Calculus
Applications of the Derivative
Rolle’s Theorem
Problems 1-2
Problems 3-7
Rolle’s theorem states that any real differentiable function that attains equal values at the endpoints of an interval must have in this interval at least one stationary point, that is a point, at which the first derivative is zero. Geometrically, this means that the tangent to the graph of the function at this point is horizontal (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.
Figure 1.
Figure 2.
In a strict form this theorem was proved in \(1691\) by the French mathematician Michel Rolle \(\left(1652-1719\right)\) (Figure \(2\)).
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 \(\xi \in \left( {a,b} \right),\) in which the derivative of the function \(f\left( x \right)\) is zero:
\[f’\left( \xi \right) = 0.\]
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 \(\xi\) of the interval \(\left( {a,b} \right),\) i.e. there exists a local extremum at the point \(\xi.\) Then by Fermat’s theorem, the derivative at this point is equal to zero:
\[f’\left( \xi \right) = 0.\]
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 on problem description to see solution.
✓ Example 1
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 2
Check the validity of Rolle’s theorem for the function
\[f\left( x \right) = {x^2} – 6x + 8.\]
✓ Example 3
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 4
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 5
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 6
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 7
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 1.
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 \(\xi \lt {x_0}.\)
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 2.
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.
Calculate the derivative:
\[f’\left( x \right) = {\left( {{x^2} – 6x + 8} \right)^\prime } = 2x – 6.\]
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 \(3\)).
Figure 3.
Problems 1-2
Problems 3-7