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.
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'(x) (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, it follows that f'(c) = 0.
Rolle’s theorem: Solved Examples
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
Since f(x) is a polynomial, it is continuous and differentiable everywhere. In addition f(-2)=0=f(0) Therefore, f(x) satisfies three hypothesis of Rolle's theorem.
Since, given function is \( f(x)=x^2+2x\)
Differentiating the function w.r. to x, we get \( f'(x) = 2x+2\)
or \( f'(x) = 2(x+1)\)
According to Rolle's theorem, there exists at least one value \( c\in (-2,0)\) such that f′(c)=0
So, \( f′(c)=0\) We see that \(2(c+1)=0 \) or
c=-1 The figure is shown below
Since f(x) is a polynomial, it is continuous and differentiable everywhere. In addition f(-2)=0=f(2) Therefore, f(x) satisfies the criteria of Roll’s theorem.
Since, given function is \( f(x)=x^3-4x\)
Differentiating the function w.r. to x, we get \( f'(x) = 3x^2-4\)
According to Rolle's theorem, there exists at least one value \( c\in (-2,0)\) such that f′(c)=0
So, \( f′(c)=0\) We see that \(3c^2-4=0 \) or
\( c=\pm 1.15\)
Both points are in the interval [−2,2], and, therefore, both points satisfy the conclusion.
The figure is shown below.
\( f(x)=x^2-4x+1\) over [0,4]
\( f(x)=x^3-3x^2+2x+5\) over [0,2]
\( f(x)=x^{2/3}\) over [0,1]
\( f(x)=x^{2/3}\) over [-1,8]
\( f(x)=x+\frac{1}{x}\) over [0.5,2]
\( f(x)=\sin 2 x\) over [-1,1]
\( f(x)=x \sqrt{x+6}\) over [-6,0]
\( f(x)= \sqrt{x-1}\) over [1,3]
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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