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

Question

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

Professor PC Thomas Classes · Answer

$f'(x) = \frac{1}{3} - \frac{3}{x^2} = \frac{x^2-9}{3x^2}$. In $(-3, 3)$, $x^2-9 < 0$, so $f'(x) < 0$, strictly decreasing.