Given the functions f=x*arctanx and g=ln(1+x^2) prove that f>=g if x is in the interval [0;1]

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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We'll subtract g(x) both sides of the inequality that has to be demonstrated.

f(x) - g(x) >= 0

We'll substitute f(x) and g(x) by their expressions:

x*arctanx - ln(1+x^2) >= 0

We'll assign a new function h(x) to the difference of functions above:

h(x) = x*arctanx - ln(1+x^2)

If h(x) is positive, that means that h(x) is an increasing function. To prove that h(x) is increasing, we'll calculate it's first derivative:

h'(x) = f'(x) - g'(x)

f'(x) = (x*arctanx)'

We'll apply product rule:

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

g'(x) = [ln(1+x^2)]'

We'll apply chain rule:

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

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

We'll subtract g'(x) from f'(x);

h'(x) = arctan x + x/(1 + x^2) - 2x/(1+x^2)

We'll combine like terms:

h'(x) = arctan x - x/(1+x^2)

Since we cannot tell that h'(x) is positive, we'll differentiate once more time:

h''(x) = [arctan x - x/(1+x^2)]'

h''(x) = 1/(1+x^2) - (1+x^2 - 2x^2)/(1+x^2)^2

We'll combine like terms:

h''(x) = 1/(1+x^2) - (1 - x^2)/(1+x^2)^2

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

We'll combine like terms fromĀ  numerator:

h"(x) = 2x^2/(1+x^2)^2

Since h"(x) > 0 for any value of x from (0 ; 1), we'll conclude that h'(x) is also positive for values of x from (0 ; 1) and h(x) is an increasing function over the range [0 ; 1].

If h(x) >= 0, then f(x) - g(x) >= 0 => f(x) >= g(x) q.e.d.

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