\(A=\frac{4x+1}{x^2+3}\Leftrightarrow Ax^2+3A=4x+1\)
\(\Leftrightarrow Ax^2+3A-4x-1=0\)
\(\Leftrightarrow Ax^2-4x+\left(3A-1\right)=0\)
\(\Delta'=4-A\left(3A-1\right)\ge0\Rightarrow4-3A^2+A\ge0\)
\(\Rightarrow3-3A^2+1+A\ge0\Leftrightarrow3\left(1+A\right)\left(1-A\right)+\left(1+A\right)\ge0\Rightarrow\left(1+A\right)\left(4-3A\right)\ge0\Rightarrow-1\le A\le\frac{4}{3}\)