On the interval (1, 3),the function f(x) = 3x | 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 with respect to $x$:$f'(x) = \frac{d}{

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Strictly increasing

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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 with respect to $x$:$f'(x) = \frac{d}{dx}(3x + 2x^{-1}) = 3 - 2x^{-2} = 3 - \frac{2}{x^2}$.For the interval $(1, 3)$,we examine the sign of $f'(x)$:Since $1 This implies $\frac{1}{9} Multiplying by $2$,we get $\frac{2}{9} Now,$f'(x) = 3 - \frac{2}{x^2}$.Since $\frac{2}{x^2}  3 - 2 = 1$.Thus,$f'(x) > 0$ for all $x \in (1, 3)$.Since the derivative is strictly positive on the interval,the function $f(x)$ is strictly increasing.

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

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