Question

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

Answer 1

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

Explanation:

Conventional range for `arctan ( x - 1 )` is `( -pi/2, pi/2 )`

`rArr x in ( - pi/2 + 1, pi/2 + 1 ) = ( - 0.571, 2.571 )`.

Correspondingly, `y in ( 1.147, 5.162 )`.

The graph has terminals, at `( - 0.571, 1.147 ) and ( 2.571, 5.162 )`

`y' = 2 ( arctan ( x - 1 ) +x/( ( x - 1 )^2 + 1 )) = 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.`larr, uarr and rarr`, beyond

terminals that are created by limits imposed on `tan^(-1)`/arctan

values.

Continued in the 2nd answer, .
.

Answer 2

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

Explanation:

Please see
https://socratic.org/questions/defining-the-wholesome-inverse-operator-sin-1-by-y-sin-1-x-k-pi-1-k-sin-1-x-y-in

My calculator displays `tan^(-1)(tan (- 120^o))` as `.- 60^o`.

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

I expect this happen soon.

Here, I define `y = 2x (tan)^( -1 )( x - 1 )`, with piecewise y,

for `(x - 1) in (k pi - pi / 2, k pi + pi / 2 ), k = 0, +-1, +-2, +-3, ...`

`rArr x in (k pi - pi / 2 +1, k pi + pi / 2 + 1 )`..

Here, the inverse `x = 1 + tan( y / (2 x ))` is used, for graph.

Graph of `y = 2x (tan)^( -1 ) ( x - 1 ),`

using the wholesome inverse `x - 1= tan( y / (2 x ))`.

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 `rarr uarr rarr darr`. Now, you would agree with

me.

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

Respectively, `y / x to` odd `pi`, as `x, y to oo`
graph{(x - 1 - tan( y / (2 x )))(x-1 +0y)=0}.