+9519 Ok, so the function is not in the simplest form yet, let's simplify it.
Note that \(\arctan a + \arctan b = \arctan \left(\dfrac{a + b}{1 - ab}\right)\), so we have:
\(\arctan x + \arctan(1 - x) = \arctan\left(\dfrac1{1 - x(1 - x)}\right) = \arctan\left(\dfrac1{x^2 -x +1}\right)\)
What is the range of \(\dfrac1{x^2 - x + 1}\)? We can find that by completing the squares:
\(x^2 - x + 1 = \left(x - \dfrac12\right)^2 + \dfrac34\\ \), so the range of x^2 - x + 1 is \(\left[\dfrac34, \infty\right)\).
Therefore, it means that the range of \(\dfrac1{x^2 - x + 1}\) is \(\left(0, \dfrac43\right]\).
(If you don't understand interval notations, that means \(\dfrac34 \leq x^2 - x + 1\) and \(0<\dfrac1{x^2 -x + 1} \leq \dfrac43\).)
Note that arctan is a strictly increasing function. As \(0<\dfrac1{x^2 -x + 1} \leq \dfrac43\), we have \(\arctan 0 < \arctan\left(\dfrac1{x^2 - x + 1}\right) \leq \arctan\left(\dfrac43\right)\), so \(0 < \arctan\left(\dfrac1{x^2 - x + 1}\right) \leq \arctan\left(\dfrac43\right)\)
The range of the function is \(\left(0, \arctan\left(\dfrac43\right)\right]\).
