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

The curve is $$y = \frac{1}{2}x^4 - 5x^3 + 18x^2 - 19x$$. The slope of the curve is given by the first derivative: $$\frac{dy}{dx} = 2x^3 - 15x^2 + 36x - 19$$.

To find the maximum slope, we differentiate the slope function and set it to zero: $$\frac{d^2y}{dx^2} = 6x^2 - 30x + 36 = 6(x^2 - 5x + 6) = 6(x - 2)(x - 3) = 0$$.

This gives critical points at $$x = 2$$ and $$x = 3$$. To determine which gives a maximum, we compute the third derivative: $$\frac{d^3y}{dx^3} = 12x - 30$$.

At $$x = 2$$: $$\frac{d^3y}{dx^3} = 24 - 30 = -6 < 0$$, so the slope has a local maximum at $$x = 2$$. At $$x = 3$$: $$\frac{d^3y}{dx^3} = 36 - 30 = 6 > 0$$, so the slope has a local minimum at $$x = 3$$.

The value of $$y$$ at $$x = 2$$ is $$y = \frac{1}{2}(16) - 5(8) + 18(4) - 19(2) = 8 - 40 + 72 - 38 = 2$$.

Therefore the maximum slope occurs at the point $$(2, 2)$$.