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.

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Solved Problems

Click a problem to see the solution.

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 \(c = – 1.\)

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 \(c:\)

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

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 \(c:\)

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

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, \(b = 1.\)

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 \(b = 2.\)