Rolle’s Theorem, Proof And Examples

Rolle's Theorem, Proof And Examples

Rolle’s Theorem Statement: let a function f be

  1. Continuous on the interval \left[a\;,\;b\right].
  2. Differentiable on the open interval \rbrack a,b\lbrack .
  3. f(a)=f(b)

then there exist at least one point c\in\rbrack a,b\lbrack such that f'(c)=0.

Proof:

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].

Rolle's Theorem, Rolle theorem

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

f(c+h)\;\leq\;f(c)

and f(c-h)\;\leq\;f(c)

Since f(c)=M=sup\;f on \left[a\;,\;b\right].

Hence

\frac{f(c+h)\;-\;f(c)}h\;\leq\;0....................(2)

and

\frac{f(c-h)\;-\;f(c)}h\;\leq\;0....................(3)

i.e,\;\;\;\;\;\;\;\;\;\;\frac{f(c+(-h))-\;f(c)}h\geq\;0

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.

Example:

Verify Rolle’s Theorem for

f(x)=1-x^\frac23 on \left[-1\;,\;1\right]

Solution:

If fis continuous on \left[-1\;,\;1\right] and f(-1)=0=f(1)

f'(x)=\frac{-2}3\;x^\frac{-1}3

f'(x)=\;\frac{-2\;x^{\displaystyle\frac{-1}3}}3

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.

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Azhar Ali

Azhar Ali

I graduated in Mathematics from the University of Sargodha, having master degree in Mathematics.

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