Page 33 - Ranger SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 2 Differentiation

Example 12 For turning points, dy = 0
dx
2
The diagram below shows a graph of 3x – 4x + 1 = 0
function y = (x – 1)(x – 3). (3x – 1)(x – 1) = 0
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1
y x = or x = 1
3
1 1 3 1 2 1
When x = , y =   – 2   + + 3
3
3
3
3
A B
0 x = 85
27
2
3
Find When x = 1, y = 1 – 2(1) + 1 + 3
(a) the gradient of the tangent at B where = 3
the curve intersects the x-axis,  1 85 
,
(b) the equation of the tangent at B, Thus, the turning points are 3 27 and
(c) the equation of the normal at B. (1, 3).
To determine the type of the turning points,
Solution
dy dy 1 dy 1
(a) y = (x – 1)(x – 3) dx at x = 0 dx at x = 3 dx at x = 2
= x – 4x + 3
2
dy + 0
= 2x – 4 –
dx
Using B(3, 0), the gradient of the 1 85
,
tangent at B Hence,  3 27  is a maximum point.
= 2(3) – 4
= 2 dy di x = 1 dy di x = 1 dy di x = 3
(b) Equation of the tangent at B dx 2 dx dx 2
y – 0 = 2(x – 3) – 0 +
y = 2x – 6
1
(c) Gradient of the normal at B = –
2 Hence, (1, 3) is a minimum point.
Equation of the normal at B
1
y – 0 = – (x – 3)
2
1 3 Alternative Method
y = – x +
2 2 Using the second derivative to determine the
type of turning points:
Form 5 Example 13 When x = , 2 2 = 6  – 4
2
d y
2 = 6x − 4
dx
Find the coordinates of the turning points
1 d y
1
3
3 dx
on the curve y = x – 2x + x + 3. Hence,
2
3
determine the type of the turning points. 1 85 = –2  0
Hence,  ,  is a maximum point.
3 27
Solution d y
2
When x = 1, 2 = 6(1) – 4
y = x – 2x + x + 3 dx = 2  0
3
2
dy = 3x – 4x + 1 Hence, (1, 3) is a minimum point.
2
dx
186
02 Ranger Mate Tambahan Tg5.indd 186 25/02/2022 9:23 AM