\(f\left(x\right)=\dfrac{4x-3}{x^2+1}=\dfrac{x^2+1-x^2+4x-4}{x^2+1}=1-\dfrac{\left(x-2\right)^2}{x^2+1}\le1\)
\(f\left(x\right)_{max}=1\) khi \(x-2=0\Rightarrow x=2\)
\(f\left(x\right)=\dfrac{-4x^2-4+4x^2+4x+1}{x^2+1}=-4+\dfrac{\left(2x+1\right)^2}{x^2+1}\ge-4\)
\(f\left(x\right)_{min}=-4\) khi \(2x+1=0\Rightarrow x=-\dfrac{1}{2}\)