Find the intervals in which the graph of tan inverse 1is concave upward and concave downward

Answer 1

Just from looking at the graph, I'd say
upward: (∞,0)
downward: (0,∞)
But let's check f"(x) to be sure
f(x) = arctan(x)
f'(x) = 1/(x^2+1)
f"(x) = -2x/(x^2+1)^2
So we see that I was right at first.
f"(x) > 0 on (-∞,0)
f"(x) < 0 on (0,∞)