Prove that the function is bijective

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!

• Well...$e^x$ is strictly increasing but it is not a bijection. Your argument is good, but incomplete,
– lulu
Commented Nov 28, 2018 at 11:50
• You are implicitly using the Intermediate Value Theorem in your argument. If you want to show the function reaches every value in the interval $[-M, M]$, you first need to show that it reaches $M$ and $-M$. Commented Nov 28, 2018 at 11:53
• Your argument applies without change to the function $\arctan(x)$. But $\arctan(x)$ isn't bijective! So you must have gone wrong somewhere. Commented Nov 28, 2018 at 13:59

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 $$a y_0$$ and $$f(a) .

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

• Is $x sinx$ bijective over $\mathbb{R}$ ?
– user612946
Commented Nov 28, 2018 at 13:10
• No, $x\sin x$ has a lot of zeros.
– Fred
Commented Nov 28, 2018 at 14:01

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.

• Strict monotonicity is a necessary condition for $f(x)$ to be bijective, but not sufficient?
– user612946
Commented Nov 28, 2018 at 12:03
• Over which domain is $xsinx$ bijective? I don't think it is bijective over $\mathbb{R}$ ,as it has zeroes in $\mathbb{R}$ for multiple $x$ in $\mathbb{R}$
– user612946
Commented Nov 28, 2018 at 12:32
• @Samurai You're right, and Kavi is wrong. Bijective, continuous functions must be monotonic as bijective must be one-to-one, so the function cannot attain any particular value more than once. Commented Nov 28, 2018 at 16:34
• Thank you @Shufflepants :-))
– user612946
Commented Nov 28, 2018 at 16:35
• But of course, it is possible to have a discontinuous function that is not monotonic but bijective. Though, it must be monotonic on every interval on which it is continuous. Commented Nov 28, 2018 at 16:39