Question

How do you find the inflection points for the function `f(x)=x^3+x`?

Answer 1

`f(x)=x^3+x`

By taking derivatives,

`f'(x)=3x^2+1`

`f''(x)=6x=0 Rightarrow x=0`,

which is the `x`-coordinate of a possible inflection point. (We still need to verify that `f` changes its concavity there.)

Use `x=0` to split `(-infty,\infty)` into `(-infty,0)` and `(0,infty)`.

Let us check the signs of `f''` at sample points `x=-1` and `x=1` for the intervals, respectively.
(You may use any number on those intervals as sample points.)

`f''(-1)=-6<0 Rightarrow f` is concave downward on `(-infty,0)`

`f''(1)=6>0 Rightarrow f` is concave upward on `(0,infty)`

Since the above indicates that `f` changes its concavity at `x=0`, `(0,f(0))=(0,0)` is an inflection point of `f`.

I hope that this was helpful.