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.

The theorem states as follows:

Rolle's Theorem

For 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:

Imgur

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 $y=\lvert x\rvert.$

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.

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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.

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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 $0< x < 1:$ $4x^3 + 3x^2 + 2x - 3 = 0 .$

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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 $0< x < 1:$

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

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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 $0<x<1.$ $_\square$