# Rolle’s Theorem: A Special Case Of Lagrange’s Mean Value Theorem

CONTENT:

Let’s try to follow this simple example in order to approach Rolle’s Theorem. If a car travels smoothly down a straight level road with an average velocity of 50 mi/hr, we would expect the speedometer reading to be exactly 50 mi/hr at least once during the trip. After all, if the car’s velocity were always above 50 mi/hr, the average velocity would also be above that level, and the same reasoning applies if the velocity were always below 50 mi/hr. More generally, if s(t) is the car’s position at a time ‘to during a trip over the time interval atb, then there should be a time t = c when the velocity s'(c) equals the average velocity between times t=a & t=b. That is, for some c with a<c<b,

$\dfrac{s\left( b\right) -s\left( a\right) }{b-a}=s’\left( c\right)$

Where expression on the right side is representing instantaneous velocity and the one on the left side is showing an average velocity.

This example illustrates a result called the “Mean Value Theorem” for derivatives (abbreviated MVT) that is fundamental in the study of calculus. Our proof of the MVT is based on the following special case, which is named for the French Mathematician, Michel Rolle (1652-1719), who gave a version of the result in an algebra text published in 1690.

The key to the proof of Mean Value Theorem is the following result, which is really, just the MVT in the special case where f(a) = f(b). In terms of our car example, Rolle’s theorem says that if a moving car begins and ends at the same place, then somewhere during this journey, it must reverse direction, since

$\dfrac{f\left( b\right) – f\left( a\right) }{b-a}=0$

for f(a) = f(b) .

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## Rolle’s theorem statement: What are its three conditions?

Let a function ‘f’ be;

i) Continuous ona closed interval [a,b]

ii) Differentiable on the open interval ]a, b[

iii) f(a) = f(b)

then there exists at least one point ‘c’ ϵ ]a,b[ such that f'(c) = 0.

## Rolle’s Theorem Proof

If f(x) =0 for all x between ‘a’ and ‘b’, then f'(x)=0 for all x between ‘a’ and ‘b’ (the derivative of a constant function is zero) and the theorem is true.

But if f(x) ≠ 0 everywhere between ‘a’ and ‘b’, then either it is positive someplace, or negative someplace, or both. In any case, the function will then have a maximum positive value or a minimum negative value, or both as per the following theorem;

Suppose that f(x) is continuous for all x in the closed interval [a,b]. Then ‘f’ has a minimum value ‘m’ (infimum) and a maximum value (supremum) ‘M’ on [a,b]. That is, there are numbers α & β in [a,b] such that f(α) = m and f(β) = M, and m≤ f(x) ≤ M for all x in [a,b]

That is, it has an extreme value at a point ‘c’ where f(c) is negative (in the case of a minimum) or f(c) is positive (in the case of a maximum). In either case, c is neither a nor b, since;

f(a) = f(b) =0 f'(c) ≠ 0

Therefore c lies between a and b showing that the derivative must be zero at x=c:

f'(c) = 0 for some c, where a<c<b

## Geometric Interpretation Of Rolle’s Theorem

Rolle’s theorem has a simple geometrical interpretation. If ‘f’ is continuous on [a,b] and differentiable on ]a,b[ such that f(a) = f(b), then there is a point ‘c’ ϵ ]a,b[ where the tangent line to the graph of y= f(x) is parallel to the x-axis. There may be more than one point on the graph where the tangent lines are parallel to the x-axis as shown in the figure below.

## Examples: How you find ‘c’ in Rolle’s thorem?

### Discuss the validity of Rolle’s theorem. Find ‘c’ wherever possible, such that f'(c) =0

Q1) f(x) = 1-x2/3 on [-1, 1]

Solution:-

‘f’ is continuous on [-1, 1] and f(-1)=0=f(1)

\begin{aligned}\Rightarrow f’\left( x\right) =-\dfrac{2}{3}x^{-\dfrac{1}{3}}=\dfrac{-2}{3x^{\dfrac{1}{3}}}\\ \Rightarrow f’\left( 0\right) =\dfrac{-2}{0}=\infty \end{aligned}

which is inderterminant and so ‘f’ is not differentiable at 0  ϵ ]-1, 1[. Hence Rolle’s theorem fails for the given function.

Q2) f(x) = x2 – 3x + 2 on [1,2]

Solution:-

\begin{aligned}\Rightarrow f\left( 1\right) =\left( 1\right) ^{2}-3\left( 1\right) +2=-3+3=0\\ f\left( 2\right) =\left( 2\right) ^{2}-3\left( 2\right) +2=4-6+2=0\end{aligned}

Thus, f(1) = f(2) =0

Moreover, f(x) is continuous in [1,2] and differentiable on ]1,2[. Hence all the conditions of Rolle’s theorem are satisfied. Therefore, there must exist a point ‘c’ in ]1,2[ such that f'(c) =0

Now f'(x) = 2x-3

So, f'(c) = 2c-3

\begin{aligned}\Rightarrow 2c-3=0\\ \Rightarrow c=\dfrac{3}{2}\end{aligned}

Hence Rolle’s Theorem is valid and c= 3/2.

Q3) f(x) = sin2 x on [0, π]

Solution:-

\begin{aligned}\Rightarrow f\left( 0\right) =\sin ^{2}\left( 0\right) =0\\ f\left( \pi \right) =\sin ^{2}\left( \pi \right) =0\end{aligned}

Thus, f(o) = 0 = f( π )

Moreover, f(x) is continuous in [0, π ] and differentiable on ]0, π [. Hence all conditions of Rolle’s theorem are satisfied.

There must exist point ‘c’ in the interval ]0, π [ such that, f'(c) = 0

\begin{aligned}\Rightarrow f’\left( x\right) =2\sin x\cos x\\ -\Rightarrow f’\left( c\right) =2\sin c\cos c=\sin 2c=0\\ \Rightarrow 2c=\sin ^{-1}0=0,\pi \\ \Rightarrow c=0,\dfrac{\pi }{2}\end{aligned}

Thus,

$c=\dfrac{\pi }{2}.$

As , 0 ∉ ]0, π [. Hence Rolle’s theorem is valid.

### Q4) f(x) = 1-x3/4 on [-1,1]

Solution:-

\begin{aligned}\Rightarrow f\left( -1\right) =1-\left( -1\right) ^{\dfrac{3}{4}}=1-\left( -1\right) =2\\ f\left( 1\right) =1-\left( 1\right) ^{\dfrac{3}{4}}=1-1=0\end{aligned}

Thus, f(-1) ≠f(1) and one of the conditions of Rolle’s theorem is not satisfied, so as the theorem isn’t valid therefore we can’t calculate the value of ‘c’.

### Q5) f(x) = (1-x2)/(1+x2) on [-1,1]

Solution:-

f(-1) = 0 = f(1). f(x) is continuous on [-1,1] and differentiable on ]-1,1[.

$\Rightarrow f’\left( x\right) =\dfrac{\left( 1+x^{2}\right) \left( -2x\right) -\left( 1-x^{2}\right) \left( 2x\right) }{\left( 1+x^{2}\right) ^{2}}=\dfrac{-4x}{\left( 1+x^{2}\right) ^{2}}$

if x=0;

\begin{aligned}\Rightarrow f’\left( c\right) =\dfrac{-1+c}{\left( 1+c^{2}\right) ^{2}}=0\\ \Rightarrow -4c=0\\ \Rightarrow c=0\end{aligned}

Thus, the said theorem holds for the given function and c=0.

### Q6) f(x) = x(x+3) e-x/2 on [-3,0]

Solution:-

\begin{aligned}\Rightarrow f\left( -3\right) =-3\left( -3+3\right) e^{-\dfrac{\left( -3\right) }{2}}=-3\left( 0\right) e^{3/2}=0\\ \Rightarrow f\left( 0\right) =0\left( 0+3\right) e^{\dfrac{-0}{2}}=0\end{aligned}

Thus f(-3) =0= f(0).

Moreover, f(x) is continuous on [-3,0] and differentiable on ]-3,0[. By the said theorem, there must exist a point ‘c’ in the interval ]-3,0[ such that f'(c) =0. Now,

\begin{aligned}\Rightarrow f’\left( x\right) =\left( x^{2}+3x\right) \left( \dfrac{-1}{2}e^{-\dfrac{x}{2}}\right) +e^{-\dfrac{x}{2}}\left( 2x+3\right) \\ \Rightarrow f’\left( x\right) =e^{-\dfrac{x}{2}}\left[ \dfrac{-x^{2}-3x}{2}+2x+3\right] \\ \Rightarrow f’\left( x\right) =-\dfrac{1}{2}e^{-\dfrac{x}{2}}\left( x^{2}-x-6\right) \end{aligned}

f‘(c) = 0 gives,

\begin{aligned}\Rightarrow c^{2}-c-6=0\\ \Rightarrow \left( c-3\right) \left( c+2\right) =0\\ \Rightarrow c=3,-2\end{aligned}

But only c= -2 lies in the interval ]-3,0[

Hence the given theorem is valid and c= -2.

### Q7) f(x) = 2+(x-1)3/2 on [0,2]

Solution:-

\begin{aligned}\Rightarrow f\left( 0\right) =2+\left( 0-1\right) ^{\dfrac{3}{2}}=2+\left( -1\right) ^{\dfrac{3}{2}}=2+\sqrt{\left( -1\right) ^{3}}\\ =2+\sqrt{-1}=2+i\\ \Rightarrow f\left( 2\right) =2+\left( 2-1\right) ^{\dfrac{3}{2}}=2+\left( 1\right) ^{\dfrac{3}{2}}=2+1=3\end{aligned}

Thus f(0) ≠ f(2)

As one of the three conditions isn’t satisfied, thus the said theorem isn’t valid.

## Real Life Applications of Rolle’s theorem

There are several examples that can help us understand how Rolle’s theorem contributes in handling multifarious real-world tasks, here I have tried to compile a few as under.

#### 1) A couple of days back I was just looking at the government statistics of COVID-19 delta cases

issued by the “Center of Diseases Control & Prevention” USA-CDC, the graph of number of days vs number of new cases looks something like this.

In this graph, I observed that the number of cases registered on June 1st and July 10th was the same, besides, the graph is beautifully continuous and differentiable as well within the interval and beyond;

Let us consider a=June 1st and b= July 10th and the curve in the graph is representing a certain function f(x), and we intend to find out a certain day ‘c’ within the interval [a,b] when the number of registered cases remains constant

i.e. no change has been recorded then indeed Rolle’s theorem is confidently going to play its role and through

f'(c) =0

we can find that day.

#### 2) Rolle’s theorem also helps a lot in finding the maximum height of a projectile trajectory,

lets try to understand with the following example;

The equation of trajectory of a projectile is y=10x-(9/5​)x2,(m). Find the position when the maximum height is reached?

Now here the very first thing I am going to determine is the interval [a,b] such that f(a) = f(b) and of course in this case the lower bound ‘a’ is “launching point” and upper bound ‘b’ is “landing point” of the projectile. Look at the graph of the projectile trajectory;

Here y = f(x), launching and landing point are actually x-intercepts of the trajectory which can be determined by equating f(x) =0, i.e.

\begin{aligned}\Rightarrow y=f\left( x\right) =0\\ \Rightarrow 10x-\dfrac{5}{9}x^{2}=0\\ \Rightarrow 5x\left( 2-\dfrac{x}{9}\right) =0\\ \Rightarrow 5x=0 , 2-\dfrac{x}{9}=0\end{aligned}

Thus;

x=0 and x=18

Now in order to find the position of the maximum height we have to trace its stationary point (extrema) which is of course “maxima”, a point where the slope of its tangent is zero.

We can clearly observe that the above curve is continuous with [0,18] and differentiable within ]0, 18[, so there certainly exists a point ‘c’ ϵ ]0, 18[ where f'(c) =0 and this ‘c’ is actually the position where the projectile is experiencing maximum height.

$y=f\left( c\right) =10c-\dfrac{5}{9}c^{2}$

Differentiating above equation w.r.t.’c’ we get;

\begin{aligned}\Rightarrow \dfrac{dy}{dc}=\dfrac{d}{dc}\left( 10c-\dfrac{5}{9}c^{2}\right) =0\\ \Rightarrow 10-\dfrac{10c}{9}=0\\ \Rightarrow 10\left( 1-\dfrac{c}{9}\right) =0\\ \Rightarrow 1-\dfrac{c}{9}=0\Rightarrow c=9\end{aligned}

i.e. at c=9m, the projectile is going to achieve a maximum height of;

$y=f\left( 9\right) =10\left( 9\right) -\dfrac{5}{9}\left( 9\right) ^{2}=90-45=45m$

#### 3) How about involving Rolle’s masterpiece in the construction of domes?

Though our medieval Muslim civilization was rather older and richer than this french Mathematician work; still this french cheese made up to step in making the architectural calculation easier smoothly shifting it from 2D to 3D.

Take a look at “Sultan Ahmed Masjid” commonly famous as “Blue Mosque”, Istanbul’s last imperial masjid constructed between 1609-1617 is indeed a Hallmark of the great architectural ability of the Ottomans.

Domes are designed not just to enhance the exquisiteness of any edifice but also to maintain a good sound echo, heights of the domes play a vital role in amplifying sound waves and turning their direction so every individual would be able to listen clearly.

Using this french theorem we can easily determine the maximum height of the domes and arches and could preserve the monuments’ stability and architectural beauty.

## Conclusion

Algebraic equations’ analysis has led to two core statements in the theory of functions which are;

• Theorem of root interval
• Theorem of root of a derivative

Which later created Mean Value Theorem.

In various Russian sources, several authors call “Theorem of root interval” as “Second Rolle’s Theorem” or “Theorem of Bolzano Cauchy”. Further, inside Russia, Mean Value Theorem is famous as “The second theorem of Bolzano-Cauchy”.

Rolle’s Theorem took approximately three hundred years to take into final form. It has indeed placed tremendous significance in differential geometry, functional analysis, mechanics, etc.

Nikolai Nikolaevich Luzin (1883-1950), a Russian mathematician, with the help of Rolle’s theorem proved his theorems of osculating curve and center of curvature. Besides, this theorem was further extended to continuous functions and supplemented their properties. As per Luzin,

This theorem underlies the theoretical development of differential and integral calculus.”

Peace

Maria Khan

Founder & Teacher MMC