Rolle’s Theorem

Rolle's Theorem

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

Geometric interpretation

There is a point c on the interval (a, b) 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 $II$ $(1114-1185)$ mentioned it in his writings.

In a strict form this theorem was proved in $1691$ by the French mathematician Michel Rolle $(1652-1719)$ (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 $$ (where $$ is the absolute value of $$) on the closed interval $$ This function does not have derivative at $$ Though $$ is continuous on the closed interval $$ there is no point inside the interval $$ at which the derivative is equal to zero. The Rolle's theorem fails here because $$ is not differentiable over the whole interval $$

   

3. The linear function $$ is continuous on the closed interval $$ and differentiable on the open interval $$ The derivative of the function is everywhere equal to $$ on the interval. So the Rolle's theorem fails here. This is explained by the fact that the $$ condition is not satisfied (since $$)

   

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 $$ is continuous on a closed interval $$ then it attains the least upper and greatest lower bounds on this interval.

Fermat's Theorem

Let a function $$ be defined in a neighborhood of the point $$ and differentiable at this point. Then, if the function $$ has a local extremum at $$ then

Consider now Rolle's theorem in a more rigorous presentation. Let a function $$ be continuous on a closed interval $$ differentiable on the open interval $$ and takes the same values at the ends of the segment:

Then on the interval $$ there exists at least one point $$ in which the derivative of the function $$ is zero:

Proof.

If the function $$ is constant on the interval $$ then the derivative is zero at any point of the interval $$ i.e. in this case the statement is true.

If the function $$ is not constant on the interval $$ then by the Weierstrass theorem, it reaches its greatest or least value at some point $$ of the interval $$ i.e. there exists a local extremum at the point $$ Then by Fermat's theorem, the derivative at this point is equal to zero:

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

Solution.

First of all, we need to check that the function $$ satisfies all the conditions of Rolle's theorem.

$$ $$ is continuous in $$ as a quadratic function;

$$ It is differentiable everywhere over the open interval $$

$$ Finally,

So we can use Rolle's theorem.

To find the point $$ we calculate the derivative

and solve the equation $$

Thus, $$ for $$

Solution.

First we determine whether Rolle's theorem can be applied to $$ on the closed interval $$

The function is continuous on the closed interval $$

The function is differentiable on the open interval $$ Its derivative is

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

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

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:

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 $$

Differentiate:

Solve the equation and find the value of $$

Solution.

We factor the polynomial:

It is now easy to see that the function has two zeros: $$ (coincides with the value of $$) and $$

Since the function is a polynomial, it is everywhere continuous and differentiable. So this function satisfies Rolle's theorem on the interval $$ Hence, $$

Solution.

If we consider the auxiliary function

we see that it has the following zeros:

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

It is obvious that the function $$ is everywhere continuous and differentiable as a cubic polynomial. Consequently, it satisfies all the conditions of Rolle's theorem on the interval $$ So $$

Solution.

In addition to $$ the first equation has the root $$ Consequently, the function $$ satisfies the conditions of Rolle's theorem:

The second equation is obtained by differentiating the first equation:

According to Rolle's theorem, there is an interior point $$ on the interval $$ where the derivative is zero. Consequently, the point $$ is a solution of the second equation where $$

Solution.

The given quadratic function has roots $$ and $$ that is

The by Rolle's theorem, there is a point $$ in the interval $$ where the derivative of the function $$ equals zero.

It is equal to zero at the following point $$

It can be seen that the resulting stationary point $$ belongs to the interval $$ (Figure $$).

Solution.

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

Hence, the derivative must be equal to zero at any point $$ Differentiate the given function:

This shows that the derivative is zero at $$ Thus, $$ where $$

Note that this function describes the upper semicircle of radius $$ centered at the origin (Figure $$). It does not have finite derivatives at the endpoints of the interval $$ i.e. this function is not differentiable at $$ This, however, does not preclude the application of Rolle's theorem, because the latter requires the function to be differentiable on the open interval $$