In the interval $(-3,3)$ the function $f(x)=\frac{x}{3}+\frac{3}{x}, x \neq 0$ is :
In the interval $(-3,3)$ the function $f(x)=\frac{x}{3}+\frac{3}{x}, x \neq 0$ is :
increasing
decreasing
neither increasing nor decreasing
partly increasing and partly decreasing
Solution
$\because \quad f(x)=\frac{x}{3}+\frac{3}{x}$
$f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^2}$
It is clear that $f^{\prime}(x)$ is less than zero in the interval $(-3,3)$.
Thus $f(x)$ is decreasing in the interval $(-3,3)$.