For the function

(b) $\sqrt{3}$

We have

$f(x)=x+\frac{1}{x}=\frac{x^{2}+1}{x}$

Clearly, $f(x)$ is continuous on $[1,3]$ and derivable on $(1,3)$.

Thus, both the conditions of Lagrange's theorem are satisfied.

Consequently, there exists $c \in(1,3)$ such that

$f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}=\frac{f(3)-f(1)}{2}$

Now, $f(x)=\frac{x^{2}+1}{x}$

$f^{\prime}(x)=\frac{x^{2}-1}{x^{2}}, f(1)=2, f(3)=\frac{10}{3}$

$\therefore f^{\prime}(x)=\frac{f(3)-f(1)}{2}$

$\Rightarrow \frac{x^{2}-1}{x^{2}}=\frac{4}{6}$

$\Rightarrow \frac{x^{2}-1}{x^{2}}=\frac{2}{3}$

$\Rightarrow 3 x^{2}-3=2 x^{2}$

$\Rightarrow x=\pm \sqrt{3}$

Thus, $c=\sqrt{3} \in(1,3)$ such that $f^{\prime}(c)=\frac{f(3)-f(1)}{3-1}$