Find local max, min, concavity, and inflection points For the function
$f(x)=\frac{x^2}{x^2+3}$
Find the intervals on which f(x) is increasing or decreasing.
Find the points of local maximum and minimum of f(x).
Find the intervals of concavity and the inflection points of f(x).
$f′(x)=\frac{6x}{(x^2+3)^2}$
$f''(x)=\frac{-18(x^2-1)}{(x^2+3)^3}$ 
 A: Sign analysis of $f'$
$f'(x)=0$ when $x=0$ so partition the number line as $(-\infty,0)\cup(0,\infty)$. 
$(-\infty,0)$: Pick a test point in this interval, say $x=-1$. Note $f'(-1)<0$ so $f'(x)<0$ on this interval. Hence $f$ is decreasing on this interval.
$(0,\infty)$: Pick a test point, say $x=1$. Then $f'(1)>0$ so $f'(x)>0$ on this interval. Hence $f$ is increasing on this interval.
Since $f$ went from decreasing to increasing at $x=0$, a local min occurs at $x=0$.

Sign analysis of $f''$
$f''(x)=0$ when $x=\pm 1$ so partition the number line as $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$. 
$(-\infty,-1)$: Pick a test point in this interval, say $x=-2$. Then $f''(-2)<0\implies f''(x)<0\implies f\text{ concave down}$ on this interval
$(-1,1)$: Pick a test point in this interval, say $x=0$. Then $f''(0)>0\implies f''(x)>0\implies f\text{ concave up}$ on this interval
$(1,\infty)$: Pick a test point in this interval, say $x=1$. Then $f''(1)<0\implies f''(x)<0\implies f\text{ concave down}$ on this interval
Since $f$ changed concavity at $x=\pm 1$, these are inflection points.

Indeed a plot of $y=f(x)$ bears out the information above:

