Solution

The given function, $f(x)=x^{2}+2 x-8$, being a polynomial function, is continuous in $[-4$, $2]$ and is differentiable in $(-4,2)$.

$f(-4)=(-4)^{2}+2 \times(-4)-8=16-8-8=0$

$f(2)=(2)^{2}+2 \times 2-8=4+4-8=0$

$\therefore f(-4)=f(2)=0$

$\Rightarrow$ The value of $f(x)$ at -4 and 2 coincides.

Rolle’s Theorem states that there is a point $c \in(-4,2)$ such that $f^{\prime}(c)=0$

$f(x)=x^{2}+2 x-8$

$\Rightarrow f^{\prime}(x)=2 x+2$

$\therefore f^{\prime}(c)=0$

$\Rightarrow 2 c+2=0$

$\Rightarrow c=-1$, where $c=-1 \in(-4,2)$

Hence, Rolle’s Theorem is verified for the given function.

Solution

By Rolle’s Theorem, for a function $f:[a, b] \to \mathbf{R}$, if

(a) $f$ is continuous on $[a, b]$

(b) $f$ is differentiable on $(a, b)$

(c) $f(a)=f(b)$

then, there exists some $c \in(a, b)$ such that $f^{\prime}(c)=0$

Therefore, Rolle’s Theorem is not applicable to those functions that do not satisfy any of the three conditions of the hypothesis.

(i) $f(x)=[x]$ for $x \in[5,9]$

It is evident that the given function $f(x)$ is not continuous at every integral point.

In particular, $f(x)$ is not continuous at $x=5$ and $x=9$

$\Rightarrow f(x)$ is not continuous in $[5,9]$.

Also, $f(5)=[5]=5$ and $f(9)=[9]=9$

$\therefore f(5) \neq f(9)$

The differentiability of $f$ in $(5,9)$ is checked as follows.

Let $n$ be an integer such that $n \in(5,9)$.

The left hand limit of $f$ at $x=n$ is,

$\lim _{h \to 0^{-}} \dfrac{f(n+h)-f(n)}{h}=\lim _{h \to 0^{-}} \dfrac{[n+h]-[n]}{h}=\lim _{h \to 0^{-}} \dfrac{n-1-n}{h}=\lim _{h \to 0^{-}} \dfrac{-1}{h}=\infty$

The right hand limit of $f$ at $x=n$ is,

$\lim _{h \to 0^{+}} \dfrac{f(n+h)-f(n)}{h}=\lim _{h \to 0^{+}} \dfrac{[n+h]-[n]}{h}=\lim _{h \to 0^{+}} \dfrac{n-n}{h}=\lim _{h \to 0^{+}} 0=0$

Since the left and right hand limits of $f$ at $x=n$ are not equal, $f$ is not differentiable at $x$ $=n$

$\therefore f$ is not differentiable in $(5,9)$.

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for $f(x)=[x]$ for $x \in[5,9]$.

The converse of Rolle’s Theorem would state that if there exists some $ c $ in $(a, b)$ such that $ f’(c) = 0 $, then $ f $ must be continuous on $[a, b]$, differentiable on $(a, b)$, and $ f(a) = f(b) $.

However, in this case, since $ f(x) = [x] $ is not differentiable at any point in $(5, 9)$, we cannot find any $ c $ in $(5, 9)$ such that $ f’(c) = 0 $.

Therefore, the converse of Rolle’s Theorem does not apply here either.

(ii) $f(x)=[x]$ for $x \in[-2,2]$

It is evident that the given function $f(x)$ is not continuous at every integral point.

In particular, $f(x)$ is not continuous at $x=-2$ and $x=2$

$\Rightarrow f(x)$ is not continuous in $[-2,2]$.

Also, $f(-2)=[-2]=-2$ and $f(2)=[2]=2$

$\therefore f(-2) \neq f(2)$

The differentiability of $f$ in $(-2,2)$ is checked as follows.

Let $n$ be an integer such that $n \in(-2,2)$.

The left hand limit of $f$ at $x=n$ is,

$\lim _{h \to 0^{-}} \dfrac{f(n+h)-f(n)}{h}=\lim _{h \to 0^{-}} \dfrac{[n+h]-[n]}{h}=\lim _{h \to 0^{-}} \dfrac{n-1-n}{h}=\lim _{h \to 0^{-}} \dfrac{-1}{h}=\infty$

The right hand limit of $f$ at $x=n$ is,

$\lim _{h \to 0^{+}} \dfrac{f(n+h)-f(n)}{h}=\lim _{h \to 0^{+}} \dfrac{[n+h]-[n]}{h}=\lim _{h \to 0^{+}} \dfrac{n-n}{h}=\lim _{h \to 0^{+}} 0=0$

Since the left and right hand limits of $f$ at $x=n$ are not equal, $f$ is not differentiable at $x$ $=n$

$\therefore f$ is not differentiable in $(-2,2)$.

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for $f(x)=[x]$ for $x \in[-2,2]$.

For the given function $ f(x) = [x] $ on the interval $[-2, 2]$, we observe that:

• The function is not continuous at integer points.
• The function is not differentiable at integer points.
• The function values at the endpoints are not equal.

Therefore, neither Rolle’s Theorem nor its converse can be applied to this function. The function $ f(x) = [x] $ serves as an example where the conditions of Rolle’s Theorem are not met, and thus, the theorem does not hold.

(iii) $f(x)=x^{2}-1$ for $x \in[1,2]$

It is evident that $f$, being a polynomial function, is continuous in $[1,2]$ and is differentiable in $(1,2)$.

$f(1)=(1)^{2}-1=0$

$f(2)=(2)^{2}-1=3$

$\therefore f(1) \neq f(2)$

It is observed that $f$ does not satisfy a condition of the hypothesis of Rolle’s Theorem.

Hence, Rolle’s Theorem is not applicable for $f(x)=x^{2}-1$ for $x \in[1,2]$.

The converse of Rolle’s theorem is not generally true. That is, if there exists some $ c \in (a, b) $ such that $ f’(c) = 0 $, it does not necessarily imply that $ f(a) = f(b) $.

Let’s find the derivative of $ f(x) $:

$ f’(x) = 2x $

Now, let’s check if there is any $ c \in (1, 2) $ such that $ f’(c) = 0 $:

$ f’(c) = 2c = 0 $

$\Rightarrow c = 0 $

However, $ c = 0 $ is not in the interval $(1, 2)$. Therefore, there is no $ c \in (1, 2) $ such that $ f’(c) = 0 $.

Solution

It is given that $f:[-5,5] \to \mathbf{R}$ is a differentiable function.

Since every differentiable function is a continuous function, we obtain

(a) $f$ is continuous on $[-5,5]$.

(b) $f$ is differentiable on $(-5,5)$.

Therefore, by the Mean Value Theorem, there exists $c \in(-5,5)$ such that

$ \begin{aligned} & f^{\prime}(c)=\dfrac{f(5)-f(-5)}{5-(-5)} \\ & \Rightarrow 10 f^{\prime}(c)=f(5)-f(-5) \end{aligned} $

It is also given that $f^{\prime}(x)$ does not vanish anywhere.

$ \begin{aligned} & \therefore f^{\prime}(c) \neq 0 \\ & \Rightarrow 10 f^{\prime}(c) \neq 0 \\ & \Rightarrow f(5)-f(-5) \neq 0 \\ & \Rightarrow f(5) \neq f(-5) \end{aligned} $

Hence, proved.

Solution

The given function is $f(x)=x^{2}-4 x-3$

$f$, being a polynomial function, is continuous in $[1,4]$ and is differentiable in $(1,4)$

whose derivative is $2 x-4$.

$f(1)=1^{2}-4 \times 1-3=-6, f(4)=4^{2}-4 \times 4-3=-3$

$\therefore \dfrac{f(b)-f(a)}{b-a}=\dfrac{f(4)-f(1)}{4-1}=\dfrac{-3-(-6)}{3}=\dfrac{3}{3}=1$

Mean Value Theorem states that there is a point $c \in(1,4)$ such that $f^{\prime}(c)=1$

$f^{\prime}(c)=1$

$\Rightarrow 2 c-4=1$

$\Rightarrow c=\dfrac{5}{2}$, where $c=\dfrac{5}{2} \in(1,4)$

Hence, Mean Value Theorem is verified for the given function.

Solution

The given function $f$ is $f(x)=x^{3}-5 x^{2}-3 x$

$f$, being a polynomial function, is continuous in $[1,3]$ and is differentiable in $(1,3)$

whose derivative is $3 x^{2}-10 x-3$.

$f(1)=1^{3}-5 \times 1^{2}-3 \times 1=-7, f(3)=3^{3}-5 \times 3^{2}-3 \times 3=-27$

$\therefore \dfrac{f(b)-f(a)}{b-a}=\dfrac{f(3)-f(1)}{3-1}=\dfrac{-27-(-7)}{3-1}=-10$

Mean Value Theorem states that there exist a point $c \in(1,3)$ such that $f^{\prime}(c)=-10$

$ \begin{aligned} & f^{\prime}(c)=-10 \\ & \Rightarrow 3 c^{2}-10 c-3=-10 \\ & \Rightarrow 3 c^{2}-10 c+7=0 \\ & \Rightarrow 3 c^{2}-3 c-7 c+7=0 \\ & \Rightarrow 3 c(c-1)-7(c-1)=0 \\ & \Rightarrow(c-1)(3 c-7)=0 \\ & \Rightarrow c=1, \dfrac{7}{3}, \text{ where } c=\dfrac{7}{3} \in(1,3) \end{aligned} $

Hence, Mean Value Theorem is verified for the given function and $c=\dfrac{7}{3} \in(1,3)$ is the only point for which $f^{\prime}(c)=0$

Solution

Mean Value Theorem states that for a function $f:[a, b] \to \mathbf{R}$, if

(a) $f$ is continuous on $[a, b]$

(b) $f$ is differentiable on $(a, b)$

then, there exists some $c \in(a, b)$ such that $f^{\prime}(c)=\dfrac{f(b)-f(a)}{b-a}$

Therefore, Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis. (i) $f(x)=[x]$ for $x \in[5,9]$

It is evident that the given function $f(x)$ is not continuous at every integral point.

In particular, $f(x)$ is not continuous at $x=5$ and $x=9$

$\Rightarrow f(x)$ is not continuous in $[5,9]$.

The differentiability of $f$ in $(5,9)$ is checked as follows.

Let $n$ be an integer such that $n \in(5,9)$.

The left hand limit of $f$ at $x=n$ is,

$\lim _{h \to 0^{-}} \dfrac{f(n+h)-f(n)}{h}=\lim _{h \to 0^{-}} \dfrac{[n+h]-[n]}{h}=\lim _{h \to 0^{-}} \dfrac{n-1-n}{h}=\lim _{h \to 0} \dfrac{-1}{h}=\infty$

The right hand limit of $f$ at $x=n$ is,

$\lim _{h \to 0^{+}} \dfrac{f(n+h)-f(n)}{h}=\lim _{h \to 0^{+}} \dfrac{[n+h]-[n]}{h}=\lim _{h \to 0^{+}} \dfrac{n-n}{h}=\lim _{h \to 0^{+}} 0=0$

Since the left and right hand limits of $f$ at $x=n$ are not equal, $f$ is not differentiable at $x$ $=n$

$\therefore f$ is not differentiable in $(5,9)$.

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for $f(x)=[x]$ for $x \in[5,9]$.

(ii) $f(x)=[x]$ for $x \in[-2,2]$

It is evident that the given function $f(x)$ is not continuous at every integral point.

In particular, $f(x)$ is not continuous at $x=-2$ and $x=2$

$\Rightarrow f(x)$ is not continuous in $[-2,2]$.

The differentiability of $f$ in $(-2,2)$ is checked as follows.

Let $n$ be an integer such that $n \in(-2,2)$.

The left hand limit of $f$ at $x=n$ is,

$\lim _{h \to 0^{-}} \dfrac{f(n+h)-f(n)}{h}=\lim _{h \to 0^{-}} \dfrac{[n+h]-[n]}{h}=\lim _{h \to 0^{-}} \dfrac{n-1-n}{h}=\lim _{h \to 0^{-}} \dfrac{-1}{h}=\infty$

The right hand limit of $f$ at $x=n$ is,

$\lim _{h \to 0^{+}} \dfrac{f(n+h)-f(n)}{h}=\lim _{h \to 0^{+}} \dfrac{[n+h]-[n]}{h}=\lim _{h \to 0^{+}} \dfrac{n-n}{h}=\lim _{h \to 0^{+}} 0=0$

Since the left and right hand limits of $f$ at $x=n$ are not equal, $f$ is not differentiable at $x$ $=n$

$\therefore f$ is not differentiable in $(-2,2)$.

It is observed that $f$ does not satisfy all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is not applicable for $f(x)=[x]$ for $x \in[-2,2]$.

(iii) $f(x)=x^{2}-1$ for $x \in[1,2]$

It is evident that $f$, being a polynomial function, is continuous in $[1,2]$ and is differentiable in $(1,2)$.

It is observed that $f$ satisfies all the conditions of the hypothesis of Mean Value Theorem.

Hence, Mean Value Theorem is applicable for $f(x)=x^{2}-1$ for $x \in[1,2]$.

It can be proved as follows.

$f(1)=1^{2}-1=0, f(2)=2^{2}-1=3$

$\therefore \dfrac{f(b)-f(a)}{b-a}=\dfrac{f(2)-f(1)}{2-1}=\dfrac{3-0}{1}=3$

$f^{\prime}(x)=2 x$

$\therefore f^{\prime}(c)=3$

$\Rightarrow 2 c=3$

$\Rightarrow c=\dfrac{3}{2}=1.5$, where $1.5 \in[1,2]$