What are the points of inflection, if any, of f(x)= (x^2+x)/(x^2+1) ?

Jun 15, 2016

The $x$-coordinates of the three inflection points are $x = - 1 , 2 \pm \sqrt{3}$ (explanation to come)

Explanation:

By the Quotient Rule, the derivative is

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

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

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

The second derivative is (also using the Chain Rule),

$f ' ' \left(x\right)$

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

$= \frac{\left({x}^{2} + 1\right) \cdot \left(- 2 x + 2\right) - \left(- {x}^{2} + 2 x + 1\right) \cdot 4 x}{{\left({x}^{2} + 1\right)}^{3}}$

$= \frac{- 2 {x}^{3} + 2 {x}^{2} - 2 x + 2 + 4 {x}^{3} - 8 {x}^{2} - 4 x}{{\left({x}^{2} + 1\right)}^{3}}$

$= \frac{2 {x}^{3} - 6 {x}^{2} - 6 x + 2}{{\left({x}^{2} + 1\right)}^{3}}$.

Inflection points occur where $f ' ' \left(x\right)$ changes sign (also note that $f ' ' \left(x\right)$ is defined for all $x$). The $x$-values where $f ' ' \left(x\right) = 0$ occur where $2 {x}^{3} - 6 {x}^{2} - 6 x + 2 = 0 \setminus \Leftrightarrow {x}^{3} - 3 {x}^{2} - 3 x + 1 = 0$.

You can either use your calculator or the Rational Root Theorem to help you find that $x = - 1$ is a root of this (once you know it's a root, it's easy to check: ${\left(- 1\right)}^{3} - 3 {\left(- 1\right)}^{2} - 3 \left(- 1\right) + 1 = - 1 - 3 + 3 + 1 = 0$.)

Next, either use long division of polynomials or synthetic division to see that ${x}^{3} - 3 {x}^{2} - 3 x + 1 = \left(x + 1\right) \left({x}^{2} - 4 x + 1\right)$.

The quadratic formula gives the other two roots of $f ' ' \left(x\right)$ as

$x = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \frac{1}{2} \cdot \sqrt{12} = 2 \pm \sqrt{3}$

You can most easily see that $f ' ' \left(x\right)$ changes sign at these three values by graphing it. But, since $f ' ' \left(x\right)$ is continuous for all $x$, you can just check its sign on various intervals by plugging in sample points.

Therefore, the $x$-coordinates of the three inflection points of $f \left(x\right)$ are $x = - 1 , 2 \pm \sqrt{3}$.

By plugging these numbers into $f \left(x\right)$ and doing various manipulations, you can find that the full rectangular coordinates of the three inflection points are: $\left(- 1 , 0\right)$, $\left(2 + \sqrt{3} , \frac{3}{4} + \frac{\sqrt{3}}{4}\right)$, and $\left(2 - \sqrt{3} , \frac{3}{4} - \frac{\sqrt{3}}{4}\right)$.