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!
Best Answer
There are at several flaws in your argument:
No, it does not. $f(x) = x$ and $g(x) = -x$ is a simple counterexample.
No, that conclusion does always hold. “Injective” means that $j(x_1)=j(x_2) \implies x_1 = x_2$.
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.