For the function f(x) = x + frac{1}{x},x in [ | vedclass

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(B) The Mean Value Theorem states that there exists at least one $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = x + \frac{

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$\sqrt{3}$

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(B) The Mean Value Theorem states that there exists at least one $c \in (a, b)$ such that $f'(c) = \frac{f(b) - f(a)}{b - a}$.Given $f(x) = x + \frac{1}{x}$ on $[1, 3]$,we have $a = 1$ and $b = 3$.$f(1) = 1 + \frac{1}{1} = 2$.$f(3) = 3 + \frac{1}{3} = \frac{10}{3}$.$f'(x) = 1 - \frac{1}{x^2}$.Applying the theorem: $1 - \frac{1}{c^2} = \frac{\frac{10}{3} - 2}{3 - 1}$.$1 - \frac{1}{c^2} = \frac{\frac{4}{3}}{2} = \frac{2}{3}$.$\frac{1}{c^2} = 1 - \frac{2}{3} = \frac{1}{3}$.$c^2 = 3 \Rightarrow c = \sqrt{3}$ (since $c \in (1, 3)$).

For the function $f(x) = x + \frac{1}{x}$,$x \in [1, 3]$,the value of $c$ for the Mean Value Theorem is:

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