Ta có $y'=\frac{\left( 2x-2 \right)\left( x-1 \right)-\left( {{x}^{2}}-2x+3 \right)}{{{\left( x-1 \right)}^{2}}}$

$=\frac{{{x}^{2}}-2x-1}{{{\left( x-1 \right)}^{2}}}=0\Leftrightarrow \left[ \begin{align} & x=1+\sqrt{2} \\ & x=1-\sqrt{2} \\ \end{align} \right.$

Do đó

$\underset{\left[ 2;4 \right]}{\mathop{\min }}\,f\left( x \right)=f\left( 1+\sqrt{2} \right)=2\sqrt{2};\,\underset{\left[ 1;4 \right]}{\mathop{\max }}\,f\left( x \right)=f\left( 4 \right)=\frac{11}{3}$