The maximum slope of the curve y=12x4−5x3+18x2−19x  occurs at the point

The given equation of the curve is y=12x4−5x3+18x2−19x            Differentiate both sidesm=dydx=2x3−15x2+36x−19dmdx=6x2−30x+36=6x2−5x+6=6x−2x−3If dmdx=0 ⇒x−2=0,x−3=0d2mdx2=62x−5<0 for x=2so the slope has maximum value at x=2, when x=2 then y=8−40+72−38=2Therefore, the point is 2,2