$\text{f}(\text{x})=\text{x}+\frac{1}{\text{x}}\text{ on }[1,3]$
f(x) attains a unique value for each
$\text{x}\in[1,3],$ so it is continuous$\text{f}'(\text{x})=1-\frac{1}{\text{x}^2}$ is defined for each $\text{x}\in(1,3)$
⇒ f(x) is differentiable in (1,3), so Lagrange's mean value theorem is applicable, so there exist a point $\text{c}\in(1,3)$ such that
$\text{f}'(\text{c})=\frac{\text{f}(3)-\text{f}(1)}{3-1}$
$\Rightarrow1-\frac{1}{\text{c}^2}=\frac{\Big(3+\frac{1}{3}-(1+1)\Big)}{2}$
$\Rightarrow1-\frac{1}{\text{c}^2}=\frac{\frac{10}{3}-2}{2}$
$\Rightarrow1-\frac{1}{\text{c}^2}=\frac{4}{3\times2}$
$\Rightarrow1-\frac{2}{3}=\frac{1}{\text{c}^2}$
$\Rightarrow\text{c}^2=3$
$\text{c}=\sqrt{3}\in(1,3)$ such that $\text{f}'(\text{c})=\frac{\text{f}(1)-\text{f}(0)}{1-0}$
So, Lagrange's mean value theorem is verified.
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