Page 24 - Focus SPM 2022 - Additional Mathematics

Additional Mathematics SPM Chapter 1 Functions
4
(b) y g[f(x)] = g 1 x + 3 2 , x ≠ –3
6 4
5 = – 3
4 g 1 4 2
3 x + 3
2 = x + 3 – 3
1 = x, x ≠ –3
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x
0 1 2 3 4 5 6 Since fg(x) = x, x in the domain of g,
Solution and gf(x) = x, x in the domain of f,
(a) The graph of function f does not have an inverse therefore, g is the inverse of f.
function. f is not a one-to-one function.
(b) The graph of function g has an inverse function. Try Questions 6 – 7 in ‘Try This! 1.3’ Form 4
g is a one-to-one function.
(III) If f and g are inverse functions of each other,
Try Question 4 – 5 in ‘Try This! 1.3’ then
• The domain of f is equal to the range of g,
and
(II) f and g are inverse functions of each other if • The domain of g is equal to the range of f.
and only if
fg(x) = x, x in domain of g, 27
and gf(x) = x, x in domain of f.
Given f and g are inverse functions of each other such
1. Note that in the diagram below, fg(x) = x and that f(x) = AB 2
x + 1 – 5, x > –1 and g(x) = (x + 5) – 1,
gf(x) = x. x > –5. Write
g f (a) the range of f, (b) the range of g.

Solution
x g(x) x f(x)
(a) y > –5. The range of f is the domain of g.
(b) y > –1. The range of g is the domain of f.
f g
y
Since fg(x) and gf(x) are both similar to the
identity function x, hence f and g are mutually 3
inverse functions. g 2
Range 1
of g x
26 –6 –5 –4 –3 –2 –1 –1 0 1 2 3 4 5 6
By using the properties of inverse functions, show Domain of g –2 f
4 4 –3
that the functions f(x) = , x ≠ –3 and g(x) = – 3, –4 Range
x + 3 x of f
x ≠ 0 are mutually inverse functions. –5 Domain of f
Solution SPM Tips
2
f [g(x)] = f 1 4 – 3 , x ≠ 0
x
4 Domain f Range
of f
of f
=
2
1 4 – 3 + 3 x f(x)
x
4 Range Domain
= –1 f –1 –1
4 of f of f
x
= x, x ≠ 0 Try Question 8 in ‘Try This! 1.3’

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