# How do you determine whether the function f(x)= x/ (x^2+2) is concave up or concave down and its intervals?

Refer to explanation

#### Explanation:

We have that $f \left(x\right) = \frac{x}{{x}^{2} + 2}$ calculating its second derivative we find that

${d}^{2} f \frac{x}{{d}^{2} x} = \frac{2 x \left({x}^{2} - 6\right)}{{x}^{2} + 2} ^ 3$.

So we need to see how the signs change of $2 x \left({x}^{2} - 6\right)$ as x goes
from $- \infty$ to $+ \infty$

So from $\left(- \infty , - \sqrt{6}\right]$ we have that $f ' ' \left(x\right) < 0$

from $\left[- \sqrt{6} , 0\right]$ we have that $f ' ' \left(x\right) > 0$

from $\left[0 , \sqrt{6}\right]$ we have that $f ' ' \left(x\right) < 0$

from $\left[\sqrt{6} , + \infty\right)$ we have that $f ' ' \left(x\right) > 0$

In order to determine concavity we use the following theorem

Concavity Theorem:

If the function $f$ is twice differentiable at $x = c$, then the graph of f is concave upward at (c;f(c)) if $f ' ' \left(c\right) > 0$ and concave downward if $f ' ' \left(c\right) < 0$ .