I found this problem in a book and I cannot solve it. Find $a, b \in \mathbb{R}$ sucht that $f:\mathbb{R} \rightarrow [-1, 1], \ f(x) = \frac{ax + b}{x^2 + 1}$ is surjective.

I have tried to do it the usual way: for every $y \in [-1, 1]$, there is a $x \in \mathbb{R}$ such that $f(x) = \frac{ax + b}{x^2 + 1} = y$ and solved the equation: $$x = \frac{a \pm \sqrt {a^2 - 4y^2 + 4b}}{2y}$$
Now, for every $x \in \mathbb{R}$, it exists a $y\in \mathbb{R}$, such that $\frac{a \pm \sqrt{a^2 - 4y^2 + 4b}}{2y}$, but I could not continue with this idea.

Another idea was set $f(x) \ge -1$ and $f(x) \le 1$. I get 2 quadratics to be greater than $0$, which means $\Delta \le 0$ and such I get a system of 2 quadratic inequations in $a$. But this does not guarantee the fact that the whole interval $[-1, 1]$ is covered.

Have you got any idea?