# How do you find the inflection points of the graph of the function: (x+1)/(x^(2)+1)?

Jul 1, 2015

The inflection points are:
$\left(2 - \sqrt{3} , f \left(2 - \sqrt{3}\right)\right)$ and $\left(1 , 1\right)$ and $\left(2 + \sqrt{3} , f \left(2 + \sqrt{3}\right)\right)$

#### Explanation:

An inflection point is a point on the graph at which the concavity changes.
We investigate concavity by looking at the sign of $f ' ' \left(x\right)$.

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

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

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

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

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

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

(Whew! That's a lot of typing. And we're not done yet.)

Now to determine where the concavity changes, we need to determine where the sign of $f ' '$ changes. A function can change signs at $x$ values where the graph crosses the $x$ axis (the function has a value of $0$), or where the function is discontinuous (for "nice" functions, that is where the function does not exist).

$f ' ' \left(x\right) = \frac{2 \left({x}^{3} + 3 {x}^{2} - 3 x - 1\right)}{{x}^{2} + 1} ^ 3$ has no discontinuities in the real numbers. (${x}^{2} + 1 = 0$ has no real solution).

So the only places at which $f ' '$ might change sign must be zeros of $f ' '$.
Since the denominator is never $0$, $f ' ' \left(x\right) = 0$ exactly where the numerator is $0$. That is, we need to solve:

${x}^{3} + 3 {x}^{2} - 3 x - 1 = 0$

Because this is a four term polynomial, it makes sense to try factoring by grouping. But it doesn't work.

Even without the rational roots theorem, it seems reasonable to try $x = 1$ and/or $x = - 1$. We see the $x = 1$ is a solution. (Or observe that the sum of the coefficients is $0$, so $x = 1$ is a root of the equation.)
Since $x = 1$ is a solution, $x - 1$ must be a factor.

If you remember polynomial division (long or synthetic) use that. Otherwise you're stuck with factoring 'from scratch' to get:

${x}^{3} + 3 {x}^{2} - 3 x - 1 = \left(x - 1\right) \left({x}^{2} + 4 x + 1\right) = 0$

$x = 1$ or ${x}^{2} + 4 x + 1 = 0$

The second polynomial (the quadratic) won't factor nicely (we just can't catch a break with this problem), so we solve by either completing the square or the quadratic formula, to get:

$x = 2 \pm \sqrt{3}$

So the partitions of the number line we need to consider are:

(-oo, 2-sqrt3), (2-sqrt3, 1), (1, 2+sqrt3), "and (2+sqrt3, oo)

(Since $1 < \sqrt{3} < 2$, we know that $2 - \sqrt{3} < 1$)

(Yes, Socratic, I know this answer is getting pretty long. And it's going to get longer yet.)

Using $f ' ' \left(x\right) = \frac{2 \left(x - 1\right) \left({x}^{2} + 4 x + 1\right)}{{x}^{2} + 1} ^ 3$ We find that the sign does in fact change at each zero. (Either use test numbers or observe that none of the 3 zeros are multiple zeros.)

The inflection points are:
$\left(2 - \sqrt{3} , f \left(2 - \sqrt{3}\right)\right)$ and $\left(1 , 1\right)$ and $\left(2 + \sqrt{3} , f \left(2 + \sqrt{3}\right)\right)$

If you want or need to actually evaluate $f \left(2 - \sqrt{3}\right)$ and $f \left(2 + \sqrt{3}\right)$ go ahead without me.