Difference of equal derivative functions value

Question

$f_1(x) = \text{tan}^{-1}(x)$

$f_2(x) = \text{tan}^{-1}\big(\frac{x-1}{x+1}\big)$

Where

$f_1^\prime(x) = f_2^\prime(x)=\frac{1}{x^2+1}$

Therefore

$f_1(x) - f_2(x) = \theta$

How to find values of $\theta$ mathematically. I have solved it graphically and the answers are $\frac{\pi}{4}$ and $-\frac{3\pi}{4}$ in my opinion. Thanks in advance.

Answer 1

In $(-1,\infty)$ put $x =0$ to get $\theta= f_1(x)-f_2(x) =0-\tan^{-1} (-1)=0-(-\frac {\pi} 4)=\frac {\pi} 4$ .

In $(-\infty, -1) $ note that $\theta =\lim_{x \to -\infty} [f_1(x)-f_2(x)]=-\frac {\pi} 2- (\frac {\pi} 4)=-\frac {3\pi} 4$.

Answer 2

Since $f_2(x)$ is nondifferentiable at $x=-1$, if you are integrating it, I think you technically need to integrate it piecewise, resulting in: $F'_2(x) = \arctan(x) + $C1 for $x > -1$ and $F_2'(x)= \arctan(x) + $C2 for $x < -1 $ with C1 and C2 constants.

From there, you can just find C1 = $f_1(x) - f_2(x)$ for some $x > -1$ and C2 = $f_1(x) - f_2(x)$ for some $x < -1$.

Answer 3

Hint:

Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$

$$\arctan x+\arctan(-1)=\begin{cases} \arctan\frac{x-1}{1-(-1)x} &\mbox{if } x\cdot(-1)<1\\ \pi+\arctan\frac{x-1}{1-(-1)x} & \mbox{if } x\cdot(-1)>1\\\text{sign}(x)\cdot\dfrac\pi2 & \mbox{if } x\cdot(-1)=1\end{cases} $$