the entire function is constant in this interval. So, it follows that $f'(x)=0$ at any value of $x$ in $(a,b)$.

2. Now suppose that either the minimum or the maximum (or both) occur in $(a,b)$. Then we can assume that one of these stationary points occur at a point $c$ with $a<c<b$. Since $f(x)$ is differentiable in $(a,b)$, we can conclude that $f'(c)=0$.

Therefore, in both cases we have shown that Rolle's Theorem holds.

Here is a worksheet containing problems asking you to find values of $c$ that satisfy Rolle's Theorem. These are like the example you encountered in Steven Kwon's explanation right above. Answers are at the bottom (but solve the problems first :P ). Good luck!