I have the function

$\displaystyle f(x)=\frac{x}{x^2+2} $

$\displaystyle f'(x)=\frac{2-x^2}{(x^2+2)^2} $

$\displaystyle f''(x)=\frac{2x(x^2-6)}{(x^2+2)^3} $

finding increasing and decreasing.

I set the first derivative $\displaystyle \frac{2-x^2}{(x^2+2)^2}=0 $

and get $\displaystyle +\sqrt{2}, -\sqrt{2} $

They create 3 intervals

1st interval

$\displaystyle -\infty$ to $\displaystyle -\sqrt{2}$

second interval

$\displaystyle -\sqrt{2}$ to $\displaystyle \sqrt{2}$

third interval

$\displaystyle \sqrt{2}$ to $\displaystyle \infty$

I plugin in values from in between these intervals into the first derivative and try to find where it is positive and where it is negative. The positive places are where the function is increasing and negatives are the places where it is decreasing. Right?

My results

$\displaystyle (-\infty,-\sqrt{2})U(\sqrt{2}, \infty)$ are decreasing

$\displaystyle (-\sqrt{2},\sqrt{2})$ are increasing

Next I set the second derivative as zero and undefined.

for this I can see i got a lot of intervals

$\displaystyle x=0$ and $\displaystyle x=+\sqrt{6} and x=-\sqrt{6}$

1st interval

$\displaystyle -\infty$ to $\displaystyle -\sqrt{6}$

second interval

$\displaystyle -\sqrt{6}$ to $\displaystyle 0$

third interval

$\displaystyle 0$ to $\displaystyle \sqrt{6}$

forth interval

$\displaystyle \sqrt{6}$ to $\displaystyle \infty$

Here I will have to do the same thing as I did when finding the increasing and decreasing but in this case negatives are concave down and positives are concave up.

So My results yield

$\displaystyle (-\infty,-\sqrt{6})U(0,\sqrt{6})$ are concave down

$\displaystyle (-\sqrt{6},0)U(\sqrt{6},\infty)$ are concave up (as they are positive)

are my answers are correct? Is there a quick way to understand that from the graph? Does that has be from the original graph?