Rolle’s Theorem

Rolle’s theorem states that if a function $ƒ(x)$ is

continuous in the closed interval $[a,b]$
differentiable in the open interval $(a,b)$
$ƒ(a)=ƒ(b)$

then there exists at least one point $c\in (a,b)$ such that $ƒ'(c)=0$.

Geometrical Interpretation of Rolle’s theorem

The first condition of Rolle’s theorem says that the function $ƒ(x)$ has a continuous graph in the interval $a≤x≤b$. By second condition, the graph has tangents at every point in $a<x<b$ and the last condition is $ƒ(a)=ƒ(b)$ i.e. the ordinates at $x=a$ and $x=b$ are equal. If all the conditions of Rolle’s theorem are satisfied, then there exists at least one point on the graph $(a<c<b)$ where the tangent is parallel to $\text{X-axis}$. If any one of the conditions of Rolle’s theorem is not satisfied, then the conclusion of Rolle’s theorem will not be true.

Verify Rolle’s theorem for the function $ƒ(x)=3x^2-4$ in $[-1,1]$.

Here, \[ƒ(x)=3x^2-4\]

Since ƒ(x) is polynomial, it is continuous in the closed interval $[-1,1]$.
$ƒ'(x)=6x$ which exists for all $x\in(-1,1)$. Hence, $ƒ(x)$ is differentiable in open interval $(-1,1)$.
\[ƒ(-1)=3(-1)^2-4=-1\] \[ƒ(1)=3(1)^2-4=-1\] \[\therefore ƒ(-1)=ƒ(1)\]

Hence, all the conditions of Rolle’s theorem are satisfied. So, there exists at least a point $c\in(-1,1)$ such that \[ƒ'(c)=0\] \[6c=0\] \[c=0\in(-1,1)\] Therefore, Rolle’s theorem is verified.

Verify Rolle’s theorem for the function $ƒ(x)=x(x-1)^2$ in $[0,1]$.

Here, \[ƒ(x)=x(x-1)^2=x^3-2x^2+1\]

Since ƒ(x) is polynomial, it is continuous in the closed interval $[0,1]$.
$ƒ'(x)=3x^2-4x+1$ which exists for all $x\in(0,1)$. Hence, $ƒ(x)$ is differentiable in open interval $(0,1)$.
\[ƒ(0)=0\] \[ƒ(1)=0\] \[\therefore ƒ(0)=ƒ(1)\]

Hence, all the conditions of Rolle’s theorem are satisfied. So, there exists at least a point $c\in(0,1)$ such that \[ƒ'(c)=0\] \[3c^2-4c+1=0\] \[(3c-1)(c-1)=0\] \[\text{Either, }c=\frac{1}{3}\text{ or }c=1\]\[c=\frac{1}{3}\in(0,1)\] Therefore, Rolle’s theorem is verified.

Verify Rolle’s theorem for the function $ƒ(x)=\cos 2x$, $x\in[-π,π]$.

Here, \[ƒ(x)=\cos 2x\]

$ƒ(x)=\cos 2x$ which exists for all $x\in[-π,π]$. So, $ƒ(x)$ is continuous in the closed interval $[-π,π]$.
$ƒ'(x)=-2\sin 2x$ which exists for all $x\in(-π,π)$. Hence, $ƒ(x)$ is differentiable in open interval $(-π,π)$.
\[ƒ(-π)=\cos 2(-π)=\cos 2π=1\] \[ƒ(π)=\cos 2π=1\] \[\therefore ƒ(-π)=ƒ(π)\]

Hence, all the conditions of Rolle’s theorem are satisfied. So, there exists at least a point $c\in(-π,π)$ such that \[ƒ'(c)=0\] \[-2\sin 2c=0\] \[\sin 2c=0\] \[\sin 2c=\sin π\] \[2c=π\] \[c=\frac{π}{2}\in(-π,π)\] Therefore, Rolle’s theorem is verified.

Verify Rolle’s theorem for the function $ƒ(x)=\sqrt{25-x^2}$, $x\in[-5,5]$.

Here, \[ƒ(x)=\sqrt{25-x^2}\]

$ƒ(x)=\sqrt{25-x^2}$ which exists for all $x\in[-5,5]$. So, $ƒ(x)$ is continuous in the closed interval $[-5,5]$.
$ƒ'(x)=\frac{-x}{\sqrt{25-x^2}}$ which exists for all $x\in(-5,5)$. Hence, ƒ(x) is differentiable in open interval $(-5,5)$.
\[ƒ(-5)=\sqrt{25-(-5)^2}=0\] \[ƒ(5)=\sqrt{25-5^2}=0\] \[\therefore ƒ(-5)=ƒ(5)\]

Hence, all the conditions of Rolle’s theorem are satisfied. So, there exists at least a point $c\in(-5,5)$ such that \[ƒ'(c)=0\] \[\frac{-c}{\sqrt{25-c^2}}=0\]\[c=0\in(-5,5)\] Therefore, Rolle’s theorem is verified.

Using Rolle’s theorem, find a point on the curve represented by the function $ƒ(x)=6x-x^2$ in $[0,6]$.

Here, \[ƒ(x)=6x-x^2\]

$ƒ(x)$ is polynomial. So, it is continuous in the closed interval $[0,6]$.
$ƒ'(x)=6-2x$ which exists for all $x\in(0,6)$. Hence, $ƒ(x)$ is differentiable in open interval $(0,6)$.
\[ƒ(0)=0\] \[ƒ(6)=6×6-6^2=0\] \[\therefore ƒ(0)=ƒ(6)\]

Hence, all the conditions of Rolle’s theorem are satisfied. So, there exists at least a point $c\in(0,6)$ such that \[ƒ'(c)=0\] \[6-2c=0\]\[c=3\in(0,6)\] $c=3$ is the $\text{x-coordinate}$ of the point at which the tangent drawn is parallel to the $\text{X-axis}$. \[\therefore ƒ(3)=6×3-3^2=9\] Hence, the required point is $(3,9)$

Using Rolle’s theorem, find a point on the curve represented by the function $ƒ(x)=\sin x$, $x\in[0,π]$.

Here, \[ƒ(x)=\sin x\]

$ƒ(x)=\sin x$ which exists for all $x\in[0,π]$. So, $ƒ(x)$ is continuous in the closed interval $[0,π]$. [Trigonometric Functions]
$ƒ'(x)=\cos x$ which exists for all $x\in(0,π)$. Hence, $ƒ(x)$ is differentiable in open interval $(0,π)$.
\[ƒ(0)=\sin 0=0\] \[ƒ(π)=\sin π=0\] \[\therefore ƒ(0)=ƒ(π)\]

Hence, all the conditions of Rolle’s theorem are satisfied. So, there exists at least a point $c\in(0,π)$ such that \[ƒ'(c)=0\] \[\cos c=0\] \[\cos c=\cos\frac{π}{2}\] \[c=\frac{π}{2}\in(0,π)\] $c=\frac{π}{2}$ is the $\text{x-coordinate}$ of the point at which the tangent drawn is parallel to the $\text{X-axis}$. \[\therefore ƒ\left(\frac{π}{2}\right)=\sin\frac{π}{2}=1\] Hence, the required point is $(\frac{π}{2},1)$

Show that Rolle’s theorem can not be applied for the function $ƒ(x)=1-x^{\frac{2}{3}}$ in $[-1,1]$.

Here, \[ƒ(x)=1-x^{\frac{2}{3}}\]

$ƒ(x)=1-x^{\frac{2}{3}}$ exists for all $x\in[-1,1]$. So, it is continuous in $[-1,1]$.
$ƒ'(x)=\frac{-2}{3x^{\frac{1}{3}}}$ which does not exist at $x=0\in(-1,1)$.

Since second condition is not satisfied, Rolle’s theorem is not applicable for $ƒ(x)$ in $[-1,1]$.