Is the function arctanx  covex or concave in the positive real set.

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neela | High School Teacher | (Level 3) Valedictorian

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f(x) = arc tanx .

f(0) = 0.

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

Therefore arc tan x is monotonously increasing.

f''(x) =  (1/1+x^2)' = -2x/(1+x^2)^2 is  <  0 for all x> 0.

Also we know the inequality  (sinx)/x < 1 < tanx/x for all x.

Therefore x < tanx for all x.

Therefore  arc tanx < x.........................(1)

Therefore f'(x) is a decreasing function as f"(x) < 0 for all x > 0.... (2)

Thereforefrom (1) and (2) , we conclude that the given function is concave down wards. Or the given function is convex upward.

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

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In order to prove that f(x) is concave or convex, 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.

If the second derivative is positive, then the graph of f(x) is convex.

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 positive real set of numbers.

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