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

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(B) Given $f(x) = \frac{x}{3} + \frac{3}{x}$.First,we find the derivative $f'(x)$:$f'(x) = \frac{1}{3} - \frac{3}{x^2} = \frac{x^2 - 9}{3x^2}$.For the

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decreasing

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(B) Given $f(x) = \frac{x}{3} + \frac{3}{x}$.First,we find the derivative $f'(x)$:$f'(x) = \frac{1}{3} - \frac{3}{x^2} = \frac{x^2 - 9}{3x^2}$.For the function to be decreasing,we need $f'(x) Since $3x^2 > 0$ for all $x 
eq 0$,the sign of $f'(x)$ depends on the numerator $x^2 - 9$.We have $x^2 - 9 Since $f'(x) < 0$ for all $x \in (-3, 3) \setminus \{0\}$,the function $f(x)$ is decreasing on 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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