The maximum slope of the curve  $\mathrm{y}=\frac{1}{2}{\mathrm{x}}^4-5{\mathrm{x}}^3+18{\mathrm{x}}^2-19\mathrm{x}$ occurs at the point: 

$$\begin{gathered} \mathrm{y}=\frac{{\mathrm{x}}^4}{2}-5{\mathrm{x}}^3+18{\mathrm{x}}^2-19\mathrm{x} \\ \frac{\mathrm{dy}}{\mathrm{dx}}=2{\mathrm{x}}^3-15{\mathrm{x}}^2+36\mathrm{x}-19 \\ \frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}=6{\mathrm{x}}^2-30+36 \\ \frac{\mathrm{dy}}{\mathrm{dx}} \max \mathrm{or} \min \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{d^{2}y}{d{x}^2}=0 \\ 6x^2-30x+36=0 \\ {x}^2-5x+6=0 \\ \left(x-2\right)\left(x-3\right)=0 \\ x=2,3 \end{gathered}$$

$$\begin{gathered} \mathrm{x}=2 \mathrm{y}=\frac{16}{2}-40+72-38 \\ =80-40-38 \\ =2 \\ \therefore \mathrm{Point} \left(2, 2\right) \end{gathered}$$