Rolle’s Theorem Statement: let a function [katex]f[/katex] be
- Continuous on the interval [katex]\left[a\;,\;b\right][/katex].
- Differentiable on the open interval [katex]\rbrack a,b\lbrack [/katex].
- [katex]f(a)=f(b)[/katex]
then there exist at least one point [katex]c\in\rbrack a,b\lbrack[/katex] such that [katex]f'(c)=0[/katex].
Proof:
since [katex]f[/katex] is contineous on [katex]\left[a\;,\;b\right][/katex] , it is bounded their attains its bounds.
Let [katex]M[/katex] and [katex]m[/katex] be the supremum and infimum of [katex]f[/katex] on [katex]\left[a\;,\;b\right][/katex].
then their either [katex]M=m[/katex] or [katex]M\neq m[/katex]
If [katex]M=m[/katex] then the funtion is constant on [katex]\left[a\;,\;b\right][/katex] so its derivatives vanishes at each point of the interval.
Hence the theorem is true in this case.
If [katex]M \neq m[/katex] at least one of them is different from the equal values [katex]f(a)[/katex] and [katex]f(b)[/katex].
Suppose [katex]M \neq f(a)=f(b)…………….(1)[/katex]
Since [katex]f[/katex] attains its supremum on [katex]\left[a\;,\;b\right][/katex], there is a point [katex]c\in\left[a\;,\;b\right][/katex] such that [katex]f(c)=M[/katex] , But then because of [katex](1), \;c[/katex] must be diffrent from [katex]a[/katex] and [katex]b[/katex]. Thus [katex]c\in\rbrack a,b\lbrack[/katex].
Let [katex]h[/katex] be a positive real number such that [katex]c-h[/katex] and [katex]c+h[/katex] both lie in [katex]c\in\rbrack a,b\lbrack[/katex]. Then
[katex]f(c+h)\;\leq\;f(c)[/katex]
and [katex]f(c-h)\;\leq\;f(c)[/katex]
Since [katex]f(c)=M=sup\;f[/katex] on [katex]\left[a\;,\;b\right][/katex].
Hence
[katex]\frac{f(c+h)\;-\;f(c)}h\;\leq\;0………………..(2)[/katex]
and
[katex]\frac{f(c-h)\;-\;f(c)}h\;\leq\;0………………..(3)[/katex]
[katex]i.e,\;\;\;\;\;\;\;\;\;\;\frac{f(c+(-h))-\;f(c)}h\geq\;0[/katex]
Taking [katex]limit\;h\rightarrow0[/katex], we obtain from [katex](2)[/katex] and [katex](3)[/katex]
[katex]f'(c)\leq\;0 \;\;and\;\; f'(c)\geq\;0[/katex]
showing that [katex]f'(c)=0[/katex].
Geometrical Interpretation of Rolle’s Theorem:
Rolle’s Theorem has a simple geometrical interpretation. If f is contineous on [katex]\left[a\;,\;b\right][/katex] and Differentiable on the open interval [katex]\rbrack a,b\lbrack [/katex] such that [katex]f(a)=f(b)[/katex] then there is a point [katex]c\in\rbrack a,b\lbrack [/katex] where the tangent line to the graph of [katex]y=f(x)[/katex] is parallel to the [katex]x-axis[/katex]. There may be more then one point on the graph where the tangent lines are parallel to the [katex]x-axis[/katex] as in figure.
for physical illustration of Rolles theorem, let a stone by thrown by the ground into the air. Suppose the Hight of the stone after time [katex]t[/katex] is [katex]s=f(t)[/katex] The stone will hit the ground after some time [katex]T[/katex]. Then clearly [katex]f(0)=0 and f(T)=0[/katex] the function [katex]s=f(t)[/katex] satisfies the condition of Rolles Theorem on the interval [katex]\left[0\;,\;t\right][/katex]. Hence the certain [katex]t_0 \in \rbrack 0,T\lbrack [/katex], the velocity of the stone is zero [katex]i.e,\;\; f'(t_0)=0[/katex]. We know that it indeed happens.
Example:
Verify Rolle’s Theorem for
[katex]f(x)=1-x^\frac23[/katex] on [katex]\left[-1\;,\;1\right][/katex]
Solution:
If [katex]f[/katex]is continuous on [katex]\left[-1\;,\;1\right][/katex] and [katex]f(-1)=0=f(1)[/katex]
[katex]f'(x)=\frac{-2}3\;x^\frac{-1}3[/katex]
[katex]f'(x)=\;\frac{-2\;x^{\displaystyle\frac{-1}3}}3[/katex]
Thus [katex]f'(x)=\;\frac20=[/katex] intermediate and so [katex]f[/katex] is not differentiable at [katex]0\in\rbrack -1,1\lbrack [/katex].
Hence Rolles Theorem fail for the given condition.