# Rolle’s Theorem Statement with Proof & Geometrical Interpretation

Rolle’s Theorem with Proof
Rolle’s Theorem with Proof

Overview

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Rolle’s theorem states if a differentiable function achieves equal values at two different points then it must possess at least one fixed point somewhere between them that is, a position where the first derivative i.e the slope of the tangent line to the graph of the function is zero. Mean Value Theorem is one of the commonly used theorems in mathematics learning literature. It is one of the most powerful tools used to explain several theorems in differential and integral calculus. Rolle’s Theorem is a specific case of the mean value theorem and was given by Michel Rolle, a French mathematician.

Rolle’s theorem states if a differentiable function achieves equal values at two different points then it must possess at least one fixed point somewhere between them that is, a position where the first derivative i.e the slope of the tangent line to the graph of the function is zero. Rolle’s Theorem is a special case of the mean value theorem which meets certain requirements. However, Lagrange’s mean value theorem is itself the mean value theorem also called the first mean value theorem. In general, one can assume mean as the average of the given data/numbers. But in integrals, the method of calculating the mean value of two different functions is different.

Let us understand Lagrange’s mean value theorem in calculus before we study Rolle’s theorem.

The mean value theorem is applied in a mathematical study dealing with a sort of average helpful for estimates and for building other theorems, such as the primary theorem of calculus. The mean value theorem states that the slope of a line joining any two points on a smooth curve is equivalent to the slope of some line tangent to the curve at a location between the two points.

That is the Mean Value Theorem establishes a link between the slope of a tangent line to a curve and the secant line by points on a curve at the endpoints of an interval. It states that:

$$f^{^{\prime}}\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$$

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Rolle’s theorem states if f(x) is a real-valued function, defined in the closed interval [a, b] such that:

Informally, Rolle’s theorem says that if the outputs of a differentiable function f are equal at the end positions of an interval, then there should be an internal point c where $$f’(c) = 0$$.

Other than being beneficial in determining the mean-value theorem, Rolle’s theorem is sometimes used, as it builds only the existence of a solution and not its value.

Rolle’s theorem can be said as Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b), such that f(a) = f(b), where a and b are some real characters. Then there is some c in (a, b) such that $$f’(c) = 0$$.

If:

Then Rolle’s theorem does not hold good. Rolle’s theorem fails even if any of the three conditions fails to follow the function.

In the above graph, the curve y = f(x) is continuous within x = a and x = b and at every position, within the interval. It is likely to form a tangent and ordinate corresponding to the abscissa and if there is a minimum of one tangent to the curve that is parallel to the x-axis.

Algebraically, the theorem describes us that if f (x) is expressing a polynomial function in x and the two roots of equation f(x) = 0 are x = a and x = b, then there is a minimum of one root of equation f‘(x) = 0 existing between these values.

The opposite of Rolle’s theorem is not valid and it is also probable that there is more than one value of x, for which the theorem holds good but there is a clear chance of the presence of one such value.

Learn about Rolle’s Theorem and Lagrange’s mean Value Theorem

Let us start by considering whether all of the conditions are satisfied. So, our discussion here links only to functions: that is continuous, that is differentiable and have f(a) = f(b).

When a function satisfies Rolle’s Theorem, the place where f′(x)=0 happens is a maximum or a minimum value, i.e., extreme. The Extreme Value Theorem says that if a function is continuous, then it is confirmed to hold both a maximum and a minimum point in the interval. Now, there are two basic possibilities for the function.

Case 1:

For a constant function, the graph is a horizontal line section. In this particular case, every point satisfies Rolle’s Theorem as the derivative is zero everywhere.

Case 2:

As the function is not constant, it must switch directions to start and end at the corresponding y-value. This implies somewhere inside the interval the function will either hold a minimum (Graph A), a maximum (Graph B) or both minimum and maximum value (Graph C).

So, now we need to confirm that at this interior extreme the derivative must approach zero.

If f′>0 we identify that the function is increasing. But it cannot increase since we are at its maximum point.

If f′<0 we recognize that function is declining, which signifies it is larger just a little to the left of where we are present. But we are at the function’s maximum state, so it could not exceed the value further. Since f′ exists, but is not larger than zero, and is not even smaller than zero, the only possibility that persists is that of f′=0. Hence proved.

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Example 1: Consider the following statements:

1. Rolle’s theorem ensures that there is a point on the curve, the tangent at which is parallel to the x-axis.
2. Lagrange’s mean value theorem ensures that there is a point on the curve, the tangent at which is parallel to the y-axis.
3. Cauchy mean value theorem can be deduced from Lagrange’s mean value theorem.
4. Rolle’s man value theorem can be deduced from Lagrange’s mean value theorem.

Which of the above statement(s), is/are true?

ROLLE’S THEOREM:

If f(x) function defined in [a, b] such that:

1. f(x) is continuous in [a, b]

2. f’(x) exist for every point in ]a, b[

3. f(a) = f(b)

Then there exist at least one point x = c ∈ ]a, b[ such that f’(c) = 0

LAGRANGE’S MEAN VALUE THEOREM:

If f(x) is a function defined in [a, b] such that

1. f(x) is continuous in [a, b]

2. f’(x) exist for every point in ]a, b[

Then there exist at least one point x = c ∈ ]a, b[, such that

$$f’\left( c \right) = \frac{{f\left( b \right) – f\left( a \right)}}{{b – a}}$$

CAUCHY’S MEAN VALUES THEOREM:

If f(x) and g(x) are function defined in [a, b] such that

1. f(x) and g(x) are continuous in [a, b]

2. f’(x) and g(x) both exist in ]a, b[

3. g’(x) goes not vanish at ay point in ]a, b[

Then there exist at least one point x = c ∈ ]a, b[ such that

$$\frac{{f\left( b \right) – f\left( a \right)}}{{g\left( b \right) – g\left( a \right)}} = \frac{{f’\left( c \right)}}{{g’\left( c \right)}}$$

Example 2:Which of the following is not true about Rolle’s Theorem?

Explanation:

Rolle’s Theorem states that if f (x) is a function that satisfies:

(i) f (x) is continuous on the closed interval [a, b]

(ii) f (x) is differentiable on the open interval (a, b)

(iii) f (a) = f (b)

then there exists a point c in the open interval (a, b) such that f’ (c) = 0.

Example 3: When Rolle’s Theorem states is verified for f (x) on [a, b] then there exists c such that

Explanation:

(Rolle’s Theorem) : Let f : [a, b] → R be continuous on [a, b] and differentiable on (a, b), such that f (a) = f (b), where a and b are some real numbers. Then there exists some c in (a, b) such that f′ (c) = 0.

Example 4: If Rolle’s theorem holds good for a function f(x) in an interval [a,b], then which of the following statement is not true.

Concept:

Rolle’s Theorem:

Let f(x) be defined in [a,b] such that

(i) f(x) is continuous in [a,b]

(ii) f(x) is differentiable in (a,b)

(iii) f(a) = f(b)

then there exists at least one point c ∈ (a,b) such that f'(c) = 0

Lagrange’s Mean value theorem:

Let f(x) is a function define on [a ,b] such that,

Then, there exist a real number C ∈ (a , b) such that, According to Lagrangian Mean Value Theorem,

f'(c) = $$\frac{{f\left( b \right) ~- ~f\left( a \right)}}{{b ~-~ a}}$$

Example 5: Rolle’s Theorem is not applicable to f (x) = [x] in (4, -4) because,
Where [.] is GIF function

Concept:

Rolle’s theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b) then,

f′(c) = 0 for some c ∈ [a, b].

Calculation:

Explanation:

f (- 4) $$\rm \neq$$ f (4)

Example 6: Which of the following is true for the function f (x) = x + |x| in the interval [-2, 2] for Rolle’s theorem?

Concept:

Rolle’s Theorem states that if f (x) is a function that satisfies:

(i) f (x) is continuous on the closed interval [a, b]

(ii) f (x) is differentiable on the open interval (a, b)

(iii) f (a) = f (b)

then there exists a point c in the open interval (a, b) such that f’ (c) = 0.

Calculation:

The fiven function is not differentiable but continous function in the interval [-2, 2].

Example 7: In which of the following interval is f (x) = –$$\rm x^2$$ satisfy Rolle’s Theorem?

Concept:

Rolle’s Theorem states that if f (x) is a function that satisfies:

(i) f (x) is continuous on the closed interval [a, b]

(ii) f (x) is differentiable on the open interval (a, b)

(iii) f (a) = f (b)

then there exists a point c in the open interval (a, b) such that f’ (c) = 0.

Calculation:

Given: f (x) = -$$\rm x^{2}\$$

When we put any value from the given option the given function does not satisfy the condition f (a) $$\rm \neq$$ f (b).

Hence, Option 4 is correct.

Example 8: The value of ‘c’ in Rolle’s Theorem for the function f(x) = $$\rm cos \frac{x}{2}\$$ on [π, 3π]:

Concept:

Rolle’s theorem:

Rolle’s theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that,

f(a) = f(b), then, for some c ∈ [a, b]

f′(c) = 0

Calculation:

The given function is f(x) = $$\rm \cos \dfrac{x}{2}$$ on [π, 3π].

f(π) = $$\rm \cos \dfrac{\pi}{2}$$ = 0 and f(3π) = $$\rm \cos \dfrac{3\pi}{2}$$ = 0.

Since, f(π) = f(3π), there must exist a c ∈ [π, 3π] such that f'(c) = 0.

f'(x) = $$\rm \dfrac{d}{dx}\left (\cos \dfrac{x}{2} \right )=-\dfrac{1}{2}\left (\sin \dfrac {x}{2} \right )$$

⇒ f'(c) = $$\rm -\dfrac{1}{2}\left (\sin \dfrac {c}{2} \right )$$ = 0

⇒ $$\rm \sin \dfrac {c}{2}$$ = 0

⇒ $$\rm \dfrac {c}{2}$$ = nπ

⇒ c = 2nπ, where n is an integer.

We want c ∈ [π, 3π], therefore c = 2π.

Example 9: What is the value of c if it satisfies Rolle’s Theorem for function f (x) = $$\rm f (x) = x^2 + 3x – 12$$, x $$\rm \epsilon\$$ [-6, 3].

Concept:

Rolle’s Theorem states that if f (x) is a function that satisfies:

(i) f (x) is continuous on the closed interval [a, b]

(ii) f (x) is differentiable on the open interval (a, b)

(iii) f (a) = f (b)

then there exists a point c in the open interval (a, b) such that f’ (c) = 0.

Calculation:

(i) f (x) is continuous on [-6, 3]

(ii) f'(x) = 2x + 3, f’ (x) exists in the interval (-6, 3) so, f (x) is differentiable

(iii) f (-6) = f (3)

All three conditions of Rolle’s Theorem Satisified.

Therefore, there exists at least one c $$\rm \epsilon$$ [-6, 3] such that f’ (c) = 0

which implies 2c + 3 = 0 = c = – 3/2 = -1.5

Example 10: The value of ‘c’ in Rolle’s Theorem for the function f(x) = $$\rm cos \dfrac{x}{2}$$ on [π, 3π]:

Concept:

Rolle’s theorem states that if a function f(x) is continuous in the closed interval [a, b] and differentiable on the open interval (a, b) such that f(a) = f(b), then f′(c) = 0 for some c ∈ [a, b].

Calculation:

The given function is f(x) = $$\rm \cos \dfrac{x}{2}$$ on [π, 3π].

f(π) = $$\rm \cos \dfrac{\pi}{2}$$ = 0 and f(3π) = $$\rm \cos \dfrac{3\pi}{2}$$ = 0.

Since, f(π) = f(3π), there must exist a c ∈ [π, 3π] such that f'(c) = 0.

f'(x) = $$\rm \dfrac{d}{dx}\left (\cos \dfrac{x}{2} \right )=-\dfrac{1}{2}\left (\sin \dfrac {x}{2} \right )$$

⇒ f'(c) = $$\rm -\dfrac{1}{2}\left (\sin \dfrac {c}{2} \right )$$ = 0

⇒ $$\rm \sin \dfrac {c}{2}$$ = 0

⇒ $$\rm \dfrac {c}{2}$$ = nπ

⇒ c = 2nπ, where n is an integer.

We want c ∈ [π, 3π], therefore c = 2π.

Example 11: Which of the following is not true about Rolle’s Theorem?

Solution:

Explanation: Rolle’s Theorem states that if f (x) is a function that satisfies:

then there exists a point c in the open interval (a, b) such that f’ (c) = 0.

This implies that option 1 is incorrect.

Example 12: $$\text{ Consider}\ f\left(x\right)=x^2+x-6\ \ \text{range}\ \left[-3,\ 2\right]$$
Check for Rolle’s theorem;

It can be seen that$$\ f\left(x\right)$$ is a polynomial function therefore it is continuous.

$$\text{in the range}\ \left[-3,\ 2\right]\ \text{and differentiable in the range}\ \left(-3,\ 2\right)$$

Now, we will check for the third condition:

$$f\left(-3\right)=0\ \text{ and }\ f\left(2\right)=0\ \text{ therefore }\ f\left(-3\right)=f\left(2\right) \text{and hence all the three conditions are satisfied.}$$

$$f^{^{\prime}}\left(x\right)=2x+1\Rightarrow f^{^{\prime}}\left(c\right)=2c+1=0\ \Rightarrow c=-\frac{1}{2}$$

$$\text{Therefore there exist}\ -\frac{1}{2}\text{in the interval}\ \left(-3,\ 2\right)\ \text{such that}\ f’(-\frac{1}{2})=0$$

Example 13: For interval (0, 2) check if the function f (x) = $$-x^2$$ satisfies Rolle’s Theorem?

Solution: Rolle’s Theorem states that if f (x) is a function that satisfies:

then there exists a point c in the open interval (a, b) such that f’ (c) = 0.

Calculation:

Given: f (x) = $$-x^2$$

When we put any value from the given option the given function does not satisfy the condition

$$f(a)\ne f(b)$$

$$\text{i.e}\ f(a)=f(0)=-0$$

$$f(b)=f(2)=-4$$

$$\text{Therefore}\ f(a)\ne f(b)\Rightarrow f(0)\ne f(2)$$

Hence the given function f (x) = $$-x^2$$ does not satisfy Rolle’s Theorem.

 If you are checking Rolle’s Theorem article, also check the related maths articles: Apollonius Theorem Inverse trigonometric functions Properties of triangles Conditional probability Diagonal matrix Exterior Angle Theorem