# How do you graph f(x) = 2x arctan(x-1)?

Jul 15, 2018

By viewing my lengthy answer , the reader can understand why this question remained unanswered, for years.

#### Explanation:

Conventional range for $\arctan \left(x - 1\right)$ is $\left(- \frac{\pi}{2} , \frac{\pi}{2}\right)$

$\Rightarrow x \in \left(- \frac{\pi}{2} + 1 , \frac{\pi}{2} + 1\right) = \left(- 0.571 , 2.571\right)$.

Correspondingly, $y \in \left(1.147 , 5.162\right)$.

The graph has terminals, at $\left(- 0.571 , 1.147\right) \mathmr{and} \left(2.571 , 5.162\right)$

$y ' = 2 \left(\arctan \left(x - 1\right) + \frac{x}{{\left(x - 1\right)}^{2} + 1}\right) = 0$,

for transcendental x = 0.5327.

See graph, revealing this approximation.

Graph for y':
graph{y-arctan(x-1)-x/((x-1)^2+1)=0 [.532 0.533 -.01 0.01]}

The turning point is near ( 0.5327, -0.4186 )#

See graph of the given equation that reveals this,

but not the terminals.
graph{(y-2x arctan(x-1))((x-0.53)^2+(y+0.418)^2-0.01)=0}

Now see graph for $y > 0$
graph{y-2x arctan(x-1)=0 [-1 3 0 6]}
Now, see graph for $y < 0$.
graph{y-2x arctan(x-1)=0 [0 1 -0.6 0]}

Sliding these graphs,

you could see graph extending.$\leftarrow , \uparrow \mathmr{and} \rightarrow$, beyond

terminals that are created by limits imposed on ${\tan}^{- 1}$/arctan

values.

Continued in the 2nd answer, .
.

Jul 15, 2018

Continuation, for the 2nd part. I would review my answer and improve/correct, if necessary. Please avoid editing my answer.

#### Explanation:

My calculator displays ${\tan}^{- 1} \left(\tan \left(- {120}^{o}\right)\right)$ as $. - {60}^{o}$.

In this 21st century, it is not impossible to return$- {120}^{o}$.

I expect this happen soon.

Here, I define $y = 2 x {\left(\tan\right)}^{- 1} \left(x - 1\right)$, with piecewise y,

for $\left(x - 1\right) \in \left(k \pi - \frac{\pi}{2} , k \pi + \frac{\pi}{2}\right) , k = 0 , \pm 1 , \pm 2 , \pm 3 , \ldots$

$\Rightarrow x \in \left(k \pi - \frac{\pi}{2} + 1 , k \pi + \frac{\pi}{2} + 1\right)$..

Here, the inverse $x = 1 + \tan \left(\frac{y}{2 x}\right)$ is used, for graph.

Graph of $y = 2 x {\left(\tan\right)}^{- 1} \left(x - 1\right) ,$

using the wholesome inverse $x - 1 = \tan \left(\frac{y}{2 x}\right)$.

This includes the 1st part graph, for the conventional $\arctan$.

graph{x - 1 - tan( y / (2 x ))=0}.

Is the graph cumbersone? I do not think so.

Slide the graph $\rightarrow \uparrow \rightarrow \downarrow$. Now, you would agree with

me.

Interestingly, x = 1, for y = even $\pi$. See graph to see these cuts.

Respectively, $\frac{y}{x} \to$ odd $\pi$, as $x , y \to \infty$
graph{(x - 1 - tan( y / (2 x )))(x-1 +0y)=0}.