Question

Question #d6112

Answer 1

`(-1, ln2)` and `(1,ln2)`

Explanation:

`y=ln(1+x^2)`

`y' = (2x)/(1+x^2)`

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

`y''` is defined everywhere and is `0` at `+-1`

Testing the sign of `y''` we find that the sign changes at `-1` and at `1`. So the inflection points of `y` are at `x=-1` and `x=1`.

To finish finding the points, find the corresponding `y` values.