I have to determine whether $x\sin x$
Is a surjective function, in $\Bbb R$.
My solution:
Let $f(x)=x\sin x$.
$f(\frac{\pi}{2})=\frac{\pi}{2}$
$f(-\frac{\pi}{2})=-\frac{\pi}{2}$
Therefore in the closed interval $[-\frac{\pi}{2},\frac{\pi}{2}]$, $f(a)\cdot f(b)$<$0$, where $a=-\frac{\pi}{2}$ and $b=\frac{\pi}{2}$.
From the IVT therorem, there is a root to $f(x)$ in that interval.
Since $\sin$ is a Periodic function, I can always find a closed interval which satisfies IVT.
Therefore, $f(x)$ is surjective.
I don't know how to formalize the solution and whether or not I'm right.
Best Answer
$$\lim_{n\to \infty }f\left(\frac{\pi}{2}+2n\pi\right)=+\infty \quad \text{and}\quad \lim_{n\to\infty }f\left(-\frac{\pi}{2}-2n\pi\right)=-\infty.$$
Therefore, by intermediate value theorem, it's surjective.