find the interval on which the graph of f(x)=ln(x^2+1) is concave upward.

Answer 1

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

f"=2/(x^2+1)- 4x^2/(x^2+1)^2 check that.

now, when is f" zero?

f"=0=> 2=4x^2/(x^2+1)

2x^2=x^2+1
x=+-1 So pick three points, say x=-2, x=0, and x=2 and see what f" is, is it positive?

Check my work, it is easy to make an error on a keyboard.