Prove that $f(x)=x\sin (x)$ is surjective I have to prove that  $f(x)=x\sin (x)$ is surjective in $\Bbb R$.
I thought that I will use the mean value theorem, however I'm finding it hard to do it with sine function. I thought of splitting the function into two indexes, so I could use below $0$ and above $0$ values.
Any ideas?
Thanks,
Alan
 A: By surjective I understand $R \rightarrow R$.
We construct sequences $a_n = \frac{\pi}{2}+2\pi n$ and $b_n = -\frac{\pi}{2}-2\pi n$.
$$\lim_{n \rightarrow \infty}f(a_n)=\infty$$
$$\lim_{n \rightarrow \infty}f(b_n)=-\infty$$
EDIT: Surjectivity follows from the Intermediate value theorem using the previous limits and the fact, that $f(x)$ is continuous.
A: Hint: What is $f(x)$ if $x=(2k+\frac12)\pi$? What if  $x=(2k+\frac32)\pi$?
A: Let $n \in \mathbb{N}$ and consider
$$f\left(\frac{\pi}{2} + 2 \pi n \right) = \frac{\pi}{2} + 2 \pi n$$
This is an increasing subsequence contained in the image of the function.
Let $y \geq 0$. By the Archimedean Principle
$$ \exists n \in \mathbb{N} : \frac{y}{2\pi}  - \frac{1}{4}\leq n + 1$$
This is equivalent to the statement that
$$ \exists n \in \mathbb{N} : y \leq \frac{\pi}{2} + 2 \pi (n+1)$$
Note that $f(0) = 0$. Now let $y \in \mathbb{R} : y \geq 0$ we have
$$ 0 \leq y \leq \frac{\pi}{2} + 2 \pi (n + 1) \iff f(0) \leq y \leq f\left(\frac{\pi}{2} + 2 \pi (n+1) \right)$$
By the intermediate value property
$$ \exists x \in [0, \frac{\pi}{2} + 2 \pi (n+ 1)] : f(x) = y$$
So we have concluded that the function is surjective for $y \geq 0$.
Finally in order to deal with the $y < 0$ case we use a similar proof but instead consider the subsequence
$$ f\left(-\frac{\pi}{2} + 2 \pi n \right) = -\frac{\pi}{2} + 2 \pi n$$
