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

##### 1 Answer

The inflection points are:

#### Explanation:

An inflection point is a point on the graph at which the concavity changes.

We investigate concavity by looking at the sign of

# = (-2(x+1)(x^2+1) + 4x(-x^2-2x+1))/(x^2+1)^3#

# = (-2(x^3+x^2+x+1)+4x(-x^2-2x+1))/(x^2+1)^3#

# = (2(x^3+3x^2-3x-1))/(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

So the only places at which **might** change sign must be zeros of

Since the denominator is never

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

Since

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

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:

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

(Since

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

Using

The inflection points are:

If you want or need to actually evaluate