Page 13 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 1 Functions
(g) Domain is x  R. (c) y
Range is y > 0. y = f(x) (3, 7)
This absolute value function has a vertex V at (2, 0). All (2, 5)
values of y are greater or equal to 0.
y (1, 3)
6 (0, 1)
x
0
4
State a corresponding range for each domain
2
given.
x
–6 –4 –2 0 2 4 6 8 10 Solution
(a) When the domain is R, all values are possible
–2
for y.
Form 4
The range is y  R.
This function represents the (b) When x is limited to x . 0 (positive values only),
(h) Domain is r . 0. volume of a sphere, so the
Range is V . 0. radius, r, must be positive. all values of y are greater than 1.
The range is y . 1.
Try Question 7 in ‘Try This! 1.1’ (c) The range is {1, 3, 5, 7}.

7 Try Question 9 in ‘Try This! 1.1’

For each of the following graphs of functions,
determine the domain and range.
Sketching graphs of absolute value
(b)
(a) Penerbitan Pelangi Sdn Bhd. All Rights Reserved.
y
y
13 18 functions
9
x

–3 0 2 5 Sketch the graph of y =  1 x – 1 for the domain
2 x 2
–9 0 1 3 7
–2 < x < 8. Hence, state the corresponding range.
Solution
(a) The domain is −3 < x < 2, the range is −9 < y < 13. Solution
(b) The domain is 1 < x < 7, the range is 2 < y < 18. Step I: Sketch the graph for y = 1 x – 1. The point
2
Try Question 8 in ‘Try This! 1.1’ of intersection at the x-axis and the y-axis
can be obtained by substituting x = 0 and

8 y = 0 into y = 1 2 x – 1.
The following diagrams show the representations of
y = f(x), where f(x) = 2x + 1 for different domains. y
(a) y (b) y 1
2
y = f(x) y = –x – 1
y = f(x) x
0 2
1 –1
1

x x
0 0

10