Let $f:\mathbb{R}\to\mathbb{R},f(x)=\frac {ax+b}{x^2+1}.$ Find $a,b$ such that $\Im_f=[-3,1].$
My attempt:
Found $f'(x)=-\frac {ax^2+2bx+a}{(x^2+1)^2}$ but can't solve the equation: $ax^2+2bx+a=0.$ To find points of extrema. Can you give me a hint with another approach? Or how can I use this one better?
If you solve the equation $$-3 ={ax+b\over x^2+1}$$ on $x$ then discriminant of $$3x^2+ax+b +3=0$$ must be $0$ (if you draw a line $y=-3$ it must touch a graph of $f$), so $$\boxed{ a^2-12(b+3)=0}$$ The same holds for $$1 ={ax+b\over x^2+1}\implies x^2-ax+1-b=0$$ so $$\boxed{a^2-4(1-b)=0}$$
You get a system to solve...