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C H A P T E R
7
E
Functions revisited
Objectives
PL
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To define sums and products of functions and graph these new functions.
To define composite functions and graph these new functions.
To understand and find inverse functions and relations.
To be able to recognise the general form of possible models of data.
To apply a knowledge of functions to solving problems.
7.1
Operations on functions
SA
M
In the previous six chapters, families of functions have been introduced and compositions and
inverses have been considered. In this section, we revisit these concepts with all the functions
of Mathematical Methods (CAS) at our disposal.
Example 1
For g(x) = |x| and f (x) = sin x:
a Find the rules for f ◦ g(x) and g ◦ f (x)
b Sketch the graphs of y = f ◦ g(x) and y = g ◦ f (x) for x ∈ [−2, 2] and state the range of
each of these composite functions.
Solution
f ◦ g(x) = f (g(x)) = f (|x|) = sin (|x|)
g ◦ f (x) = g( f (x)) = g(sin x) = |sin x|
a
y
b
y
y = |sin x|
y = sin (| x|)
0.8
0.5
–6
–4
–2 0
–0.5
0.6
x
2
4
0.4
6
0.2
0
Range = [–1, 1]
x
Range = [0, 1]
260
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261
Using the TI-Nspire
Define g(x) = abs(x) and f (x) = sin(x)
by using define from the Actions menu
(b 1 1).
Enter f (g(x)) and g( f (x)) as shown.
SA
M
PL
Open a Graphs & Geometry application
(c 2) and let f 1(x) = f (g(x)) and
f 2(x) = g( f (x)).
Select an appropriate window setting (b
4 1).
The graph of f (g (x)) is shown.
E
P1: FXS/ABE
Using the Casio ClassPad
Define g(x) = abs(x) and f (x) = sin x using
Interactive > Define.
Note that although there is a function b in the
b menu for entering absolute value, this
cannot be used within the Define box and you
must type abs(x).
Enter f (g(x)) and g( f (x)) to find f ◦g(x) and
g ◦ f (x).
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Essential Mathematical Methods 3 & 4 CAS
E
Enter f (g(x)) and g( f (x)) into line y1 = and
menu. Tick the box for the
y2 = in the
to produce the
graph you want and tap
graph. The screen shows f (g(x)).
Example 2
For g(x) = |x| and f (x) = loge x:
a State the maximal domain for g for f ◦ g to exist and the maximal domain for f for g ◦ f to
exist.
b Find the rules for f ◦ g(x) and g ◦ f (x)
c Sketch the graphs of y = f ◦ g(x) and y = g ◦ f (x)
Solution
PL
P1: FXS/ABE
SA
M
a For f ◦ g to exist the range of g must be a subset of the domain of f. The maximal
domain of f is R+ and hence f ◦ g is defined for x ∈ R\{0}, i.e. the maximal
domain of f ◦ g is R\{0}. For g ◦ f to exist the range of f must be a subset of the
domain of g. The maximal domain of g is R and hence g ◦ f is defined for x ∈ R+ ,
i.e. the maximal domain of g ◦ f is R+ .
b f ◦ g(x) = f (g(x)) = f (|x|) = loge |x|
g ◦ f (x) = g( f (x)) = g(loge x) = | loge x|
y
y
c
2
y = |loge x|
2
y = loge |x|
1.5
1
1
–2 –1.5 –1 –0.5 0
–1
0.5
0
x
x
0.5 1 1.5
2
–2
(1, 0)
It is sometimes useful to be able to express a given function as a composition of
two functions. This will be seen for differentiation in Chapter 9.
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Example 3
Express each of the following as the composition of two functions:
2
b h(x) = sin |x|
c h(x) = (x2 − 2)n where n is a natural number
a h(x) = ex
Solution
2
E
a h(x) = ex
b h(x) = sin |x|
x
2
Choose f (x) = e and g(x) = x
Choose f (x) = sin x and g(x) = |x|
Then h(x) = f ◦ g(x)
Then h(x) = f ◦ g(x)
2
n
c h(x) = (x − 2) where n is a natural number
Choose f (x) = xn and g(x) = x2 − 2
Then h(x) = f ◦ g(x)
This is not the only possible choice of functions, but the ‘natural’ choice is made.
Example 4
PL
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1
Let f (x) = e2x and g(x) = √ for x ∈ R+ .
x
Find f −1 , g−1 , f ◦ g, g ◦ f, (f ◦ g)−1 , (g ◦ f )−1 , f −1 o g−1 and g−1 o f −1
Solution
1
loge x, x ∈ R+
2
1
g−1 (x) = 2 , x ∈ R+
x −1
1
√2
f ◦ g(x) = f √
= e2x 2 = e x , x ∈ R+
x
1
2x
g ◦ f (x) = g(e ) = x = e−x , x ∈ R
e
√2
For (f ◦ g)−1 , let x = e y 2
2
2
Then loge x = √ and y =
y
loge (x)
2
−1
2
(f ◦ g) (x) =
, x ∈ R+
loge (x)
SA
M
f −1 (x) =
−1
g of
−1
1
loge x
(x) = g
2
2
2
=
, x ∈ R+
loge (x)
−1
1
f o g (x) = f
x2
1
1
= loge 2
2
x
= −loge x, x ∈ R+
Note that (g ◦ f )−1 (x) = f −1 o g−1 (x) and (f ◦ g)−1 (x) = g−1 o f −1 (x)
1
For (g ◦ f ) , let x = y and y = −loge x
e
(g ◦ f )−1 (x) = −loge x, x ∈ R+
−1
−1
−1
−1
Exercise 7A
1 For each of the following, where g(x) = |x|:
i Find the rules f ◦ g(x) and g ◦ f (x)
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ii Find the range of y = f ◦ g(x) and y = g ◦ f (x) (and state the maximal domain for
each of the composite functions to exist).
a f (x) = 3 sin 2x
b f (x) = −2 cos 2x
c f (x) = ex
d f (x) = e2x − 1
e f (x) = −2ex − 1
f f (x) = loge 2x
h f (x) = −loge x
g f (x) = loge (x − 1)
2 Express each of the following as the composition of two functions:
3
h(x) = e x
b h(x) = cos |2x|
2
n
d h(x) = cos (x2 )
h(x) = (x − 2x) where n is a natural number
2
2
4
f h(x) = (x − 1)
g h(x) = loge x2
h(x) = cos x
h(x) = |cos 2x|
i h(x) = (x2 − 2x)3 − 2(x2 − 2x)
E
a
c
e
h
√
3 The functions f and g are defined by f: R → R, f (x) = e4x and g: R + → R, g(x) = 2 x
Find each of the following:
a g ◦ f (x)
PL
P1: FXS/ABE
b (g ◦ f )−1 (x)
c f ◦ g−1 (x)
4 The functions f and g are defined by f: R → R, f (x) = e−2x and g: R → R, g(x) = x3 + 1
a Find the inverse function of each of these functions.
b Find the rules f ◦ g(x) and g ◦ f (x) and state the range of each of these composite
functions.
1
x +1
b Solve the equation f (x) = f −1 (x) for x.
5 The function f is defined by f: (−1, ∞) → R, f (x) =
SA
M
a Find f −1 .
6 The functions f and g are defined by f: (−1, ∞) → R, f (x) = loge (x + 1) and
g: (−1, ∞) → R, g(x) = x2 + 2x
a Define f −1 and g−1 , giving their rules and domains.
b Find the rule for f ◦ g(x)
7 The functions f and g are defined by f: (0, ∞) → R, f (x) = loge x and g: (0, ∞) → R,
1
g(x) = . Find f ◦ g(x) and simplify f (x) + f ◦ g(x)
x
8 The functions
g and h are defined by g: R → R, g(x) = 5x2 + 3 and h: [3, ∞) → R,
x −3
. Find h(g(x)).
h(x) =
5
9 For f (x) = (x − 4)(x −6) and g(x) = x2 − 4:
a Find f (g(x)) and g( f (x))
b Solve the equation g( f (x)) − f (g(x)) = 158 for x.
10 For f (x) = 4 − x2 , solve the equation f ( f (x)) = 0 for x.
11 For f (x) = ex − e−x , show that:
a
f (−x) = − f (x)
b [ f (x)]3 = f (3x) − 3 f (x)
12 Let g: R → R, loge g(x) = ax + b. Given that g(0) = 1 and g(1) = e6 , find a and b and
hence g(x).
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7.2
Inverse relations
In Chapter 1, we showed that for one-to-one functions an inverse function can be found. In
fact, the inverse of any relation may be defined. The inverse relation is not a function, however,
unless the initial relation is a one-to-one function.
Recall that a relation S is a set of ordered pairs. If the set S is described as {(a, b)}, i.e.
S = {(a, b)}, then the inverse relation T is described as {(b, a) : (a, b) ∈ S}.
For the relation
{(1, 2), (1, 3), (4, 2), (4, 3), (1, 5)}
the inverse relation is
{(2, 1), (3, 1), (2, 4), (3, 4), (5, 1)}
E
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PL
In general the inverse relation can be found by reversing the first and second coordinates.
For example, the function f: R → R, f (x) = x2 can be written as {(x, y): y = x 2 }.
The inverse relation is {(x, y): x = y 2 }.
The function f: R → R, f (x) = x2 + 2 can be written as {(x, y): y = x2 + 2}, and the inverse
relation is {(x, y): x = y2 + 2}.
The relation {(x, y): x2 + y2 = 4} has inverse {(x, y): y2 + x2 = 4}, which is the same set.
√
√
The relation {(x, y): y = ± x} has inverse {(x, y): x = ± y}, which is equal to
{(x, y): y = x2 }.
The functions of the families
A, B, b ∈ R, A = 0
h(x) = A|x + b| + B
A, B, b ∈ R, A = 0
SA
M
f (x) = A(x + b)2 + B
A
g(x) =
+B
(x + b)2
A, B, b ∈ R, A = 0
are many-to-one functions and therefore the corresponding inverse relations are not functions.
Example 5
Find the inverse relation of the function f (x) = 4(x + 1)2 and sketch the graph of both
relations on the one set of axes.
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y
Solution
Consider x = 4(y + 1)2
x
Solving for y:
= (y + 1)2
4
x
∴ y+1=±
4
x
and
y = −1 ±
4
(0, 4)
(4, 0)
(–1, 0)
0
E
The inverse relation is:
1√
x
(x, y): y = −1 ±
2
y=x
(0, –1)
The graphs of the original function and its inverse relation do not meet.
If the function f is restricted to form a new function
PL
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f ∗ : [−1, ∞) → R, f ∗ (x) = 4(x + 1)2
f ∗ is a one-to-one function and therefore the inverse is also a function.
Example 6
Find the inverse of the function h(x) = e|x|
SA
M
Solution
The function is not one-to-one. The range of h(x) = e|x|
is [1, ∞). The domain of the inverse is [1, ∞).
Let y = e|x| . Consider x = e|y|
Therefore loge x = |y| and y = ± loge x for x ≥ 1
Example 7
2
Find the inverse of the function with the rule h(x) = x 3 + 16
Solution
2
2
h(x) = x 3 + 16. Consider x = y 3 + 16.
2
Then y 3 = x − 16 and y2 = (x − 16)3 .
Therefore y = ± (x − 16)3
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x
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Using the TI-Nspire
E
£
Complete as shown.
Use the up arrow (£ ) to move up to the
answer and use the right arrow ( ) to
display the remaining part of the answer.
PL
P1: FXS/ABE
Using the Casio ClassPad
Enter and highlight x = y ∧ (2/3) + 16, then tap
Interactive – Equation/inequality– solve and
ensure the variable is set to y.
SA
M
Example 8
Find the inverse of the function g: R\{1} → R, g(x) =
1
(x − 1)2
Solution
The function is not one-to-one. The range of
g is R+ . Therefore the domain of the inverse
relation is R+ .
1
Consider x =
(y − 1)2
1
1
Therefore (y − 1)2 = and y − 1 = ± √
x
x
1
Therefore y = 1 ± √
x
y
y=x
y=1
x
0
x=1
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Exercise 7B
1 Let f : [a, ∞) → R, f (x) = (x − 1)2 + 2
a
b
c
e
Find the smallest value of a such that f is one-to-one.
With this value of a, state the range of f.
Sketch the graph of f.
d Find f −1 and state the domain and range of f −1 .
Sketch the graph of f and f −1 on the one set of axes.
E
P1: FXS/ABE
2 Find the inverse relation of each of the following functions:
a f (x) = 2x 2
d g(x) = (x + 1)2 − 1
b f (x) = 2x 2 − 1
e h(x) = 2(x − 1)2 + 1
c
f
f (x) = (x + 1)2
f (x) = 1 − (x − 1)2
PL
3 Find the inverse relation of each of the following functions, and state its domain:
1
a g: R\{2} → R, g(x) =
(x − 2)2
1
b g: R\{1} → R, g(x) =
+1
(x − 1)2
2
c g: R\{−1} → R, g(x) =
(x + 1)2
d f (x) = |x − 1|
e f (x) = |x − 2| + 3
f f (x) = |x + 1| − 2
4 a For f (x) = 2x 5 + 6 find f −1 (x) and state the domain of f −1 .
2
b For f (x) = 2x 5 + 6 find the rule and domain for the inverse relation.
SA
M
3
7.3
Sums and products of functions and addition
of ordinates
In Chapter 1, we established that for functions f and g new functions ( f + g) and (fg) can be
defined by:
( f + g)(x) = f (x) + g(x) (f g)(x) = f (x) g(x)
dom ( f + g) = dom ( f ) ∩ dom (g) dom (f g) = dom ( f ) ∩ dom (g)
In Chapter 3, we considered graphing by addition of ordinates. The new functions that have
been defined in Chapters 4 to 6 may now be included.
Example 9
For f (x) = cos x and g(x) = e−x :
a Find the rules for (f + g)(x) and (fg)(x)
b Evaluate (f + g)(0) and (fg)(0)
Solution
a (f + g)(x) = cos x + e−x and (fg)(x) = e−x cos x
b (f + g)(0) = 1 + 1 = 2 and (fg)(0) = 1 × 1 = 1
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Example 10
y
6
Sketch the graph of
y = f (x)
5
y = ( f + g)(x)
4
where f (x) = x
and g(x) = 2 − 2x 2
3
2
2
1
Solution
y = ( f + g)(x)
x
1 2
E
–2 –1 0
–1
It can be seen that the function with rule
( f + g)(x) = 2 − x 2 is defined by the
addition of the two functions f and g.
–2
–3
–4
y = g(x)
–5
–6
Example 11
PL
P1: FXS/ABE
For f (x) = x and g(x) = e2x sketch the graph of y = ( f + g)(x)
Solution
y
SA
M
Note that ( f + g)(0) = 0 + e0 = 1 and
( f + g)(x) = 0 implies e2x + x = 0. This cannot
be solved analytically but a calculator
can be used to find that a numerical solution
is −0.43, correct to two decimal places. Also
note that as x→−∞, (f + g)(x) → x from ‘above’.
y = e2x
y=x
1
y = e2x + x
0
x
Exercise 7C
1 For f (x) = e−2x and g(x) = −2x:
a Find the rules for:
i ( f + g)
ii
b Evaluate: −1
ii
i ( f + g)
2
x 2 For f (x) = sin
and g(x) = −2x:
2
a Find the rules for:
i ( f + g)
ii
b Evaluate:
i ( f + g)(1)
ii
(fg)
−1
(fg)
2
(fg)
(fg)(1)
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3 For f (x) = cos
x 2
a Find the rules for:
i ( f + g)
b Evaluate:
i ( f + g)(0)
and g(x) = ex :
ii (fg)
ii (fg)(0)
E
4 Sketch the graph of f (x) = x 2 and g(x) = 3x + 2 on the one set of axes and hence, using
addition of ordinates, sketch the graph of y = x 2 + 3x + 2.
5 Copy and add the graph of y = ( f + g)(x) using addition of ordinates.
y
a
y
b
y = f (x)
y = g(x)
(2, 3)
(–1, 1)
PL
P1: FXS/ABE
y = f (x)
0
y = g(x)
(1, 1)
x
0
x
(–1, –1)
(1, –1)
6 For each of the following, sketch the graph of f + g
f : R → R, f (x) = x 2 , g: R → R, g(x) = 3
√
f : R → R, f (x) = x 2 + 2x, g: R + ∪ {0} → R, g(x) = x
√
f : R → R, f (x) = −x 2 , g: R + ∪ {0} → R, g(x) = x
SA
M
a
b
c
7 Sketch the graph of f (x) = e−2x and g(x) = −2x on the one set of axes and hence, using
addition of ordinates, sketch the graph of y = e−2x − 2x
8 Sketch the graph of f (x) = 2e2x and g(x) = x + 2 on the one set of axes and hence, using
addition of ordinates, sketch the graph of y = 2e2x + x + 2
9 For each of the following, find the rule and sketch the graph of:
i ( f + g)(x)
ii ( f g)(x)
a f (x) = |x| and g(x) = x
c f (x) = −|x| and g(x) = x
7.4
b
f (x) = |x| and g(x) = −x
Identities with function notation
Many of the properties which have been investigated for the functions introduced in the first
six chapters may be expressed using function notation.
For example the rules for logarithms
loge (x)
loge (y) = loge (x y)
+
x
= loge (x) − loge (y)
loge
y
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271
can be written in the following way if f = loge :
f (x) + f (y) = f (x
y)
x
f (x) − f (y) = f
y
Example 12
E
a For the function with rule f (x) = 2x, verify that f (x + y) = f (x) + f (y) for all x and y.
b For the function with rule f (x) = x + 2 verify that f (x + y) = f (x) + f (y) for all x
and y.
Solution
a f (x + y) = 2(x + y) = 2x + 2y = f (x) + f (y)
b f (x + y) = (x + y) + 1 = x + 1 + y + 1 − 1 = f (x) + f (y) − 1
If f (x) + f (y) − 1 = f (x) + f (y) then −1 = 0, which is a contradiction.
Example 13
If f (x) =
PL
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1
show that f (x) + f (y) = (x + y) f (x y) for x and y non-zero real numbers.
x
Solution
1
y+x
1
1
+ =
= (x + y) ×
= (x + y) f (x y)
x
y
xy
xy
SA
M
f (x) + f (y) =
Example 14
For the function f (x) = e x , give the functional identities for f (x + y) and f (x − y).
Solution
f (x + y) = e x+y = e x e y = f (x) f (y)
ex
f (x)
f (x − y) = e x−y = y =
e
f (y)
Example 15
For the function f (x) = cos x, give a counterexample to show that f (x + y) = f (x) + f (y)
for all x and y.
Solution
For x = 0 and y = :
f (0 + ) = f () = −1
f (0) + f () = 1 + −1 = 0
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Exercise 7D
1 For f (x) = 2x, find an equivalent expression for f (x − y) in terms of f (x) and f (y).
2 For f (x) = 2x + 3, show that f (x + y) can be written in the form f (x) + f (y) + a and
give the value of a.
E
3 A function g satisfies the property that [g(x)]2 = g(x). Find the possible values of g(x).
4 For the function with rule f (x) = |x|, give a counterexample to show that
f (x + y) = f (x) + f (y) for all x and y.
PL
5 For the function with rule f (x) = sin x, give a counterexample to show that
f (x + y) = f (x) + f (y) for all x and y.
6 For the function with rule f (x) =
1
, show that f (x) + f (y) = (x 2 + y 2 ) f (x y)
x2
7 a For the function with rule h(x) = x 2 , give a counterexample to show that
h(x + y) = h(x) + h(y) for all x and y.
b Show that h(x + y) = h(x) + h(y) implies x = 0 or y = 0
8 Show that for g(x) = 23x , g(x + y) = g(x)g(y)
SA
M
9 Show that the functions with rule of the form f(x)= x n where n is a natural number
x
f (x)
satisfy the identities f (x y) = f (x) f (y) and f
=
y
f (y)
10 For the function with rule f (x) = ax where a ∈ R\{0, 1}, give a counterexample to show
that f (x y) = f (x) f (y) for all x and y.
7.5
Families of functions and solving
literal equations
This section demonstrates a different use of parameters. They can be used to discuss families
of relations.
Here are some familiar families of relations:
f
f
f
f
:
:
:
:
R
R
R
R
→
→
→
→
R, f (x) = |mx|, where m ∈ R.
R, f (x) = ax 3 where a ∈ R\{0}.
R, f (x) = |mx + 2| where m ∈ R.
R, f (x) = kemx where m ∈ R\{0} and k ∈ R\{0}.
Their use makes it possible to describe general properties.
What can be said in general about each of these? The family of functions of the form
f : R → R, f (x) = |mx + 2| where m ∈ R is explored in the following example.
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Example 16
For f : R → R, f (x) = |mx + 2| where m ∈ R + :
a Find the x-axis intercept.
b For which values of m is the x-axis intercept less than −2?
c Find the inverse relation of f.
d Find the equation of the line perpendicular to the graph at the point with coordinates (0, 2).
E
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Solution
SA
M
PL
2
2
a |mx + 2| = 0 implies mx = −2 and x = − . The x-axis intercept is − .
m
m
2
2
b If − < −2, then
>2
m
m
Multiply both sides of the inequality by m (m > 0).
2 > 2m
Therefore m < 1
Therefore the x-axis intercept is less than 2 for 0 < m < 1
c Consider x = |my + 2| and solve for y.
−2
If my + 2 ≥ 0, i.e. y ≥
m
then x = my + 2
x −2
my = x − 2 and y =
m
x
2
Therefore y =
−
m
m
−2
If my + 2 < 0, i.e. y <
m
then x = −my − 2
y=−
2
x
−
m
m
1
.
m
1
The equation is determined as y − 2 = − x
m
1
and the gradient–intercept form is y = − x + 2
m
d The perpendicular line has gradient −
The following example demonstrates how literal equations can be formed and solved with the
exponential and logarithmic functions discussed in the previous two chapters.
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Essential Mathematical Methods 3 & 4 CAS
Example 17
Solve each of the following for x. All constants are positive reals.
b loge (x − a) = b
c loge (cx − a) = 1
a aebx − c = 0
Solution
b loge (x − a) = b
x − a = ex
x = ex + a
Example 18
c loge (cx − a) = 1
cx − a = e
cx = a + e
a+e
x=
c
E
c
a c
bx = loge
a 1
c
x = loge
b
a
a ebx =
PL
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The graph of a quadratic function passes through the points (1, 6) and (2, 4). Find the
coefficients of the quadratic rule in terms of c, the y-axis intercept of the graph.
Solution
SA
M
Let f (x) = ax 2 + bx + c be a function in this family. Then f (1) = 6 and f (2) = 4
The following equations are obtained:
a + b + c = 6 and 4a + 2b + c = 4
Solving these gives:
20 − 3c
c−8
b=
a=
2
2
The equation of the quadratic in terms of c is:
c − 8 2 20 − 3c
y=
x +
x +c
2
2
Example 19
A transformation
is defined
equation
by the
matrix
AX
+ C = X , where
5 0
5
x
x
A=
,C =
,X =
and X =
where k is a non-zero real number.
0 k
2
y
y
a Solve the equation for X.
1
under this transformation.
x
c Find the value of k if the image passes through the origin.
b Find the image of the curve with equation y =
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a AX + C = X
Therefore X = A−1 (X − C)
1

0 5
 x −5
=
1  y− 2
0
k

1
(x − 5)


= 5

1 (y − 2)
k
E
Solution
x − 5
y − 2
and y =
5
k
5
y − 2
= b The image has equation
k
x −5
Therefore x =
and
PL
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y=
5k
+2
x −5
c If the graph passes through the origin, 0 = −k + 2 and k = 2
Exercise 7E
SA
M
1 Consider f : R → R, f (x) = mx − 4 where m ∈ R\{0}.
Find the x-axis intercept.
For which values of m is the x-axis intercept less than or equal to 1?
Find the inverse function of f.
Find the coordinates of the point of intersection of the graph of y = f (x) with the
graph of y = x
e Find the equation of the line perpendicular to the line at the point with coordinates
(0, −4).
a
b
c
d
2 Consider f : R → R, f (x) = −2x + c where c ∈ R.
Find the x-axis intercept.
For which values of c is the x-axis intercept less than or equal to 1?
Find the inverse function of f.
Find the coordinates of the point of intersection of the graph of y = f (x) with the
graph of y = x
e Find the equation of the line perpendicular to the line at the point with coordinates
(0, c).
a
b
c
d
3 Consider the family of quadratics with rule of the form y = x 2 − bx where b is a
non-zero real number.
a Find the x-axis intercepts.
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b Find the coordinates of the vertex of the parabola.
c i Find the coordinates of the points of intersection of the graph of y = x 2 − bx
with the line y = −x in terms of b.
ii For what value(s) of b is there one intersection point?
iii For what value(s) of b are there two intersection points?
a The graph of f (x) = x 2 is translated to the graph of y = f (x + h). Find the possible
values of h if f (1 + h) = 8
b The graph of f (x) = x 2 is transformed to the graph of y = f (ax). Find the possible
values of a if the graph of y = f (ax) passes through the point with coordinates (1, 8).
c The quadratic with equation y = ax 2 + bx has vertex with coordinates (1, 8). Find
the values of a and b.
√
5 Consider the family of functions with rule of the form f (x) = 2a − x, where a is a
positive real number.
PL
E
4
SA
M
a State the maximal domain of f.
b Find the coordinates of the point of intersection of the graph of y = f (x) with the
graph of y = x
c For what value of a does the line with equation y = x intersect the graph of y = f (x)
at the point with coordinates (1, 1)?
d For what value of a does the line with equation y = x intersect the graph of y = f (x)
at the point with coordinates (2, 2)?
e For what value of a does the line with equation y = x intersect the graph of y = f (x)
at the point with coordinates (c, c) where c is a positive real number?
6 Consider the function with rule f (x) = |x 2 − ax|.
a
b
c
d
State the coordinates of the x-axis intercepts.
State the coordinates of the y-axis intercept.
Find the maximum value of the function in the interval [0, a].
Find the possible values of a for which the point (−1, 4) lies on the graph of y = f (x)
7 Solve each of the following for x. All constants are positive reals.
b c loge (x + a) = b
a −aebx + c = 0
c loge (cx − a) = 0 d eax+b = c
8 Consider the functions with rule of the form f (x) = c loge (x − a) where a is a positive
constant.
a
b
c
d
State the equation of the vertical asymptote.
State the coordinates of the x-axis intercept.
State the coordinates of the point where the graph crosses the line y = 1
If the graph of the function crosses the line y = 1 when x = 2, find the value of c in
terms of a.
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9 Consider the family of functions with rule f (x) = e x−1 − b where b > 0.
a State the equation of the horizontal asymptote.
b State the coordinates of the x-axis intercept.
c Give the values of b for which the x-axis intercept is:
i at the origin
ii a negative number
E
10 The graph of a quadratic function passes through the points (−1, 6) and (1, 4). Find the
coefficients of the quadratic rule in terms of c, the y-axis intercept of the graph.
PL
11 The graph of a cubic function passes through the points (−1, 6), (1, −2) and (2, 4). Find
the coefficients of the cubic rule in terms of d, the y-axis intercept of the graph.
20 − 3c
c−8
2
x +
x + c. Find the values of c
12 A quadratic function has rule y =
2
2
for which:
a the graph of y = f (x) touches the x-axis
b the graph of y = f (x) has two distinct x-axis intercepts
13 The graph of a cubic function passes through the points (−2, 8), (1, 1) and (3, 4). Find the
coefficients of the quadratic rule in terms of d, the y-axis intercept of the graph.
SA
M
14 A transformation
is defined
by the matrix
equation AX
+
C = X , where
−4 0
3
x
x
A=
,C=
,X=
and X =
where k is a non-zero real
0 k
2
y
y
number.
a Solve the equation for X.
1
under this transformation.
x
c Find the value of k if the image passes through the origin.
−4
0
,
15 A transformation is defined by the matrix equation AX + C = X , where A =
0 2
a
x
x
C=
,X=
, and X =
where a is a non-zero real number.
−2
y
y
b Find the image of the curve with equation y =
a Solve the equation for X.
b Find the image of the curve with equation y = 2x under this transformation.
c Find the value of a if the image passes through the origin.
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Chapter summary
g ◦ f : X → R, where g ◦ f (x) = g( f (x))
E
Let f and g be functions such that dom f ∩ dom g = Ø. The sum, f + g, and the product, fg,
as functions on dom f ∩ dom g are defined by:
r (f + g)(x) = f (x) + g(x)
r fg(x) = f (x) · g(x)
In general for the composition of g with f to be defined, the range of f ⊆ domain of g. When
this composition (or composite function) of g with f is defined it is denoted g ◦ f
For functions f and g with domains X and Y respectively and such that the range of f ⊆ Y,
the composite function of g with f is defined by:
PL
If f is a one-to-one function, a new function, f −1 , called the inverse of f, may be defined by:
f −1 (x) = y if f (y) = x, for x ∈ ran f, y ∈ dom f
For a function f and its inverse f −1 :
dom f −1 = ran f
ran f −1 = dom f
The inverse of any relation may be defined. The inverse relation is not a function unless
the initial relation is a one-to-one function.
For a relation S = {(a, b)} the inverse relation is {(b, a)}.
SA
M
Review
278
Multiple-choice questions
x4 + 2
can be drawn by adding the ordinates of
x2
the graphs of two functions f and g. The rules for f and g could be:
2
2
B f (x) = x2 , g(x) = 2
A f (x) = x4 , g(x) = 2
x
x
2
4
4
2
D
f
(x)
=
x
+
2,
g(x)
= 2
C f (x) = x + 2, g(x) = x
x
E f (x) = x2 , g(x) = 2
1 The graph of the function with rule h(x) =
2 Which one of the following functions is not a one-to-one function?
1
A f: R+ → R, f (x) = 2
B f: R → R, f (x) = x3
x
C f: R → R, f (x) = 10x
D f: R+ → R, f (x) = log10 x
E f: R → R, f (x) = cos x
4
3 The function f has domain R and rule f (x) = 2 + 9. The rule of the inverse relation is:
x
2
2
2
A y=2± √
B y=3± √
C y=3± √
x −3
x
x −9
2
2
D y = ±√
E y = ±√
9−x
x −9
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279
E [0, 3]
E
5 The range of the function f: R → R, f (x) = 3|cos x| − 1 is:
A [1, 4]
B [−1, 2]
C [−4, 2]
D [−2, 1]
PL
6 For the function with equation f (x) = ex , which one of the following is not correct for all
positive real x and y?
A f (x + y) = f (x) f (y)
B f −1 (xy) = f −1 (x) + f −1 (y)
1
E f −1 (x) =
C f −1 (xy ) = y f −1 (x)
D f −1 (1) = 0
f (x)
is equal to:
7 If f (x) = cos x and g(x) = 3x2 then g f
3
2
1
3
4
A cos
B √
C 1
D
E
9
4
3
2
8 Which of the following is not an even function?
A f: R → R, f (x) = |2x|
B f: R → R, f (x) = cos2 x
C f: R → R, f (x) = cos x
D f: R → R, f (x) = 4x2 − 3
E f: R → R, f (x) = (x − 2)2
SA
M
9 It is known that the graph of the function with rule y = 2ax + cos 2x has an x-axis intercept
when x = . The value of a is:
−1
1
E
C 2
D −2
A 2
B
2
2
10 If, for x > 5, g(x) = loge (x − 5) and 2[g(x)] = g( f (x)) then f (x) is equal to:
A 5x − 8
B x2 − 10x + 30
C 5x2
D (2x − 10)2
E 2x − 2
Short-answer questions (technology-free)
1 For each of the following, where g(x) = |x|:
i Find the rules f ◦ g(x) and g ◦ f (x)
ii Find the range of y = f ◦ g(x) and y = g ◦ f (x) (and state the maximal domain for each
of the composite functions to exist).
a f (x) = 3 cos 2x
b f (x) = loge 3x
d f (x) = −loge 2x
c f (x) = loge (2 − x)
2 Express each of the following as the composition of two functions:
a h(x) = cos |2x|
b h(x) = (x2 − x)n where n is a natural number
d h(x) = −2 |sin 2x|
c h(x) = loge (sin x)
2
4
2
2
e h(x) = (x − 3x) − 2(x − 3x)
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Review
4 The function f has domain R and rule f (x) = |x| + 2. The domain and rule of the inverse
are:
A [−∞, 4] and x = |y| + 2
B [0, ∞) and x = ±y + 2
C R and x = ±y + 2
D [2, ∞) and x = |y| + 2
E R\{2} and x − y = 2
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x and g(x) = e−x :
2
a Find the rules for:
i ( f + g)
ii (fg)
3 For f (x) = 2 cos
b Evaluate:
i ( f + g)(0)
ii (fg)(0)
PL
E
4 Let f: [a, ∞) → R, f (x) = −(3x − 2)2 + 3
a Find the smallest value of a such that f is one-to-one.
b With this value of a, state the range of f.
c Sketch the graph of f.
d Find f −1 and state the domain and range of f −1 .
e Sketch the graph of f and f −1 on the one set of axes.
5 Find the inverse relation of each of the following functions:
a f (x) = (3x − 2)2 + 4
b f (x) = −2x2 + 3
Extended-response questions
1 The deficit of a government department in Ningteak, a small monarchy east of Africa, is
shown below. The deficit is continually assessed over a period of 8 years. The graph shown
below is that of the deficit over these 8 years:
SA
M
Review
280
Deficit
in millions (0, 1.8)
of Ningteak
dollars (D) (1, 1.6)
0
1
(3, 1.5)
2
3
4
5
6
7
8
Time (t)
The graph is read as follows: ‘The deficit at the beginning of the 8-year period is
$1.8 million. At the end of the third year the deficit is $1.5 million, and this is the smallest
deficit for the period, 0 ≤ t ≤ 8’.
a Find the rule for D in terms of t, assuming that it is of the form D = at 2 + bt + c
b Use this model to predict the deficit at the end of 8 years.
2 The rainfall, R, recorded during a very rainy day in North Queensland, was as follows:
At 4:00 a.m.
7.5 mm per hour
At 8:00 a.m.
9 mm per hour
At 10:00 a.m.
8 mm per hour
Assume a quadratic rule of the form R = at 2 + bt + c is applicable for 0 ≤ t ≤ 12 where
t = 0 is 4:00 a.m. Use the quadratic model to predict the rate of rainfall at 12:00 noon. At
what time was the rate of rainfall greatest?
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281
1
x −1
State the range of f and g.
Find f −1 and g−1 .
i Find g ◦ f
ii Sketch the graph of y = g ◦ f (x)
i Find (g ◦ f )−1 ii Sketch the graph of y = (g ◦ f )−1 (x)
√
4 a For f: [5, ∞) → R, f (x) = x − 3:
i Sketch the graph of y = f (x) for x ∈ [5, ∞).
ii State the range.
iii Find f −1 .
√
b For h: [4, ∞) → R, h(x) = x − p with inverse function h−1 that has domain
[1, ∞):
i Find p.
ii Find the rule for h−1 .
iii Sketch the graphs of y = h(x) and y = h−1 (x) on the one set of axes.
1
5 Let f: (0, ) → R with f (x) = sin x and g: [1, ∞) → R and g(x) =
x
a Find the range of f.
b Find the range of g.
c Give a sufficient reason why f ◦ g is defined and find f ◦ g(x)
d State, with sufficient reason, whether g ◦ f is defined.
e Find g−1 , giving its domain and range.
f Give a sufficient reason why g−1 ◦ f is defined and find g−1 ◦ f (x). Also state the domain
and range of this function.
SA
M
PL
E
a
b
c
d
6 A population of insects is determined by a rule of the form
c
t ≥0
n=
1 + ae−bt
where n is the number of insects alive at time t days.
a Consider the population for c = 5790, a = 4 and b = 0.03.
i Find the equation of the horizontal asymptotes by considering values of n as t
becomes large.
ii Find n when t = 0.
iii Sketch the graph of the function.
iv Find the exact value of t for which n = 4000.
b
i Use your calculator to find values of a, b and c such that the population growth yields
the following table:
t
1
10
100
n
1500
2000
5000
ii Sketch the graph for this population.
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Review
3 Consider f: R+ → R, f (x) = e−x and g: (−∞, 1) → R, g(x) =