# 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).

$f′(x)=\frac{6x}{(x^2+3)^2}$

$f''(x)=\frac{-18(x^2-1)}{(x^2+3)^3}$

• I know the critcal point is $x = 0$ if you plug that in to $f(x)$ and $f'(x)$ you get $0$ if you plug it into $f''(x)$ you get $f''(x) = \frac{2}{3}$ does this mean that the local min is $0$ at $x = 0$ ? – Csci319 Oct 31 '14 at 17:13
• Indeed it does. That method is called the "second derivative test" if you want to Google it. – GFauxPas Oct 31 '14 at 17:17
• but does that mean there is no max? how would I find the max? – Csci319 Oct 31 '14 at 17:47
• If the domain of your function is all real numbers, then the function has no maximum. – GFauxPas Oct 31 '14 at 18:05

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: