1. FUNCTIONS
1 Function Basics
1.1 Set notations
[a, b] = {x ∈ R : a ≤ x ≤ b}
(a, b] = {x ∈ R : a < x ≤ b}
(a, b) = {x ∈ R : a < x < b}
(a, ∞) = {x ∈ R : x > a}
R = set of real numbers
R+
= set of positive real numbers
1.2 Rule, domain and range
A function f, is defined by its rule and domain.
f : x → x + 2
rule
, x ∈ [0, ∞).
domain
f(x) = x + 2, x ≥ 0.
X f(X)
1
2
3
3
4
5
f
Domain Range
The domain Df , is the set of all possible x values.
The range Rf , is the set of all possible y values.
Remark: When we state a function, we must always state both its rule and domain.
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2. Finding Range
Sketch the graph to find Rf . Rf is the range of y-values that the graph takes.
Example 1.
Find the range of the following:
(a) f : x → x2 − 1, x ∈ [−1, 2],
−1 1 2
−1
3
x
y
∴ Rf =
(b) g : x → ex + 1, x ∈ R.
1
x
y
∴ Rg =
2 Horizontal line test
A function is said to be 1 − 1 if for every y ∈ Rf , there is only ONE x such that f(x) = y.
−1 1
1
x
y
It is not 1-1 since the line y = 1
cuts the graph twice.
x
y
It is 1-1 since every horizontal line
cuts the graph at most once.
Horizontal Line Test
f is NOT a 1 − 1 function if there is a horizontal line that cuts the graph at MORE THAN
ONE point.
f IS a 1 − 1 function if any horizontal line y = a, where a ∈ Rf , cuts the graph AT ONLY
ONE point.
Differentiation test for 1-1
f is 1 − 1 if
f′(x) > 0 for ALL x in the domain (strictly increasing functions)
or
f′(x) < 0 for ALL x in the domain (strictly decreasing functions)
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3. Example 2 (Different techniques to answer 1-1 questions).
[2012/RVHS/I/6modified]
The function f is defined by
f : x → | − x2
− 2x + 3|, x ∈ R.
(a) With the aid of a diagram, explain why f is not 1-1. [2]
(b) If the domain of f is restricted to the set {x ∈ R : x ≥ k}, state with a reason the least value of k
for which the function is 1-1. [2]
(c) By considering the derivative of f(x), prove that f is a one-one function for the domain you have
found in (b). [2]
Solution:
(a)
Show not 1-1 through graph
Sketch the graph, give the equation of a SPECIFIC horizontal line that cuts the graph in
at least 2 points.
−3 1
3
y = f(x)
x
y
Solution:
(b)
Show 1-1 through graph
Sketch the graph, explain that any horizontal line cuts the graph at only 1 point.
Least value of k is 1. When x ≥ 1, any horizontal line y = a for a ∈ [0, ∞), cuts the graph of
y = f(x) at only 1 point, hence it is 1-1.
(c)
Show 1-1 through differentiation
Show that f′(x) is either > 0 or < 0 for all x ∈ Df .
Note: This method cannot be used to show that a function is not 1-1.
For x ≥ 1, f(x) = x2 + 2x − 3.
Since f′(x) > 0 for x ≥ 1, f is a strictly increasing function, and hence it is 1-1.
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4. 3 Inverse functions
Df Rf
x y
f
f(x) = y
f−1
f−1(y) = x
Properties of inverse function
1. For f−1 to exist, f must be a 1-1 function.
2. Df−1 = Rf .
3. Rf−1 = Df .
4. (f−1)−1 = f.
Geometrical relationship between a function and its inverse
(i) The graph of f−1 is the reflection of the graph f about the line y = x.
(ii) (a, b) lies on f ⇔ (b, a) lies on f−1.
(iii) x = k is an asymptote of f ⇔ y = k is an asymptote of f−1
Remark: The notation f−1 stands for the inverse function of f. It is not the same as 1
f .
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5. Example 3 (Cambridge N2008/II/4).
The function f is defined by f : x → x2 − 8x + 17 for x > 4.
(i) Sketch the graph of y = f(x). Your sketch should indicate the position of the graph in relation
to the origin.
(ii) Show that the inverse function f−1 exists and find f−1(x) in similar form.
(iii) On the same diagram as in part (i), sketch the graph of y = f−1.
(iv) Write down the equation of the line in which the graph of y = f(x) must be reflected in order to
obtain the graph of y = f−1, and hence find the exact solution of the equation f(x) = f−1(x).
Solution:
(i)
4
1
y = f(x)
x
y
(ii)
Showing inverse exists
To show f−1 exists, we only need to show that f is 1-1.
Every horizontal line y = a for a > 1 cuts the graph of y = f(x) at only 1 point, hence it is 1-1
and f−1 exists.
[We first make x the subject:]
∴ f−1(x) =
√
x − 1 + 4, x > 1. [Change x to f−1x and y to x.]
[Note: You must state the domain of f−1 !]
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6. (iii)
Geometrical relationship between a function and its inverse
(i) The graph of f−1 is the reflection of the graph f about the line y = x.
(ii) (a, b) lies on f ⇔ (b, a) lies on f−1.
(iii) x = k is an asymptote of f ⇔ y = k is an asymptote of f−1
1 4
1
4
y = f(x)
y = f−1(x)
y = x
x
y
(iv) It must be reflected along the line y = x. Since f and f−1 intersect at the line y = x, finding
exact solution of the equation f(x) = f−1(x) is equivalent to finding
f(x) = x
x2
− 8x + 17 = x
x2
− 9x + 17 = 0
x =
9 ±
√
92 − 4(1)(17)
2
=
9 +
√
13
2
or
9 −
√
13
2
(rej as it is not in Df )
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7. Example 4 (2010/MI/Prelim/I/2c).
Function g is defined by
g : x → x2
− 3x for x ∈ R,
If the domain of g is restricted to the set {x ∈ R : x ≥ a}, find the least value of a for which g−1
exists. Hence, find g−1 and state its domain. [4]
[a = 1.5, g−1(x) = 3
2 +
√
x + 9
4, x ≥ −9
4]
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8. Example 5 (2014/PJC/Prelim/I/7modified).
It is given that
f(x) =
{
1
1+
√
x
for 0 ≤ x < 4,
−1 for 4 ≤ x < 5,
and that f(x + 5) = f(x) for all real values of x.
(i) Find f(24) and f(30).
(ii) Sketch the graph of y = f(x) for −5 ≤ x < 12.
Solution:
(i)
(ii)
−5 5 10 12
1
-1
1
3
y
x
4
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9. 4 Composite Functions
4.1 Domain, Range, Rule
Let f and g be the following functions:
g : x → x + 1 for x ≥ −1,
f : x → x − 2 for x ≥ 0.
Lets investigate the composite function fg(x).
-1
0
1
2
0
1
2
3
...
...
g
Dg Rg
What happens now if we change the domain of f to the following?
g : x → x + 1 for x ≥ −1,
f : x → x − 2 for x ≥ 1.
-1
0
1
2
0
1
2
3
...
...
g
Dg Rg
At the intermediate stage, the number 0 will have no place to map to! This is an undesired situation.
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10. fg
Diagram for function fg(x)
Dg
Rfg
Df
Rg
Properties of Composite Functions
(i) Domain fg = domain g.
(ii) Composite function fg exists ⇔ Rg ⊆ Df
Example 6.
Consider the following functions:
f : x → x2
x ∈ R
g : x →
1
x − 3
x ∈ R, x ̸= 3
Find the composition functions (i)fg, (ii) gf.
Solution:
(i)
fg(x) = f (g(x))
= f
(
1
x − 3
)
=
(
1
x − 3
)2
Domain of fg(x) = domain g = R {3}.
(ii)
gf(x) = g (f(x))
= g(x2
)
=
1
x2 − 3
Domain of gf(x) = domain f = R.
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11. Example 7.
Two functions, f, g are defined by
f : x → x2
, x ≥ 0
g : x → 2x + 1, 0 ≤ x ≤ 1
(a) Show that the composite function fg exist. Find (b) its rule and domain and (c) corresponding
range.
Solution:
(a) To show that fg exists, we need to show that Rg ⊆ Df .
1
1
3 y = g(x)
x
y
Since Dg = [0, 1], from the graph of g, we observe
that Rg = [1, 3].
Rg = [1, 3]
Df = [0, ∞)
⇒ Rg ⊆ Df
Therefore, fg exists.
(b)
f(g(x)) = f(2x + 1)
= (2x + 1)2
Dfg = Dg = [0, 1]
(c) To find Rfg,
Step 1. Draw the graph of f.
Step 2. For the domain, set it as Range g, which in this case is [1, 3]
Step 3. Find the range under this new domain.
−1 1 3
1
9
y = f(x)
x
y
Therefore, Rfg = [1, 9].
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12. Example 8 (2009/NYJC/Prelim/I/2a modified).
The functions f and g are defined by
f : x → x2
− 1, x ∈ R,
g : x →
√
x + 4, x ∈ R+
.
(i) Show that the composite function fg exists.
(ii) Define fg in similar form.
(iii) Find the range of fg.
[ii) fg(x) = x + 3, x ∈ R+ iii) (3,∞)]
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13. 4.2 ff−1
(x) and f−1
f(x)
A function given by f(x) = x is called an identity function. It maps any value to itself.
f−1f
Diagram for function f−1f(x)
a b a
x
ff−1(x) = f−1f(x) = x
Dff−1 = Df−1 .
Df−1f = Df .
Key Observations
(a) Both ff−1(x) and f−1f(x) are equal to x.
(b) ff−1(x) and f−1f(x) have different domains. (Sketch the correct domain on a graph)
Example 9.
Let the function f be defined by
f : x → x + 4, x ∈ [0, ∞).
Sketch the graphs of the functions ff−1 and, f−1f on 2 separate graphs.
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14. Example 10.
Consider the following functions:
g : x →
1
x − 3
x ∈ R, x ̸= 3
g−1
h : x →
1
x
, x ≥ 0
Find the function h.
Solution:
h = gg−1
h = g(g−1
h) = g
(
1
x
)
=
1
1
x − 3
=
x
1 − 3x
Domain h = domain g−1
h = [0, ∞)
Example 11.
Consider the following functions:
fg : x → x2
− 6x + 14 x ∈ R+
g−1
: x → x + 5, x ∈ R
Find the function f.
[f(x) = x2 + 4x + 9, x ∈ R]
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15. Example 12 (Self-inverse Functions).
The functions f and g are defined as follows:
f : x →
5 − x
1 − x
, x ∈ R, x ̸= 1.
(i) Explain why f has an inverse, f−1, and show that f−1 = f.
(ii) Evaluate f51(4).
Solution:
(i) f(x) = 5−x
1−x = 1 + 4
1−x .
f′
(x) = 4(−1)(1 − x)−2
(−1) =
4
(1 − x)2
Since f′(x) > 0 for all x ∈ R, x ̸= 1, f is 1-1 and hence has an inverse.
y = 1 +
4
1 − x
y − 1 =
4
1 − x
1
y − 1
=
1 − x
4
4
y − 1
= 1 − x
1 +
4
1 − y
= x ∴ f−1
(x) = 1 +
4
1 − x
1
1
x
y
From the graph of f, Rf = R{1}. ∴ Df−1 = R{1} = Df . Hence f = f−1.
(ii)
f(x) = f−1
(x)
f2
(x) = x
f3
(x) = f
(
f2
(x)
)
= f(x)
f4
(x) = f
(
f3
(x)
)
= f(f(x)) = f2
(x) = x
...
∴ f51
(x) = f(x)
∴ f51
(4) = f(4) =
5 − 4
1 − 4
=
−1
3
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