Find the local maximum value of the function $f\left(x\right)=(x^2-2x)\mathcal{l}nx-\frac{3}{2}{x}^2+4x+\frac{5}{2},x>0$  is

Given,  $f\left(x\right)=(x^2-2x).\mathcal{l}nx-\frac{3}{2}{x}^2+4x+\frac{5}{2}$
$\therefore f\text{'}\left(x\right)=(2x-2)\mathcal{l}nx+(x^2-2x).\frac{1}{x}-\frac{3}{2}\left(2x\right)+4.$
$=2(x-1)\mathcal{l}nx+(x-2)-3x+4=2(x-1)\mathcal{l}nx-2(x-1)=2(x-1).(\mathcal{l}nx-1)$
So,  $f\text{'}\left(x\right)=0\Rightarrow x=1,e.$

$\therefore f\left(x\right)$  has local maximum at  $x=1.$
$\Rightarrow f(x=1)=0-\frac{3}{2}+4+\frac{5}{2}=1+4=5.$