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.
Find the largest open intervals where the function is concave upward.
f(x)= x/x^2+1
1 answer
1 answer