Example 1 (Maximum Value of Quadratic Function)
(b)
y=12x−3x2When x=2,y=12(2)−3(2)2y=12
Given that y = 3x (4 – x), calculate
(a) the value of x when y is a maximum,
(b) the maximum value of y.
Solution:
(a)
y=3x(4−x)y=12x−3x2dydx=12−6xWhen y is maximum, dydx=00=12−6xx=2
(b)
y=12x−3x2When x=2,y=12(2)−3(2)2y=12
Example 2 (Determine the Turning Points and Second Derivative Test)
Find the coordinates of the turning points on the curve y = 2x3 + 3x2 – 12x + 7 and determine the nature of these turning points.Solution:
y=2x3+3x2−12x+7dydx=6x2+6x−12At turning point, dydx=0
6x2 + 6x – 12 = 0
x2 + x – 2 = 0
(x – 1) (x + 2) = 0
x = 1 or x = –2
When x = 1
y = 2(1)3 + 3(1)2 – 12(11) + 7
y = 0
(1, 0) is a turning point.
When x = –2
y = 2(–2)3 + 3(–2)2 – 12(–2) + 7
y = 27
(–2, 27) is a turning point.
d2ydx2=12x+6When x=1,d2ydx2=12(1)+6=18>0 (positive)
(–2, 27) is a turning point.
Hence, the turning point (1, 0) is a minimum point.
When x=−2,d2ydx2=12(−2)+6=−18<0 (negative)
When x=−2,d2ydx2=12(−2)+6=−18<0 (negative)
Hence, the turning point (–2, 27) is a maximum point.