9.7 Second-Order Differentiation, Turning Points, Maximum and Minimum Points

(A) Second-Order Differentiation

1. When a function y = x3 + x2 – 3x + 6 is differentiated with respect
    to x, the derivative  dy dx =3 x 2 +2x3

2. The second function   dy dx can be differentiated again with respect
    to x. This is called the second derivative of y with respect to x
    and can be written as d 2 y d x 2 .

3. Take note that   d 2 y d x 2 ( dy dx ) 2.

For example,
If y = 4x3 – 7x2 + 5x – 1,

The first derivative      dy dx =12 x 2 14x+5

The second derivative       d 2 y d x 2 =24x14

(B) Turning Points, Maximum and Minimum Points


(a) At Turning Points A and B,

(b) At Maximum Point A

(c) At Minimum Point B,


Example 1 (Maximum Value of Quadratic Function)
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( 4x ) y=12x3 x 2 dy dx =126x When y is maximum,  dy dx =0 0=126x x=2

(b) 
y=12x3 x 2 When x=2, y=12( 2 )3 ( 2 ) 2 y=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=2 x 3 +3 x 2 12x+7 dy dx =6 x 2 +6x12 At turning point,  dy dx =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.

d 2 y d x 2 =12x+6 When x=1, d 2 y d x 2 =12( 1 )+6=18>0 (positive)

Hence, the turning point (1, 0) is a minimum point.

When x=2, d 2 y d x 2 =12( 2 )+6=18<0 (negative)

Hence, the turning point (–2, 27) is a maximum point.