Calculus

Applications of the Derivative

Applications of Derivative Logo

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.

A geometric illustration of Rolle's theorem
Figure 1.

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 \(2\)).

French mathematician Michel Rolle
Fig.2 Michel Rolle (1652-1719)

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 \(\left[ {0,1} \right].\))
    The function y = {x} is not continuous everywhere
    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 \(\left( { - 1,1} \right).\)
    The function y = |x| is not differentiable everywhere
    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 \(f\left( 0 \right) \ne f\left( 1 \right).\))
    The function does not have equal values at the endpoints of the interval [0,1]
    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 \(6\)).

Application of Rolle's theorem to the function y=x^2-6x+8
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 \(\left( { - 1,1} \right).\)

Application of Rolle's theorem to the upper half-circle
Figure 7.

See more problems on Page 2.

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