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