# Rolle’s Theorem

• ### Rolle’s Theorem

Suppose that a function $$f\left( x \right)$$ is continuous on the closed interval $$\left[ {a,b} \right]$$ and differentiable on the open interval $$\left( {a,b} \right)$$. Then if $$f\left( a \right) = f\left( b \right),$$ then there exists at least one point $$c$$ in the open interval $$\left( {a,b} \right)$$ for which $$f^\prime\left( c \right) = 0.$$

### Geometric interpretation

There is a point $$c$$ on the interval $$\left( {a,b} \right)$$ where the tangent to the graph of the function is horizontal.

This property was known in the $$12$$th century in ancient India. The outstanding Indian astronomer and mathematician Bhaskara $$II$$ $$\left(1114-1185\right)$$ mentioned it in his writings.

In a strict form this theorem was proved in $$1691$$ by the French mathematician Michel Rolle $$\left(1652-1719\right)$$ (Figure $$2$$).

All $$3$$ conditions of Rolle’s theorem are necessary for the theorem to be true:

1. $$f\left( x \right)$$ is continuous on the closed interval $$\left[ {a,b} \right];$$
2. $$f\left( x \right)$$ is differentiable on the open interval $$\left( {a,b} \right);$$
3. $$f\left( a \right) = f\left( b \right).$$

### Some counterexamples

1. Consider $$f\left( x \right) = \left\{ x \right\}$$ ($$\left\{ x \right\}$$ is the fractional part function) on the closed interval $$\left[ {0,1} \right].$$ The derivative of the function on the open interval $$\left( {0,1} \right)$$ is everywhere equal to $$1.$$ In this case, the Rolle’s theorem fails because the function $$f\left( x \right)$$ has a discontinuity at $$x = 1$$ (that is, it is not continuous everywhere on the closed interval $$\left[ {0,1} \right].$$)
2. Consider $$f\left( x \right) = \left| x \right|$$ (where $$\left| x \right|$$ is the absolute value of $$x$$) on the closed interval $$\left[ { – 1,1} \right].$$ This function does not have derivative at $$x = 0.$$ Though $$f\left( x \right)$$ is continuous on the closed interval $$\left[ { – 1,1} \right],$$ there is no point inside the interval $$\left( { – 1,1} \right)$$ at which the derivative is equal to zero. The Rolle’s theorem fails here because $$f\left( x \right)$$ is not differentiable over the whole interval $$\left( { – 1,1} \right).$$
3. The linear function $$f\left( x \right) = x$$ is continuous on the closed interval $$\left[ { 0,1} \right]$$ and differentiable on the open interval $$\left( { 0,1} \right).$$ The derivative of the function is everywhere equal to $$1$$ on the interval. So the Rolle’s theorem fails here. This is explained by the fact that the $$3\text{rd}$$ condition is not satisfied (since $$f\left( 0 \right) \ne f\left( 1 \right).$$)

In modern mathematics, the proof of Rolle’s theorem is based on two other theorems − the Weierstrass extreme value theorem and Fermat’s theorem. They are formulated as follows:

### The Weierstrass Extreme Value Theorem

If a function $$f\left( x \right)$$ is continuous on a closed interval $$\left[ {a,b} \right],$$ then it attains the least upper and greatest lower bounds on this interval.

### Fermat’s Theorem

Let a function $$f\left( x \right)$$ be defined in a neighborhood of the point $${x_0}$$ and differentiable at this point. Then, if the function $$f\left( x \right)$$ has a local extremum at $${x_0},$$ then

$f’\left( {{x_0}} \right) = 0.$

Consider now Rolle’s theorem in a more rigorous presentation. Let a function $$y = f\left( x \right)$$ be continuous on a closed interval $$\left[ {a,b} \right],$$ differentiable on the open interval $$\left( {a,b} \right),$$ and takes the same values at the ends of the segment:

$f\left( a \right) = f\left( b \right).$

Then on the interval $$\left( {a,b} \right)$$ there exists at least one point $$c \in \left( {a,b} \right),$$ in which the derivative of the function $$f\left( x \right)$$ is zero:

$f’\left( c \right) = 0.$

#### Proof.

If the function $$f\left( x \right)$$ is constant on the interval $$\left[ {a,b} \right],$$ then the derivative is zero at any point of the interval $$\left( {a,b} \right),$$ i.e. in this case the statement is true.

If the function $$f\left( x \right)$$ is not constant on the interval $$\left[ {a,b} \right],$$ then by the Weierstrass theorem, it reaches its greatest or least value at some point $$c$$ of the interval $$\left( {a,b} \right),$$ i.e. there exists a local extremum at the point $$c.$$ Then by Fermat’s theorem, the derivative at this point is equal to zero:

$f’\left( c \right) = 0.$

### Physical interpretation

Rolle’s theorem has a clear physical meaning. Suppose that a body moves along a straight line, and after a certain period of time returns to the starting point. Then, in this period of time there is a moment, in which the instantaneous velocity of the body is equal to zero.

• ## Solved Problems

Click a problem to see the solution.

### Example 1

Let $$f\left( x \right) = {x^2} + 2x.$$ Find all values of $$c$$ in the interval $$\left[ { – 2,0} \right]$$ such that $$f^\prime\left( c \right) = 0.$$

### Example 2

Given the function $$f\left( x \right) = {x^2} – 6x + 5.$$ Find all values of $$c$$ in the open interval $$\left( {2,4} \right)$$ such that $$f^\prime\left( c \right) = 0.$$

### Example 3

Let $$f\left( x \right) = {x^2} + 8x + 14.$$ Find all values of $$c$$ in the interval $$\left( { – 6, – 2} \right)$$ such that $$f’\left( c \right) = 0.$$

### Example 4

Given an interval $$\left[ {a,b} \right]$$ that satisfies hypothesis of Rolle’s theorem for the function $f\left( x \right) = {x^4} + {x^2} – 2.$ It is known that $$a = – 1.$$ Find the value of $$b.$$

### Example 5

Given an interval $$\left[ {a,b} \right]$$ that satisfies hypothesis of Rolle’s theorem for the function $f\left( x \right) = {x^3} – 2{x^2} + 3.$ It is known that $$a = 0.$$ Find the value of $$b.$$

### Example 6

Prove that if the equation $f{\left( x \right) = {a_0}{x^n} + {a_1}{x^{n – 1}} + \ldots }\kern0pt{\;+\;{a_{n – 1}}x = 0\;\;\;}$ has a positive root $$x = {x_0},$$ then the equation $n{a_0}{x^{n – 1}} + \left( {n – 1} \right){a_1}{x^{n – 2}} + \ldots + {a_{n – 1}} = 0$ also has a positive root $$x = \xi,$$ where $$\xi \lt {x_0}.$$

### Example 7

Check the validity of Rolle’s theorem for the function $f\left( x \right) = {x^2} – 6x + 8.$

### Example 8

Check the validity of Rolle’s theorem for the function $f\left( x \right) = \sqrt {1 – {x^2}}$ on the segment $$\left[ { – 1,1} \right].$$

### Example 9

Check the validity of the Rolle’s theorem for the function $f\left( x \right) = \frac{{{x^2} – 4x + 3}}{{x – 2}}$ on the segment $$\left[ {1,3} \right].$$

### Example 10

Check the validity of Rolle’s theorem for the function $f\left( x \right) = \left| {x – 1} \right|$ on the segment $$\left[ {0,2} \right].$$

### Example 11

Let $$f\left( x \right) = {x^3} – 3x.$$ Find all values of $$c$$ in the interval $$\left( { – 1,2} \right)$$ such that $$f^\prime\left( c \right) = 0.$$

### Example 12

Given the function $$f\left( x \right) = {x^3} – 4x + 1.$$ Find all values of $$c$$ in the interval $$\left( {-2,2} \right)$$ such that $$f^\prime\left( c \right) = 0.$$

### Example 13

Let $$f\left( x \right) = \ln \left( {2 – {x^2}} \right).$$ Find all values of $$c$$ in the open interval $$\left( { – 1,1} \right)$$ such that $$f^\prime\left( c \right) = 0.$$

### Example 14

Determine the number of stationary points of the function $f\left( x \right) = x\left( {x – 1} \right)\left( {x – 2} \right)$ and indicate the intervals, in which they are located.

### Example 15

Determine whether the function $$f\left( x \right) = \sin \large{\frac{x}{2}}\normalsize$$ satisfies conditions of Rolle’s theorem for the interval $$\left[ {0,2\pi } \right].$$ If so find all numbers $$c$$ that satisfy the conclusion of the theorem.

### Example 16

Check the validity of Rolle’s theorem for the function $f(x) = \begin{cases} x^2, & \text{if}\;\;\; 0 \le x \le 2 \\ 6-x, & \text{if}\;\;\; 2 \lt x \le 6 \end{cases}.$

### Example 17

Determine whether the function $$f\left( x \right) = \text{sech}\,x = \large{\frac{2}{{{e^x} + {e^{ – x}}}}}\normalsize$$ satisfies conditions of Rolle’s theorem for the interval $$\left[ {-1,1 } \right].$$ If so find all numbers $$c$$ that satisfy the conclusion of the theorem.

### Example 1.

Let $$f\left( x \right) = {x^2} + 2x.$$ Find all values of $$c$$ in the interval $$\left[ { – 2,0} \right]$$ such that $$f^\prime\left( c \right) = 0.$$

Solution.

First of all, we need to check that the function $$f\left( x \right)$$ satisfies all the conditions of Rolle’s theorem.

$$1.$$ $$f\left( x \right)$$ is continuous in $$\left[ {-2,0} \right]$$ as a quadratic function;

$$2.$$ It is differentiable everywhere over the open interval $$\left( { – 2,0} \right);$$

$$3.$$ Finally,

${f\left( { – 2} \right) = {\left( { – 2} \right)^2} + 2 \cdot \left( { – 2} \right) = 0,}$

${f\left( 0 \right) = {0^2} + 2 \cdot 0 = 0,}$

$\Rightarrow f\left( { – 2} \right) = f\left( 0 \right).$

So we can use Rolle’s theorem.

To find the point $$c$$ we calculate the derivative

$f^\prime\left( x \right) = \left( {{x^2} + 2x} \right)^\prime = 2x + 2$

and solve the equation $$f^\prime\left( c \right) = 0:$$

${f^\prime\left( c \right) = 2c + 2 = 0,}\;\; \Rightarrow {c = – 1.}$

Thus, $$f^\prime\left( c \right) = 0$$ for $$c = – 1.$$

### Example 2.

Given the function $$f\left( x \right) = {x^2} – 6x + 5.$$ Find all values of $$c$$ in the open interval $$\left( {2,4} \right)$$ such that $$f^\prime\left( c \right) = 0.$$

Solution.

First we determine whether Rolle’s theorem can be applied to $$f\left( x \right)$$ on the closed interval $$\left[ {2,4} \right].$$

The function is continuous on the closed interval $$\left[ {2,4} \right].$$

The function is differentiable on the open interval $$\left( {2,4} \right).$$ Its derivative is

${f^\prime\left( x \right) = \left( {{x^2} – 6x + 5} \right)^\prime }={ 2x – 6.}$

The function has equal values at the endpoints of the interval:

${f\left( 2 \right) = {2^2} – 6 \cdot 2 + 5 }={ – 3,}$

${f\left( 4 \right) = {4^2} – 6 \cdot 4 + 5 }={ – 3.}$

This means that we can apply Rolle’s theorem. Solve the equation to find the point $$c:$$

${f^\prime\left( c \right) = 0,}\;\; \Rightarrow {2c – 6 = 0,}\;\; \Rightarrow {c = 3.}$

### Example 3.

Let $$f\left( x \right) = {x^2} + 8x + 14.$$ Find all values of $$c$$ in the interval $$\left( { – 6, – 2} \right)$$ such that $$f’\left( c \right) = 0.$$

Solution.

The function is a quadratic polynomial. Therefore it is everywhere continuous and differentiable. Calculate the values of the function at the endpoints of the given interval:

${f\left( { – 6} \right) = {\left( { – 6} \right)^2} + 8 \cdot \left( { – 6} \right) + 14 }={ 36 – 48 + 14 }={ 2,}$

${f\left( { – 2} \right) = {\left( { – 2} \right)^2} + 8 \cdot \left( { – 2} \right) + 14 }={ 4 – 16 + 14 }={ 2.}$

Since both the values are equal to each other we conclude that all three conditions of Rolle’s theorem are satisfied. So we can apply this theorem to find $$c.$$

Differentiate:

${f^\prime\left( x \right) = \left( {{x^2} + 8x + 14} \right)^\prime }={ 2x + 8.}$

Solve the equation and find the value of $$c:$$

${f^\prime\left( c \right) = 0,}\;\; \Rightarrow {2c + 8 = 0,}\;\; \Rightarrow {c = – 4.}$

### Example 4.

Given an interval $$\left[ {a,b} \right]$$ that satisfies hypothesis of Rolle’s theorem for the function $f\left( x \right) = {x^4} + {x^2} – 2.$ It is known that $$a = – 1.$$ Find the value of $$b.$$

Solution.

We factor the polynomial:

${{x^4} + {x^2} – 2 }={ \left( {{x^2} + 2} \right)\left( {{x^2} – 1} \right) }={ \left( {{x^2} + 2} \right)\left( {x – 1} \right)\left( {x + 1} \right).}$

It is now easy to see that the function has two zeros: $${x_1} = – 1$$ (coincides with the value of $$a$$) and $${x_2} = 1.$$

Since the function is a polynomial, it is everywhere continuous and differentiable. So this function satisfies Rolle’s theorem on the interval $$\left[ {-1,1} \right].$$ Hence, $$b = 1.$$

### Example 5.

Given an interval $$\left[ {a,b} \right]$$ that satisfies hypothesis of Rolle’s theorem for the function $f\left( x \right) = {x^3} – 2{x^2} + 3.$ It is known that $$a = 0.$$ Find the value of $$b.$$

Solution.

If we consider the auxiliary function

${{f_1}\left( x \right) }={ {x^3} – 2{x^2}} ={ {x^2}\left( {x – 2} \right),}$

we see that it has the following zeros:

${x_1} = 0,\;{x_2} = 2.$

The original function differs from this function in that it is shifted 3 units up. Therefore, we can write that

$f\left( 0 \right) = f\left( 2 \right) = 3.$

It is obvious that the function $$f\left( x \right)$$ is everywhere continuous and differentiable as a cubic polynomial. Consequently, it satisfies all the conditions of Rolle’s theorem on the interval $$\left[ {0,2} \right].$$ So $$b = 2.$$

Page 1
Problems 1-5
Page 2
Problems 6-17