Rolle's Theorem

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Statement: If f(x) is
(i) continuous in the closed interval [a, b];
(ii) differentiable in the open interval (a, b).
(iii) f(a)=f(b)
Then, there exists at least one number \(c\in (a, b)\) such that f '(c) = 0.

Geometrical Meaning of Roll’s Theorem

Let f(x) be a continuous function in an interval [a,b] with A=f(a) and B=f(b).
Then Role’s Theorem asserts that there exist at least one point c lying between A and B such that the tangent at that point is parallel to x-axis.
There may exist more than one points \(c_1, c_2\) and may be more between A and B, so that the tangents at those points are parallel to x-axis

This function \(f(x)=-x^2+4x+5\) meets all the criteria for Rolle's Theorem in the interval [-2,6]. It's continuous on the interval [-2, 6], differentiable in the interval (-2, 6), and f(-2) equals f(6), both equaling to each other.
As expected, there exists at least one point where the derivative is zero. In this case, it happens at x = 2, where the tangent line is horizontal.



This function \(f(x)=(x+4)^2-4(x+4)+4\) meets all the criteria for Rolle's Theorem in the interval [-6,2]. It's continuous on the interval [-6,2], differentiable in the interval (-6,2), and f(-6) equals f(2), both equaling to each other.
As expected, there exists at least one point where the derivative is zero. In this case, it happens at x = -2, where the tangent line is horizontal.



This function \(f(x)=-(x-2)^2+4(x-2)+5\) meets all the criteria for Rolle's Theorem in the interval [1,7]. It's continuous on the interval [1,7], differentiable in the interval (1,7), and f(1) equals f(7), both equaling to each other.
As expected, there exists at least one point where the derivative is zero. In this case, it happens at x = 4, where the tangent line is horizontal.



This function \(f(x)=x^3-4x\) meets all the criteria for Rolle's Theorem in the interval [-2,2]. It's continuous on the interval [-2,2], differentiable in the interval (-2,2), and f(-2) equals f(2), both equaling to each other.
As expected, there exists at least one point where the derivative is zero. In this case, it happens at x = -1.15 and x=1.15, where the tangent line is horizontal.

All three conditions of Rolle’s theorem are necessary for the theorem to be true

The condition where Rolle’s theorem fails

The condition where Rolle’s theorem is NOT applicable, of the Rolle’s theorem fails.
Discontinuous Discontinuous Not Differentiable

Rolle’s theorem fails: Counter Examples

Counter Example 1

Consider f(x)={x}, a fractional part function on the closed interval [0,1].
The derivative of the function on the open interval (0,1) is everywhere equal to 1.
Eventhough, the Rolle’s theorem fails because the function f(x)={x} has a discontinuity at x=0 and x=1
that is
the function f(x)={x} is not continuous everywhere on the closed interval [0,1])

Counter Example 2

Consider f(x)=|x|, an absolute value function on the closed interval [-1,1].
This function does not have derivative at x=0.The Rolle’s theorem fails here because f(x) is not differentiable over the whole interval (-1,1).

Counter Example 3

Consider f(x)=x, a function on the closed interval [0,1].
Though f(x) is continuous on the closed interval [-1,1], differentiable in the open interval (0,1), it does not satisfy third condition; i.e., f(0)≠f(1).Thus , the Rolle’s theorem fails .

Rolle’s theorem :History

The Roll's theorem says that, if a continuous curve passes through the same y-value (such as the x-axis) twice and has a unique tangent line (derivative) at every point of the interval, then somewhere between the endpoints it has a tangent parallel to the x-axis. The theorem was proved in 1691 by the French mathematician Michel Rolle, though it was stated without a modern formal proof in the 12th century by the Indian mathematician Bhaskara II.

Rolle’s theorem :Proof

Statement: If f(x) is
(i) continuous in the closed interval [a, b];
(ii) differentiable in the open interval (a, b).
(iii) f(a)=f(b)
Then, there exists at least one number \(c\in (a, b)\) such that f '(c) = 0.

Proof

If f(x) is constant then
f'(c) = 0 for all c∈(a, b)
Suppose there exists x ∈ (a, b) such that
f(x) > f(a)
(A similar argument can be given if f(x) < f(a)).
Then there exists c ∈ (a, b) such that f(c) is a maximum.
Since, f(x) has a local maximum at c ∈(a, b), For small (enough) h,
\(f(c + h) \le f(c)\).
If h > 0 then
\( \frac{f(c + h)-f(c)}{h} \le 0\).
Similarly,
If h < 0, then
\(\frac{f(c + h)-f(c)}{h} \ge 0\).
By elementary properties of the limit, taking limit as \(h \to 0\), we get
f'(c) = 0.

Rolle’s theorem: Solved Examples

  1. Veryfy that the function satisfy the three hypothesis of Rolle's theorem, and find all numbers c that satisfy the conclusion of Rolle's theorem
    1. \( f(x)=x^2+2x\) over [-2,0]

      Solution 👉 Click Here

    2. \( f(x)=x^3-4x\) over [-2,2]

      Solution 👉 Click Here

    3. \( f(x)=x^2-4x+1\) over [0,4]
    4. \( f(x)=x^3-3x^2+2x+5\) over [0,2]
    5. \( f(x)=x^{2/3}\) over [0,1]
    6. \( f(x)=x^{2/3}\) over [-1,8]
    7. \( f(x)=x+\frac{1}{x}\) over [0.5,2]
    8. \( f(x)=\sin 2 x\) over [-1,1]
    9. \( f(x)=x \sqrt{x+6}\) over [-6,0]
    10. \( f(x)= \sqrt{x-1}\) over [1,3]
  2. For what value of a,m, and b does the function satisfy Roll's theorem in the interval [0,2]
    \( f(x)= \begin{cases} 3 & x =0\\ -x^2+3x+a & 0 < x < 1\\ mx+b &1 \le x \le 2 \end{cases} \)

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