Integration, Long Questions (Question 3)


Question 3:
The gradient function of a curve which passes through P(2, 14) is 6x² – 12x
Find
(a) the equation of the curve,
(b) the coordinates of the turning points of the curve and determine whether each of the
      turning points is a maximum or a minimum.

Solution:
(a)
Given gradient function of a curve  dy dx =6 x 2 12x The equation of the curve, y= ( 6 x 2 12x )  dx y= 6 x 3 3 12 x 2 2 +c y=2 x 3 6 x 2 +c 14=2 ( 2 ) 3 6 ( 2 ) 2 +c, at point P ( 2,14 ) 14=8+c c=6 y=2 x 3 6 x 2 6
(b)
dy dx =6 x 2 12x At turning points,  dy dx =0 6 x 2 12x=0 6x( x2 )=0 x=0x=2 x=0,  y=2 ( 0 ) 3 6 ( 0 ) 2 6=6 x=2,  y=2 ( 2 ) 3 6 ( 2 ) 2 6=14 d 2 y d x 2 =12x12 When x=0 d 2 y d x 2 =12( 0 )12=12 <0 ( 0,6 ) is a maximum point. When x=2 d 2 y d x 2 =12( 2 )12=12 >0 ( 2,14 ) is a minimum point.