The maximum slope of the curve y = 1 2 x 4 - 5 x 3 + 18 x 2 - 19 x occurs at the point

Since, slope is maximum so,

$\frac{d^{2}y}{d{x}^2}=6x^2-30x+36=0$

$\Rightarrow x^2-5x+6=0$

$x=2,3$

at $x=2$

$\frac{d^{3}y}{d{x}^3}=12x-30$

at $x=2,\frac{d^{3}y}{d{x}^3}<0$

So, maxima

$y=\frac{1}{2}\times 16-5\times 8+18\times 4-19\times 2$

$=8-40+72-38=80-78=2$

point $\left(2,2\right)$