माध्यमान प्रमेय सत्यापित कीजिए यदि अंतराल $[a, b]$ में $f(x)=x^{3}-5 x^{2}-3 x,$ जहाँ $a=1$ और $b=3$ है। $f(c)=0$ के लिए $c \in(1,3)$ को ज्ञात कीजिए।

The given function $f$ is $f(x)=x^{2}-5 x^{2}-3 x$

$f,$ being a polynomial function, is continuous in $[1,3],$ and is differentiable in $(1,3)$

Whose derivative is $3 x^{2}-10 x-3$

$f(1)=1^{2}-5 \times 1^{2}-3 \times 1=-7, f(3)=3^{3}-3 \times 3=27$

$\therefore \frac{f(b)-f(a)}{b-a}=\frac{f(3)-f(1)}{3-1}=\frac{-27-(-7)}{3-1}=-10$

Mean Value Theorem states that there exist a point $c \in(1,3)$ such that $f^{\prime}(c)=-10$

$f^{\prime}(c)=-10$

$\Rightarrow 3 c^{2}-10 c-3=10$

$\Rightarrow 3 c^{2}-10 c+7=0$

$\Rightarrow 3 c^{2}-3 c-7 c+7=0$

$\Rightarrow 3 c(c-1)-7(c-1)=0$

$\Rightarrow(c-1)(3 c-7)=0$

$\Rightarrow c=1, \frac{7}{3}$ where $c=\frac{7}{3} \in(1,3)$

Hence, Mean Value Theorem is verified for the given function and $c=\frac{7}{3} \in(1,3)$ is the only point for which $f^{\prime}(c)=0$