hmmm, it appears that i may have been wrong. when i did it my way, i just drew a rough graph and assumed it was so *dusts hands* but now that i do it algebraically; let f(x)=1/x^2 +1
f''(x)= (2*(3*x^2-1))/(x^2+1)^3 [used wolframalpha caz i didn't want to make a mistake and i am too lazy to write it out by hand]
so it is concave up when f''(x) >0 when i solve this i get 1/root(3) and since f(x) is even hence the other side is the same
so from -infinity to -1/root(3) the graph is concave up; from -1/root(3) to 1/root(3) the graph is concave down and from 1/root(3) to infinity the graph is concave up. note that x= +- 1/root(3) are points of inflexion
Edit: this may help:
http://www.wolframalpha.com/input/?i=1/(x^2++1)