Prove that the function is bijective

Question

A function $f:\mathbb{R}\to\mathbb{R}$ is defined by, $$f(x)={x+\arctan(x)}.$$ Prove that this function is bijective.

Here's my attempt:- Since $x$ and $\arctan(x)$ are continuous, $f(x)={x+\arctan(x)}$ is also continuous. And further, $$ f'(x) = 1+ \frac1{1+x^2}\gt 0 \quad\forall x \in\mathbb{R}$$

Therefore the function is strictly increasing and hence, is strictly monotonic. Therefore this function is bijective. Would this be correct? Can you show me a better way (that doesn't involve derivative) of proving this? Thank you!

Answer 1

You only have shown that $f$ is injective.

It remains to show that $f$ is surjective: to this end let $y_0 \in \mathbb R$.

Since $f(x) \to \infty$ as $ x\to \infty$ and $f(x) \to -\infty$ as $ x\to -\infty$, there are $a,b \in \mathbb R$ such that $a<b, f(b)> y_0$ and $f(a) <y_0$.

The intermediate value theorem shows now that $f(x_0)=y_0$ for some $x_0 \in [a,b].$

Answer 2

A strictly monotonic function need not be surjective. In this case $-\pi /2 <\arctan x< \pi /2$ so $f(x) \to \infty$ as $x \to \infty$ and $f(x) \to -\infty$ as $x \to -\infty$. From this it follows that the range (which is necessarily an interval by IVP) is the whole real line.