f(x)=\frac{2}{x^{2}+3}\na) find f(x).\nb) identify the graph that displays f in blue and f in red.

Explanation:

Step1: Apply quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y'=\frac{u'v - uv'}{v^{2}}$. Here, $u = 2$, $u'=0$, $v=x^{2}+3$, and $v' = 2x$.

Step2: Calculate $f'(x)$

$f'(x)=\frac{0\times(x^{2}+3)-2\times(2x)}{(x^{2}+3)^{2}}=\frac{-4x}{(x^{2}+3)^{2}}$

Step3: Analyze the properties of $f(x)$ and $f'(x)$

$f(x)=\frac{2}{x^{2}+3}$ is an even function ($f(-x)=f(x)$), and its graph is symmetric about the y - axis. Also, $f(x)>0$ for all real $x$. $f'(x)=\frac{-4x}{(x^{2}+3)^{2}}$ is an odd function ($f'(-x)=-f'(x)$). When $x < 0$, $f'(x)>0$ (so $f(x)$ is increasing on $(-\infty,0)$), and when $x>0$, $f'(x)<0$ (so $f(x)$ is decreasing on $(0,\infty)$).

Answer:

a) $f'(x)=\frac{-4x}{(x^{2}+3)^{2}}$ b) B