(A) Second-Order Differentiation
The first derivative
1. When a function y = x3 + x2 – 3x + 6 is differentiated with respect to x, the derivative
2. The second function
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
.
3. Take note that
.
For example,
If y = 4x3 – 7x2 + 5x – 1,
The first derivative
The second derivative
(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)
(b)
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)
(b)
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:
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.
(–2, 27) is a turning point.
Hence, the turning point (1, 0) is a minimum point.
Hence, the turning point (–2, 27) is a maximum point.