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

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### 2 Answers

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).

To prove tha arc tanx is concave over the interval (0 , ifinity)

The concavity of a function is determined by the sign of second differential coefficien.

f(x) = arctanx .

Therefore tan (f(x)) = x.

Differentiate both sides:

( = 1.sec^2 f(x) ) f'(x) = 1.

f'(x) = 1/ {secf(x)}^2 = 1/1+(tanf(x))^2

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

Differentiate again:

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

f"(x) = -2x/(1+x^2)^2 which is negative for any x>0.

Therefore f(x) is concave in (0 , infinity.)