A function

The function $f(x)=1+\frac{1}{x}$ is defined on the interval $[1,3]$.

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

So, by mean value theorem there must exist at least one real number $c \in(1,3)$ such that

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

$\Rightarrow-\frac{1}{c^{2}}=\frac{\frac{4}{3}-2}{3-1}$        $\left[f(x)=1+\frac{1}{x} \Rightarrow f^{\prime}(x)=-\frac{1}{x^{2}}\right]$

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

$\Rightarrow \frac{1}{c^{2}}=\frac{1}{3}$

$\Rightarrow c^{2}=3$

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

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

Hence, a point in the interval where the given function satisfies the mean value theorem is $\sqrt{3}$.

A function $f(x)=1+\frac{1}{x}$ is defined on the closed interval $[1,3]$. A point in the interval, where the function satisfies the mean value theorem, is $\sqrt{3}$