# Rolle's Theorem

**Rolle's theorem** is one of the foundational theorems in differential calculus. It is a special case of, and in fact is equivalent to, the mean value theorem, which in turn is an essential ingredient in the proof of the fundamental theorem of calculus.

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## Summary

The theorem states as follows:

Rolle's TheoremFor any function $f(x)$ that is continuous within the interval $[a,b]$ and differentiable within the interval $(a,b),$ where $f(a)=f(b),$ there exists at least one point $\big(c,f(c)\big)$ where $f'(c)=0$ within the interval $(a,b).$

A graphical demonstration of this will help our understanding; actually, you'll feel that it's very apparent:

In the figure above, we can set any two points as $\big(a,f(a)\big)$ and $\big(b,f(b)\big)$ as long as $f(a)=f(b)$ and the function is differentiable within the interval $(a,b).$ Then, of course, there has to be a point in between where $f'=0,$ which is the red point in the diagram. Now let's take a look at the mathematical proof of this theorem.

We divide it into two cases:

(1)$f(x)$ is a constant function.If $f(x)$ is a constant function, then $f'=0$ for the whole interval. Then, of course, there exists a $c$ such that $f'(c)=0$ within the interval $(a,b).$

$$

(2)$f(x)$ is not a constant function.When $f(x)$ is not a constant function but is continuous within the interval $[a,b],$ according to the extreme value theorem, $f(x)$ must have a maximum function value and minimum function value within the interval $[a,b].$ Since $f(x)$ is not a constant function, at least one of the extrema must exist within the interval $(a,b).$

(2)-1

If $f(x)$ has its maximum function value $f(c)$ at $x=c\in(a,b),$ then for a real number $h$ whose absolute value is small enough that $a<c+h<b,$ it follows that$f(c+h)-f(c)\leq0.$

Hence we have

$\begin{aligned} \lim_{h \rightarrow 0^-} \frac{f(c+h)-f(c)}{h} \geq 0, \quad \lim_{h \rightarrow 0^+} \frac{f(c+h)-f(c)}{h} \leq 0. \end{aligned}$

Since $f(x)$ is differentiable in the interval $(a,b) ,$ according to the squeeze theorem we have

$\begin{aligned} 0 \leq \lim_{h \rightarrow 0^-} \frac{f(c+h)-f(c)}{h} = \lim_{h \rightarrow 0^+} \frac{f(c+h)-f(c)}{h} &\leq 0 \\ \Rightarrow f'(c) &= \lim_{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}\\&=0. \end{aligned}$

(2)-2

If $f(x)$ has its minimum function value $f(c)$ at $x=c\in(a,b),$ then for a real number $h$ whose absolute value is small enough that $a<c+h<b,$ it follows that$f(c+h)-f(c)\geq0.$

Hence we have

$\begin{aligned} \lim_{h \rightarrow 0^-} \frac{f(c+h)-f(c)}{h} \leq 0, \quad \lim_{h \rightarrow 0^+} \frac{f(c+h)-f(c)}{h} \geq 0. \end{aligned}$

Since $f(x)$ is differentiable in the interval $(a,b) ,$ according to the squeeze theorem we have

$\begin{aligned} 0 \leq\lim_{h \rightarrow 0^+} \frac{f(c+h)-f(c)}{h}=\lim_{h \rightarrow 0^-} \frac{f(c+h)-f(c)}{h} &\leq 0 \\ \Rightarrow f'(c) &= \lim_{h \rightarrow 0} \frac{f(c+h)-f(c)}{h}\\&=0. \end{aligned}$

Therefore, whichever case we are given, there exists a point where $f'=0$ within the interval $(a,b).\ _\square$

Obviously, for Rolle's theorem to hold, the function must be differentiable within the interval we are considering. Thus Rolle's theorem cannot be applied to functions like $y=\lvert x\rvert.$

## Example Problems

When

$f(x) = x^2-2x+1 ,$

show that $f'(x)=0$ has at least one root in the interval $0< x < 2$ using Rolle's theorem.

Observe that $f(x) = x^2-2x+1$ is continuous in the interval $[0,2]$ and differentiable in $(0,2) . \qquad (1)$

The function values of $f(x)$ at $x=0,2$ are

$\begin{aligned} f(0) &= 1\\ f(2) &= 2^2-2\cdot 2 +1 \\&= 1\\ \Rightarrow f(0)&=f(2)=1. \qquad \qquad (2) \end{aligned}$

Then from $(1)$ and $(2)$, it is confirmed that Rolle's theorem can be applied. According to Rolle's theorem, there exists a point where $f'(x)=0$ in the interval $(0 , 2). \ _\square$

When

$f(x) = \cos x +2 ,$

show that $f'(x)=0$ has at least one root in the interval $-\pi< x < \pi$ using Rolle's theorem.

Observe that $f(x) = \cos x +2$ is continuous in the interval $[-\pi,\pi]$ and differentiable in $(-\pi,\pi) . \qquad (1)$

The function values of $f(x)$ at $x=-\pi,\pi$ are

$\begin{aligned} f(-\pi) &= \cos(-\pi)+2 \\&=1, \\ f(\pi) &= \cos\pi+2 \\&= 1\\ \Rightarrow f(-\pi)&=f(\pi)=1. \qquad \qquad (2) \end{aligned}$

Then from $(1)$ and $(2)$, it is confirmed that Rolle's theorem can be applied. According to Rolle's theorem, there exists a point where $f'(x)=0$ in the interval $(-\pi , \pi). \ _\square$

Show that the following formula has at least one root in the interval $0< x < 1:$ $4x^3 + 3x^2 + 2x - 3 = 0 .$

Let $f'(x) = 4x^3 + 3x^2 +2x - 3 .$ Then

$\begin{aligned} f(x) &= x^4 + x^3 + x^2 -3x + C \\ \Rightarrow f(0) &= C \\ f(1) &= 1+1+1-3+C = C, \end{aligned}$

where $C$ is the constant of integration. Since $f(x)$ is continuous in the interval $[0,1]$ and differentiable in the interval $(0,1),$ and $f(0) = f(1) = C,$ we can apply Rolle's theorem. According to Rolle's theorem, there must be a point where $f'=0$ within the interval $(0,1).$

Therefore the given equation has at least one root in the interval $0<x<1.$ $_\square$

Show that the following formula has at least one root in the interval $0< x < 1:$

$\tan x+x-1=0 .$

Since $\cos x$ is never equal to zero within the given interval, the given equation is equivalent to

$\begin{aligned} \tan x +x -1 &=0 \\ \frac{\sin x}{\cos x} + x -1 &=0 \\ \sin x + (x-1)\cos x &= 0. \end{aligned}$

Let $f'(x) = \sin x + (x-1)\cos x .$ Then

$\begin{aligned} f(x) &= \int f'(x)\, dx \\ &= \int (\sin x + (x-1) \cos x)\, dx \\ &= (x-1)\sin x +C \\ \Rightarrow f(0) &= C \\ f(1) &= C, \end{aligned}$

where $C$ is the constant of integration. Since $f(x)$ is continuous in the interval $[0,1]$ and differentiable in the interval $(0,1),$ and $f(0) = f(1) = C,$ we can apply Rolle's theorem. According to Rolle's theorem, there must be a point where $f'=0$ within the interval $(0 , 1).$

Therefore the given equation has at least one root in the interval $0<x<1.$ $_\square$