For what intervals is

Question

g(x) = 1/x2 + 1
concave down? (Enter your answer using interval notation.)

For this question I understand that we have to find where the second derivative is negative.

So first I found the first derivative and got 2x/x^4+2x^2+1
and then I took the second derivative and ended up getting:
6x^4+4x^2-2 over x^8+4x^6+6x^4+4x^2+1 but I didn't know where to go from there.

Answer 1

what's with all these expanding polynomials? That gives no extra insight whatever

g = 1/(x^2+1)
g' = -2x/(x^2+1)^2
g" = 2(3x^2-1)/(x^2+1)^3

since x^2+1 is always positive, g" is negative where
3x^2-1 < 0
x^2 < 1/3
That is on the interval (-1/√3,1/√3)

A glance at the graph will confirm this:

http://www.wolframalpha.com/input/?i=+1%2F%28x^2%2B1%29

Answer 2

Thank you so much!