Search any question & find its solution
Question:
Answered & Verified by Expert
In the interval $(-3,3)$ the function $f(x)=\frac{x}{3}+\frac{3}{x}, x \neq 0$ is :
Options:
Solution:
2338 Upvotes
Verified Answer
The correct answer is:
decreasing
$\because \quad f(x)=\frac{x}{3}+\frac{3}{x}$
$f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^2}$
It is clear that $f^{\prime}(x)$ is less than zero in the interval $(-3,3)$.
Thus $f(x)$ is decreasing in the interval $(-3,3)$.
$f^{\prime}(x)=\frac{1}{3}-\frac{3}{x^2}$
It is clear that $f^{\prime}(x)$ is less than zero in the interval $(-3,3)$.
Thus $f(x)$ is decreasing in the interval $(-3,3)$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.