Rolle’s Theorem, Proof And Examples

Rolle’s Theorem Statement: let a function $f$ be

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

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 $\left[a\;,\;b\right]$ .

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 $f$ is 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.