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Find all numbers in the range of f(x) = arctan x + arctan (1 - x).

 May 14, 2022
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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 \(\left(0, \arctan\left(\dfrac43\right)\right]\).

 May 14, 2022

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