# Calculus

## Applications of the Derivative # 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 $$\left[ {0,1} \right].$$) Figure 3.
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 $$\left( { - 1,1} \right).$$ Figure 4.
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 $$f\left( 0 \right) \ne f\left( 1 \right).$$) Figure 5.

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 or tap 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

$$f{\left( x \right) = {a_0}{x^n} + {a_1}{x^{n - 1}} + \ldots }$$ $$+\;{a_{n - 1}}x = 0$$

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 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 $$b = 2.$$

### Example 6.

Prove that if the equation

$$f{\left( x \right) = {a_0}{x^n} + {a_1}{x^{n - 1}} + \ldots }$$ $$+\;{a_{n - 1}}x = 0$$

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 $$\xi \lt {x_0}.$$

Solution.

In addition to $$x = {x_0},$$ the first equation has the root $$x = 0.$$ Consequently, the function $$f\left( x \right)$$ satisfies the conditions of Rolle's theorem:

$f\left( 0 \right) = f\left( {{x_0}} \right) = 0.$

The second equation is obtained by differentiating the first equation:

$f'\left( x \right) = {\left( {{a_0}{x^n} + {a_1}{x^{n - 1}} + \ldots + {a_{n - 1}}x} \right)^\prime } = n{a_0}{x^{n - 1}} + \left( {n - 1} \right){a_1}{x^{n - 2}} + \ldots + {a_{n - 1}} = 0.$

According to Rolle's theorem, there is an interior point $$x = \xi$$ on the interval $$\left[ {0,{x_0}} \right]$$ where the derivative is zero. Consequently, the point $$x = \xi$$ is a solution of the second equation where $$0 \lt \xi \lt {x_0}.$$

### Example 7.

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

Solution.

The given quadratic function has roots $$x = 2$$ and $$x = 4,$$ that is

$f\left( 2 \right) = f\left( {4} \right) = 0.$

The by Rolle's theorem, there is a point $$\xi$$ in the interval $$\left( {2,4} \right)$$ where the derivative of the function $$f\left( x \right)$$ equals zero.

$f'\left( x \right) = {\left( {{x^2} - 6x + 8} \right)^\prime } = 2x - 6.$

It is equal to zero at the following point $$x = \xi:$$

$f'\left( x \right) = 0,\;\; \Rightarrow 2x - 6 = 0,\;\; \Rightarrow x = \xi = 3.$

It can be seen that the resulting stationary point $$\xi = 3$$ belongs to the interval $$\left( {2,4} \right)$$ (Figure $$6$$).

### Example 8.

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

Solution.

Let us make sure that the function has the same values at the endpoints:

$f\left( {-1} \right) = f\left( {1} \right) = 0.$

Hence, the derivative must be equal to zero at any point $$\xi \in \left( { - 1,1} \right).$$ Differentiate the given function:

$f'\left( x \right) = {\left( {\sqrt {1 - {x^2}} } \right)^\prime } = \frac{1}{{2\sqrt {1 - {x^2}} }} \cdot {\left( {1 - {x^2}} \right)^\prime } = \frac{{ - \cancel{2}x}}{{\cancel{2}\sqrt {1 - {x^2}} }} = - \frac{x}{{\sqrt {1 - {x^2}} }}.$

This shows that the derivative is zero at $$x = 0.$$ Thus, $$\xi = 0$$ where $$\xi \in \left( { - 1,1} \right).$$

Note that this function describes the upper semicircle of radius $$R = 1$$ centered at the origin (Figure $$7$$). It does not have finite derivatives at the endpoints of the interval $$\left[ { - 1,1} \right],$$ i.e. this function is not differentiable at $$x = \pm 1.$$ This, however, does not preclude the application of Rolle's theorem, because the latter requires the function to be differentiable on the open interval $$\left( { - 1,1} \right).$$