x4−5x3+18x2−19x . . . (i) First, find the slope of given curve i.e. dy∕dx, Differentiate Eq. (i),
dy
dx
=
1
2
(4x3)−5(3x2)+18(2x)−19 =2x3−15x2+36x−19 Now, let f(x)=2x3−15x2+36x−19 is slope of the curve and find its maximum value as follows, f′(x)=2(3x2)−15(2x)+36=6x2−30x+36 Equate f′(x)=0 and solve for ' x′ ', 6x2−30x+36=0 ⇒x2−5x+6=0 ⇒x2−3x−2x+6=0 ⇒(x−3)(x−2)=0 ⇒x=2 and 3 Now, f′′(x)=
d
dx
(6x2−30x+36) =12x−30 Then, f′′(2)=12(2)−30=24−30 =−6<0 and f′′(3)=12(3)−30=6>0 ∵ f′′(2)<0 , this implies 2′ is point of maxima. ∴ At x=2, slope will be maximum. Since, at x=2, slope will be maximum, then y-coordinate will be, y=
1
2
(2)4−5(2)3+18(2)2−19(2) =8−40+72−38=72−70=2 ∴ Maximum slope occurs at point (2,2).