# How do you determine whether the function f(x)= 4/(x^2+1) is concave up or concave down and its intervals?

Dec 5, 2016

$f \left(x\right)$ is concave up for $| x | > \frac{1}{\sqrt{5}}$ and concave down for $| x | < \frac{1}{\sqrt{5}}$

#### Explanation:

If a function is differentiable twice, we know that it is concave up if $f ' ' \left(x\right) > 0$ and concave down viceversa.

So let us calculate:

$f \left(x\right) = \frac{4}{{x}^{2} + 1} = 4 {\left({x}^{2} + 1\right)}^{- 2}$

$f ' \left(x\right) = - 16 x {\left({x}^{2} + 1\right)}^{- 3}$

$f ' ' \left(x\right) = 96 {x}^{2} {\left({x}^{2} + 1\right)}^{- 4} - 16 {\left({x}^{2} + 1\right)}^{- 3} = \frac{96 {x}^{2} - 16 \left({x}^{2} + 1\right)}{{\left({x}^{2} + 1\right)}^{4}} = \frac{96 {x}^{2} - 16 {x}^{2} - 16}{{\left({x}^{2} + 1\right)}^{4}} = 16 \frac{5 {x}^{2} - 1}{{\left({x}^{2} + 1\right)}^{4}}$

Evidently the denominator of $f ' ' \left(x\right)$ is always positive, so the sign of $f ' ' \left(x\right)$ is the sign of the numerator and we see that:

$f ' ' \left(x\right) < 0$ for $| x | < \frac{1}{\sqrt{5}}$

So $f \left(x\right)$ is concave up in the intervals $\left(- \infty , - \frac{1}{\sqrt{5}}\right)$ and $\left(\frac{1}{\sqrt{5}} , + \infty\right)$ and concave down for $x$ in the interval $\left(- \frac{1}{\sqrt{5}} , \frac{1}{\sqrt{5}}\right)$.

graph{4/(x^2+1) [-10, 10, -5, 5]}