Use the mean value theorem to show that x/x^2 + 1 < arctan x < x in the interval (0,∞).

Question

Use the mean value theorem to show that x/x^2 + 1 < arctan x < x in the interval (0,∞).

Answer 1

Let f(u) = arctan u on [0, x]

Then, from the mean value theorem, there is a value c on [0, x] such that f'(c) = (arctan(x)-arctan(0))/(x - 0) = arctan(x)/x

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

1/(c^2+1) = arctan(x)/x

Then, arctan(x) = x/(c^2+ 1) > x/(x^2 + 1)

Using i[0, x] as the integral from 0 to x,

As 1/(x^2+1) < 1 on (0, x] and arctan(x) = I[0, x] 1/(x^2+1) dx, arctan(x) < I[0, x] 1 dx = x

x/x^2+1 < arctan x < x