Question

Examine the applicability of Mean Value Theorem for the following function.

$f\left(x\right)=x^2-1$ for $xϵ[1,2]$

Answer 1

Mean Value Theorem holds for a function $f:[a,b]\to R$, if following two conditions holds

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

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

Then, there exists some $c\in (a,b)$ such that

$f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

Mean Value Theorem is not applicable to those functions that do not satisfy any of the two conditions of the hypothesis.

$f\left(x\right)=x^2-1$ for $x\in$ $[1,2]$
Since, $f\left(x\right)$ is a polynomial function.

Polynomial functions are continuous and differentiable everywhere.

So, $f\left(x\right)$ is continuous in $[1,2]$ and is differentiable in $(1,2)$.
Hence, $f$ satisfies the conditions of the hypothesis of Mean Value Theorem.
Hence, Mean Value Theorem is applicable for given function $f\left(x\right)$
It can be proved as follows.
$f\left(1\right)=1^2-1=0$, $f\left(2\right)=2^2-1=3$
$\therefore \frac{f\left(b\right)-f\left(a\right)}{b-a}=\frac{f\left(2\right)-f\left(1\right)}{2-1}=\frac{3-0}{1}=3$
Also, $f^{\prime }\left(x\right)=2x$
$\therefore f^{\prime }\left(c\right)=3$
$\Rightarrow 2c=3$
$\Rightarrow c=\frac{3}{2}$

$c=1.5$, where $1.5\in [1,2]$