# Examine the applicability of Mean Value Theorem

Question:

Examine the applicability of Mean Value Theorem for all three functions given in the above exercise 2.

Solution:

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

(a) f is continuous on [ab]

(b) f is differentiable on (ab)

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

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,

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

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

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

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

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 = −2 and = 2

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

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

$\lim _{h \rightarrow 0} \frac{f(n+h)-f(n)}{h}=\lim _{h \rightarrow 0} \frac{[n+h]-[n]}{h}=\lim _{h \rightarrow 0} \frac{n-1-n}{h}=\lim _{h \rightarrow 0} \frac{-1}{h}=\infty$

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

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

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

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.

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

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

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

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

$\Rightarrow 2 c=3$

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

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