In the interval (-3, 3), the function f(x) = | vedclass

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(B) Given the function $f(x) = \frac{x}{3} + \frac{3}{x}$.First, we find the derivative $f'(x)$ with respect to $x$:$f'(x) = \frac{d}{dx}(\frac{x}{3}

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decreasing

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(B) Given the function $f(x) = \frac{x}{3} + \frac{3}{x}$.First, we find the derivative $f'(x)$ with respect to $x$:$f'(x) = \frac{d}{dx}(\frac{x}{3} + \frac{3}{x}) = \frac{1}{3} - \frac{3}{x^2}$.To determine the nature of the function in the interval $(-3, 3)$, we analyze the sign of $f'(x)$:$f'(x) = \frac{x^2 - 9}{3x^2}$.Since $x^2 Since $3x^2$ is always positive for $x 
eq 0$, the derivative $f'(x) Therefore, the function $f(x)$ is decreasing in the interval $(-3, 3)$.

In the interval $(-3, 3)$, the function $f(x) = \frac{x}{3} + \frac{3}{x}, x \neq 0$ is :

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