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 \(\left[ {0,1} \right].\))
    2. The function y = {x} is not continuous everywhere
      Figure 3.
    3. 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 \(\left( { – 1,1} \right).\)
    4. The function y = |x| is not differentiable everywhere
      Figure 4.
    5. 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 \(f\left( 0 \right) \ne f\left( 1 \right).\))
    6. 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 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 \[f{\left( x \right) = {a_0}{x^n} + {a_1}{x^{n – 1}} + \ldots }\kern0pt{\;+\;{a_{n – 1}}x = 0\;\;\;}\] 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 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 9

    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 10

    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 11

    Let \(f\left( x \right) = {x^3} – 3x.\) Find all values of \(c\) in the interval \(\left( { – 1,2} \right)\) such that \(f^\prime\left( c \right) = 0.\)

    Example 12

    Given the function \(f\left( x \right) = {x^3} – 4x + 1.\) Find all values of \(c\) in the interval \(\left( {-2,2} \right)\) such that \(f^\prime\left( c \right) = 0.\)

    Example 13

    Let \(f\left( x \right) = \ln \left( {2 – {x^2}} \right).\) Find all values of \(c\) in the open interval \(\left( { – 1,1} \right)\) such that \(f^\prime\left( c \right) = 0.\)

    Example 14

    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 15

    Determine whether the function \(f\left( x \right) = \sin \large{\frac{x}{2}}\normalsize\) satisfies conditions of Rolle’s theorem for the interval \(\left[ {0,2\pi } \right].\) If so find all numbers \(c\) that satisfy the conclusion of the theorem.

    Example 16

    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 17

    Determine whether the function \(f\left( x \right) = \text{sech}\,x = \large{\frac{2}{{{e^x} + {e^{ – x}}}}}\normalsize\) satisfies conditions of Rolle’s theorem for the interval \(\left[ {-1,1 } \right].\) If so find all numbers \(c\) that satisfy the conclusion of the theorem.

    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 \(b = 2.\)

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    Problems 1-5
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    Problems 6-17