$a = 1$ અને $b = 4$ લઈ વિધેય $f(x)=x^{2}-4 x-3$ માટે $[a, b]$ પર મધ્યકમાન પ્રમેય ચકાસો.

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

$f,$ being a polynomial function, is a continuous in $[1,4]$ and is differentiable in $(1,4)$ whose derivative is $2 x-4$

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

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

Mean Value Theorem states that there is a point $c \in(1,4)$ such that

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

$\Rightarrow 2 c-4=1$

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

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