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  d y d x = 3 x 2 + 2 x 3

2. The second function   d y d x can be differentiated again with respect
    to x. This is called the second derivativeof 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 ( d y d x ) 2 .

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

The first derivative      d y d x = 12 x 2 14 x + 5

The second derivative       d 2 y d x 2 = 24 x 14


(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 = 3 x ( 4 x ) y = 12 x 3 x 2 d y d x = 12 6 x When  y  is maximum,  d y d x = 0 0 = 12 6 x x = 2

(b) 
y = 12 x 3 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 12 x + 7 d y d x = 6 x 2 + 6 x 12 At turning point,  d y d x = 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 = 12 x + 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.

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