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(A) Given the function $f(x) = x + \frac{1}{x}$ on the interval $[1, 3]$.According to the Lagrange's Mean Value Theorem $(LMVT)$,there exists at least

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$\sqrt{3}$

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(A) Given the function $f(x) = x + \frac{1}{x}$ on the interval $[1, 3]$.According to the Lagrange's Mean Value Theorem $(LMVT)$,there exists at least one $c \in (1, 3)$ such that $f'(c) = \frac{f(3) - f(1)}{3 - 1}$.First,calculate the derivative: $f'(x) = 1 - \frac{1}{x^2}$,so $f'(c) = 1 - \frac{1}{c^2}$.Next,calculate the values at the endpoints: $f(1) = 1 + \frac{1}{1} = 2$ and $f(3) = 3 + \frac{1}{3} = \frac{10}{3}$.Substitute these into the $LMVT$ formula: $1 - \frac{1}{c^2} = \frac{\frac{10}{3} - 2}{3 - 1}$.Simplify the right side: $1 - \frac{1}{c^2} = \frac{\frac{4}{3}}{2} = \frac{2}{3}$.Rearrange to solve for $c^2$: $\frac{1}{c^2} = 1 - \frac{2}{3} = \frac{1}{3}$,which implies $c^2 = 3$.Since $c \in (1, 3)$,we take the positive root: $c = \sqrt{3}$.

If the $L.M.V.T.$ holds for the function $f(x) = x + \frac{1}{x}$ on the interval $x \in [1, 3]$,then $c$ is:

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