The function f(x)=frac{x}{3}+frac{3}{x} decre | vedclass

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(A) Given,$f(x) = \frac{x}{3} + \frac{3}{x}$.On differentiating with respect to $x$,we get:$f^{\prime}(x) = \frac{1}{3} - \frac{3}{x^2}$.For the funct

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$(-3,3)$

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(A) Given,$f(x) = \frac{x}{3} + \frac{3}{x}$.On differentiating with respect to $x$,we get:$f^{\prime}(x) = \frac{1}{3} - \frac{3}{x^2}$.For the function to be decreasing,we must have $f^{\prime}(x) $\frac{1}{3} - \frac{3}{x^2} $\frac{1}{3} $x^2 Taking the square root on both sides,we get $|x| Note that at $x=0$,the function is undefined,but the interval $(-3, 3)$ is the standard solution for the inequality $x^2 < 9$.

The function $f(x)=\frac{x}{3}+\frac{3}{x}$ decreases in the interval

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