The function f(x) = 3x + frac{2}{x} on the in | vedclass

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(B) Given the function $f(x) = 3x + \frac{2}{x}$.To determine the nature of the function,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(3x + 2x^

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Strictly increasing on $(1, 3)$.

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(B) Given the function $f(x) = 3x + \frac{2}{x}$.To determine the nature of the function,we find its derivative $f'(x)$.$f'(x) = \frac{d}{dx}(3x + 2x^{-1}) = 3 - 2x^{-2} = 3 - \frac{2}{x^2}$.Simplifying the expression,we get $f'(x) = \frac{3x^2 - 2}{x^2}$.For the interval $x \in (1, 3)$,we observe that $x^2 > 1$,which implies $3x^2 > 3$.Therefore,$3x^2 - 2 > 3 - 2 = 1$,which is always positive.Since $f'(x) > 0$ for all $x \in (1, 3)$,the function $f(x)$ is strictly increasing on the interval $(1, 3)$.

The function $f(x) = 3x + \frac{2}{x}$ on the interval $(1, 3)$ is:

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