Question

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

Answer 1

The inflection points are:
`(2-sqrt3, f(2-sqrt3))` and `(1,1)` and `(2+sqrt3, f(2+sqrt3))`

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''(x)`.

`f(x) = (x+1)/(x^(2)+1)`

`f'(x) = ((1)(x^2+1)-(x+1)(2x))/(x^2+1)^2 = (-x^2-2x+1)/(x^2+1)^2`

`f''(x) = ((-2x-2)(x^2+1)^2-(-x^2-2x+1)2(x^2+1)2x)/(x^2+1)^4`

`= (-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 `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''(x) = (2(x^3+3x^2-3x-1))/(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''(x) = 0` exactly where the numerator is `0`. That is, we need to solve:

`x^3+3x^2-3x-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+3x^2-3x-1 =(x-1)(x^2+4x+1) = 0`

`x=1` or `x^2+4x+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+- sqrt3`

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 < sqrt3 < 2`, we know that `2-sqrt3 < 1`)

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

Using `f''(x) = (2(x-1)(x^2+4x+1))/(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:
`(2-sqrt3, f(2-sqrt3))` and `(1,1)` and `(2+sqrt3, f(2+sqrt3))`

If you want or need to actually evaluate `f(2-sqrt3)` and `f(2+sqrt3)` go ahead without me.