Question
Verify Lagrange's mean value theorem for the function $f(x)=x+\frac{1}{x}$ on the interval $[1,3]$.

Answer

Given that $f(x)=x+\frac{1}{x}$
The function $f(x)$ is continuous on the closed interval $[1,3]$ and differentiable on the open interval $(1,3)$, so the LMVT is applicable to the function.
Differentiate (I) w. r.t. $x$.
$f^{\prime}(x)=1-\frac{1}{x^2}$
Let $a=1$ and $b=3$
From (I),
$
\begin{aligned}
& f(a)=f(1)=1+\frac{1}{1}=2 \\
& f(b)=f(3)=3+\frac{1}{3}=\frac{10}{3}
\end{aligned}
$
Let $c \in(1,3)$ such that
$
\begin{aligned}
& f^{\prime}(c)=\frac{f(b)-f(a)}{b-a} \\
& 1-\frac{1}{c^2}=\frac{\frac{10}{3}-2}{3-1} \\
& 1-\frac{1}{c^2} \quad= \frac{\frac{4}{3}}{2}=\frac{2}{3} \\
& \therefore \quad c^2=3 \Rightarrow c= \pm \sqrt{3} \\
& \therefore \quad c=\sqrt{3} \in(1,3) \text { and } c=-\sqrt{3} \notin(1,3)
\end{aligned}
$

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