Question

Verify Mean Value Theorem, if $f\left(x\right)=x^2-4x-3$ in the interval $[a,b]$ , where $a=1$ and $b=4$

Answer 1

Verify conditions :

$f\left(x\right)=x^2-4x-3,x\in [1,4]$

Mean value theorem is satisfied if

$\left(i\right)f\left(x\right)$ is continuous in $[a,b]$

$f\left(x\right)=x^2-4x-3$ is a polynomial of degree ‘two’. so, $f\left(x\right)$ is continuous in $[1,4]$

$\left(ii\right)f\left(x\right)$ is differentiable in $(a,b)$

$f\left(x\right)=x^2-4x-3$ is a polynomial of degree ‘two’. so, $f\left(x\right)$ is differentiable in $(1,4)$

Hence, Function is satisfying the conditions of Mean value theorem.
Applying Mean value theorem
$f\left(x\right)=x^2-4x-3$

$f^{\prime }\left(x\right)=2x-4-0$
Putting $x=c,f^{\prime }\left(c\right)=2c-4$
$f\left(1\right)=(1{)}^2-4(1)-3$

$\Rightarrow f\left(1\right)=1-4-3=-6$

$f\left(4\right)=(4{)}^2-4(4)-3$

$\Rightarrow f\left(4\right)=16-16-3=-3$

By mean value theorem

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

$\Rightarrow 2c-4=\frac{f\left(4\right)-f\left(1\right)}{4-1}$

$\Rightarrow 2c-4=\frac{-3-(-6)}{3}$

$\Rightarrow 2c-4=\frac{3}{3}$
$\Rightarrow 2c-4=1$
$\Rightarrow 2c=1+4$
$\Rightarrow 2c=5$
$\Rightarrow c=\frac{5}{2}$

and $c\in (1,4)$

Thus, mean value theorem is verified