Prove that the function arctanx is concave over the interval [0,infinite].

In order to prove that f(x) is concave, we'll have to
calculate the second derivative of the function.

If the
second derivative is negative, then the graph of f(x) is
concave.

For the beginning, we'll calculate the first
derivative of
f(x):

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

Now,
we'll calculate f"(x) of the expression arctan x, or we'll calculate the first
derivative of f'(x).

f"(x) =
[f'(x)]'

Since f'(x) is a ratio, we'll apply the quotient
rule:

f"(x) =
[1'*(1+x^2)-1*(1+x^2)']/(1+x^2)^2

We'll put 1' = 0 and
we'll remove the brackets:

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

Because of the fact that denominator is
always positive, then the numerator will influence the
ratio.

Since numerator is negative over the interval
[0,infinite), f"(x)<0.

So, the function arctan x is
concave over the interval [0,infinite).