# Rolle's Theorem

**Rolle's theorem** is one of the fundamental rules that gives us insight about the graphs of differentiable functions and comes from the mean value theorem for derivatives.

#### Contents

## 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 \((c,f(c))\) where \(f'(c)=0 \) within the interval \((a,b).\) \(_\square\)

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 \((a,f(a))\) and \((b,f(b))\) 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

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

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

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

(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

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

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

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

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

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

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

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

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

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

\[ \begin{align} \tan x +x -1 &=0 \\ \frac{\sin x}{\cos x} + x -1 &=0 \\ \sin x + (x-1)\cos x &= 0. \end{align} \]

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

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

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 \(0<x<1.\) \(_\square \)