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

Aug 3, 2015

You can use the second derivative test.

#### Explanation:

The second derivative test allows you to determine the concavity of a function by analyzing the behavior of the function's second derivative around inflexion points, which are points at which ${f}^{' '} = 0$.

If ${f}^{' '}$ is positive on a given interval, then $f \left(x\right)$ will be concave up. LIkewise, if ${f}^{' '}$ 8s negative on a given interval, then $f \left(x\right)$ will be concave down.

The inflexion points will determine on which intervals the sign of the second derivative should be examined.

So, start by calculating the function's first derivative - use the quotient rule

$\frac{d}{\mathrm{dx}} \left(f \left(x\right)\right) = \frac{\left[\frac{d}{\mathrm{dx}} \left({x}^{2}\right)\right] \cdot \left({x}^{2} + 1\right) - {x}^{2} \cdot \frac{d}{\mathrm{dx}} \left({x}^{2} + 1\right)}{{x}^{2} + 1} ^ 2$

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

${f}^{'} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2 {x}^{3}}}} + 2 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{2 {x}^{3}}}}}{{x}^{2} + 1} ^ 2 = \frac{2 x}{{x}^{2} + 1} ^ 2$

Next, calculate the second derivative - use the quotient and chain rules

d/dx(f^') = ([d/dx(2x)] * (x^2 + 1)^2 - 2x * d/dx(x^2 + 1)^2)/([(x^2 + 1)^2]^2

${f}^{' '} = \frac{2 \cdot {\left({x}^{2} + 1\right)}^{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}}} - 2 x \cdot 2 \cdot \textcolor{red}{\cancel{\textcolor{b l a c k}{\left({x}^{2} + 1\right)}}} \cdot 2 x}{{x}^{2} + 1} ^ \textcolor{red}{\cancel{\textcolor{b l a c k}{4}}}$

${f}^{' '} = \frac{2 {x}^{2} + 2 - 8 {x}^{2}}{{x}^{2} + 1} ^ 3 = - \frac{2 \left(3 {x}^{2} - 1\right)}{{x}^{2} + 1} ^ 3$

Next, find the inflexion points by calculating ${f}^{' '} = 0$.

$- \frac{2 \left(3 {x}^{2} - 1\right)}{{x}^{2} + 1} ^ 3 = 0$

This is equivalent to

$3 {x}^{2} - 1 = 0 \implies {x}^{2} = \frac{1}{3}$

Take the square root of both sides to get

$\sqrt{{x}^{2}} = \sqrt{\frac{1}{3}} \implies x = \pm \frac{\sqrt{3}}{3}$

So, the function's graph has two inflexion points at $x = - \frac{\sqrt{3}}{3}$ and $x = \frac{\sqrt{3}}{3}$, so you're going to have to look at three intervals to determine its concavity.

Since ${\left({x}^{2} + 1\right)}^{3} > 0 , \left(\forall\right) x$ on $\left(- \infty , + \infty\right)$, the sign of the second derivative will depend exclusively on the sign of $3 {x}^{2} - 1$.

More specifically, if $3 {x}^{2} - 1 > 0$, then ${f}^{' '} < 0$, and if $3 {x}^{2} - 1 < 0$, then ${f}^{' '} > 0$.

• $\left(- \infty , - \frac{\sqrt{3}}{3}\right)$

On this interval, $3 {x}^{2} - 1 > 0$, which means that ${f}^{' '} < 0$ and the function is concave down.

• $\left(- \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3}\right)$

On this interval, $3 {x}^{2} - 1 < 0$, which means tha t${f}^{' '} > 0$ and the function is concave up.

• $\left(\frac{\sqrt{3}}{3} , + \infty\right)$

Once again, $3 {x}^{2} - 1 > 0$, ${f}^{' '} < 0$ and the function is concave down.

So, your function is concave down on $\left(- \infty , - \frac{\sqrt{3}}{3}\right) \cup \left(\frac{\sqrt{3}}{3} , + \infty\right)$ and concave up on $\left(- \frac{\sqrt{3}}{3} , \frac{\sqrt{3}}{3}\right)$.

The function's graph will have two inflexion points at $\left(- \frac{\sqrt{3}}{3} , f \left(- \frac{\sqrt{3}}{3}\right)\right)$ and $\left(\frac{\sqrt{3}}{3} , f \left(\frac{\sqrt{3}}{3}\right)\right)$.

graph{x^2/(x^2+1) [-4.932, 4.934, -2.465, 2.467]}