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

calculus 
bobpursley
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.
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