Question7. Find the interval where the graph of $f(x)=\frac{2 x}{4 x^{2}+1}$ is decreasing and increasing. Hence, determine the extremum points if exist.

Studdy Solution
Identify extremum points based on the sign changes of $f'(x)$:
- At $x = -\frac{1}{2}$, $f'(x)$ changes from negative to positive, indicating a local minimum. - At $x = \frac{1}{2}$, $f'(x)$ changes from positive to negative, indicating a local maximum.
The function $f(x) = \frac{2x}{4x^2 + 1}$ is decreasing on $(-\infty, -\frac{1}{2})$ and $(\frac{1}{2}, \infty)$, and increasing on $(-\frac{1}{2}, \frac{1}{2})$. There is a local minimum at $x = -\frac{1}{2}$ and a local maximum at $x = \frac{1}{2}$.