Question

Find the largest open intervals where the function is concave upward.

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

Answer 1

I think you meant:
f(x) = x/(x^2 + 1)

(or else why not just reduce it to
f(x) = 1/x + 1 ? )

for concave upwards, f ''(x) > 0

f'(x) = (1 - x^2)/(x^2 + 1)^2 using the quotient rule
and again
f''(x) = 2x(x^2-3)/(x^2+1)^3

so when is f''(x) positive ?
well, the denominator is always positive,
so we just have to look at the top

let's find the points of inflection:
2x = 0 ---> x = 0, then y = 0
x^2 - 3 = 0
x = ± √3 , then y = ±√3/4

look at the original graph
http://www.wolframalpha.com/input/?i=plot+y+%3D+x%2F%28x%5E2+%2B+1%29

opens up for x >√3/4
and -√3/4 < x < 0

the first of these being the largest open interval where it opens upwards.

Strange wording of the question.