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Q.
For the function $f(x)=x^2-4 x-3, x \in(a, b)$, where $a=1$ and $b=1$, the value of $c$ for mean value theorem where $c \in(a, b)$ is
Continuity and Differentiability
Solution:
(i) $f(x)$ is continuous in $[a, b]$
(ii) $f(x)$ is differentiable in $(a, b)$.
Then, there will be atleast one value of $c \in(a, b)$ such that $f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}$.
Here, $f(x)=x^2-4 x-3, x \in[1,4]$
which is a polynomial function, so it is continuous and derivable at all $x \in R$, therefore
(i) $f(x)$ is continuous on $[1,4]$.
(ii) $f(x)$ is derivable on $(1,4)$.
$\therefore$ Conditions of Lagrange's theorem are satisfied on $[1,4]$.
Hence, there is atleast one real number $c \in(1,4)$. Such that
$ f^{\prime}(c)=\frac{f(4)-f(1)}{4-1} \left(\therefore f^{\prime}(c)=-\frac{f(b)-f(a)}{b-a}\right) $
$\Rightarrow 2 c-4=\frac{\left(4^2-4 \times 4-3\right)-\left(1^2-4 \times 1-3\right)}{4-1}=1 $
$\Rightarrow {\left[\because f^{\prime}(x)=\frac{d}{d x}\left(x^2-4 x-3\right)=2 x-4\right]}$
$ 2 c-4=1 \Rightarrow c=\frac{5}{2} \in(1,4)$
$\therefore$ MVT is verified for $f(x)$ in $[1,4]$.