Show that $x + \arctan x =1$ has exactly one solution can some of you confirm (give feedback) to my solution to the equation in the question:
Firstly, let $f(x) = x + \arctan(x)$
In order to prove that we only have one unique solution, we first have to prove that $f(x)$ is injective.
We know that both $x$ and $\arctan(x)$ are injective functions, so does this imply that the sum of two injective functions is injective? Let's try to prove my statement.
Proof: Suppose we got a function $g(x)$ and $h(x)$, both of which are injective in their corresponding domains. Now let's define a new function $j(x) = g(x) + h(x)$. If a function is injective, it follows from the definition that if $x_1 = x_2 \Rightarrow j(x_1)=j(x_2)$. Thus:
$j(x_1) = g(x_1) + h(x_1)$ and
$j(x_2) = g(x_2) + h(x_2)$. But since $g(x)$ and $h(x)$ are injective, it follows from the definition that $j(x)$ is injective itself.
So, if the LHS is injective, it means that it's one to one. In other terms, for every $x$, it maps to one unique $y$, in this case 1.
However, we also have to study the range of $f(x)$. Since $f(x)$ is defined for all real numbers $x$, and the function is strictly increasing, we can conclude that the range contains all real numbers.
This, together with that the function is injective, concludes that the function must have exactly one solution.
Thanks!
 A: There are at several flaws in your argument:

does this imply that the sum of two injective functions is injective?

No, it does not.  $f(x) = x$ and $g(x) = -x$ is a simple counterexample.

If a function is injective, it follows from the definition that if $x_1 = x_2 \Rightarrow j(x_1)=j(x_2)$.

No, that conclusion does always hold. “Injective” means that $j(x_1)=j(x_2) \implies x_1 = x_2$.

Since $f(x)$ is defined for all real numbers x, and the function is strictly increasing, we can conclude that the range contains all real numbers.

No, you cannot. $f(x) = \arctan(x)$ is a counterexample.
But you can argue that the sum of two continuous and strictly increasing functions is again continuous and strictly increasing. If you find arguments $x$ and $y$ with $f(x) < 1$ and $f(y) > 1$ then the intermediate value theorem for continuous functions guarantees a solution of $f(x) = 1$. And a strictly increasing function is injective, to that the solution is unique.
A: 
We know that both $x$ and $\arctan(x)$ are injective functions, so does this imply that the sum of two injective functions is injective?

No.  Just take $x\mapsto-x$ and $x\mapsto x^3$ which are both injective, but $x\mapsto x^3-x$ is not.

In your case:

*

*Both functions are strictly increasing, hence their sum is also strictly increasing.  Hence the sum is injective.

*Both functions are continuous, thus their sum is.

*The image of the sum is all of $\Bbb R$, which is not hard to show (use 2. and that it is not bounded in any direction).  Hence the sum is surjective.

*From 1. and 3. we get that the sum is a bijection, which implies the claim.

A: Let $f(x) = \text{arctan}(x)$ and $g(x) = x.$ 
Here, it is understood that the range of $f(x)$ is $(-\pi/2, \pi/2).$ 
Also, $f'(x) = \frac{1}{1 + x^2}.$
Let $h(x) = f(x) + g(x).$
Then, since $f(x)$ and $g(x)$ are continuous on $\Bbb{R}$, so is $h(x).$
Further, since $f'(x)$ and $g'(x)$ are each always positive, so is $h'(x)$.
Now, consider that $h(0) = 0$ and $h(1) = 1 + \pi/4 > 1.$
Also, for $x < 0, f(x) < 0$ and $g(x) < 0$.  
Therefore, $h(x) < 0.$
For $x > 1, f(x) > \pi/4,$ and $g(x) > 1.$ 
Therefore, $h(x) > h(1) > 1.$
So, the question reduces to how many values of $x$ there are,in the interval $(0, 1)$ such that $h(x) = 1.$
By the Intermediate Value Theorem, there is at least one value of $x$ in the interval $(0,1)$ such that $h(x) = 1.$
Denote this value as $x = c.$
Then, since $h(x)$ is a strictly increasing function, 
for $0 < x < c, h(x) < h(c) = 1$ 
and 
for $c < x < 1, h(x) > h(c) = 1.$
Therefore, $c$ is the sole value in $(0,1)$ such that $f(c) = 1.$

Above analysis made strong use of the idea that since $h'(x)$ is always positive, then if $a < b$ you have that $h(a) < h(b).$  This result may be established via Rolle's Theorem.
An elaboration of this argument would be: 
suppose that $0 < a < b$ and that $h(a) \geq h(b)$.
Since the $\lim_{x\to\infty} h(x)$ is unbounded, there exists at least one value $x_0 > b$ such that $h(b) \leq h(a) < h(x_0)$.
By the intermediate value theorem, there exists a value $d$ in the interval $[b,x_0]$ such that $h(d) = h(a).$
This implies, by Rolle's theorem, that there exists a value $e$ in the interval $(a,d)$ such that $h'(e) = 0.$
This generates a contradiction, since $h'(x)$ is positive, for all $x$.
