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Relevant wiki: Application of Derivatives

$f(x)=3+\frac{mx+n-3}{1+x^2}$ for $f(x)$ to lie in $[-4,3)$ , $mx+n-3\leq0\quad \forall x\in\text{R}$ $\therefore \boxed{m=0} \text{ \& } \boxed{n\lt 3}$ $\text{as }x\rightarrow\pm\infty \text{ ; }f(x)\rightarrow 3$

which means the other extreme of $f$ lies in between $-\infty \text{ \& } +\infty$ , where $f'(x)=0$

$f'(x)=0\Rightarrow 2x(n-3)=0\Rightarrow x=0$ $\therefore f(0)=-4 \Rightarrow \boxed{n=-4}$ $\text{ thus, }\frac{m^2+n^2}{4}=4$