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FIRST YEAR CALCULUS
W W L CHEN
c


W W L Chen, 1982, 2008
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It is available free to all individuals, on the understanding that it is not to be used for financial gain,
and may be downloaded and/or photocopied, with or without permission from the author
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Chapter 2
FUNCTIONS

2
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Introduction
Let us begin with two very simple examples which everybody can understand
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1
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Consider a simple test where there are 4 questions each of which is marked 1 (correct)
or 0 (incorrect), and a student is awarded a mark equal to the number of correct answers obtained
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The rule is then
given by a function f : A → B, where
f (1111) = 4,
f (1011) = 3,
f (0111) = 3,
f (0011) = 2,

f (1110) = 3,
f (1010) = 2,
f (0110) = 2,
f (0010) = 1,

f (1101) = 3,
f (1001) = 2,
f (0101) = 2,
f (0001) = 1,

f (1100) = 2,
f (1000) = 1,
f (0100) = 1,
f (0000) = 0
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1
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The set of even natural numbers can be obtained by taking the set N of all natural
numbers and multiplying each of them by 2
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Chapter 2 : Functions

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First Year Calculus

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W W L Chen, 1982, 2008

More formally, let A and B be sets
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We write f : A → B : x 7→ f (x) or simply f : A → B
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The element f (x) is called the image of x under f
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Two functions f : A → B and g : A → B are said to be equal, denoted by f = g, if f (x) = g(x) for
every x ∈ A
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This is defined by
G = {(x, f (x)) : x ∈ A} = {(x, y) : x ∈ A and y = f (x) ∈ B}
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1
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Consider the function f : R → R defined by f (x) = 2x for every x ∈ R
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Also, we have f (1) = 2 and f (−2) = −4
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1
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Consider the function f : N → N defined by f (x) = 2x for every x ∈ N, as discusssed
in Example 2
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2
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Also, we have f (1) = 2, while it is inappropriate to discuss f (−2), since −2 does not
belong to the domain of the function
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1
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Consider the function f : R → R defined by f (x) = x2 for every x ∈ R
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Example 2
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6
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Consider the function f : R → S
defined by f (x) = x2 for every x ∈ R
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The functions in Examples 2
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5 and 2
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6 are different, although they share the same defining formula
and domain
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1
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This is a very important point in the definition of a function
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Sometimes, we make our choice to suit our precise
needs
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1
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In the previous four examples, the functions have defining formulas
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Suppose that A = {1, 2} and B = {a, b, c}
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Example 2
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8
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It follows that the distance travelled by light in
time t is given by the formula f (t) = ct
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An appropriate choice for the domain may be the
set S of all non-negative real numbers, in which case an appropriate choice for the codomain will be S
again
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Example 2
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9
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Then it is convenient to represent each point on the disc in polar coordinates r and θ, where 0 ≤ r ≤ 1
and 0 ≤ θ < 2π
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For instance, we may take B = R
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1
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Suppose that the air resistence that an object encounters is proportional to the speed
of the object
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The domain must be a set of the form [0, V ], where V is a
suitably chosen number not exceeding the speed of light
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Then we have a function f : [0, V ] → [0, R], where f (v) = kv for every v ∈ [0, V ]
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2
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Example 2
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1
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We may
first represent the altitude of land by a non-negative real number and the depth of sea by a negative
real number
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Then (x, y) ∈ [−180, 180] × [−90, 90], and we can represent
the altitude or depth at the point (x, y) by a real number which we denote by h(x, y)
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Next,
we may use some colour to denote the ranges of altitude and depth
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To produce a map, we now need to associate position of any point on earth with the colour that represents
its altitude of depth
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Suppose that A, B and C are sets and f : A → B and g : B → C are functions
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Put simply, for
every x ∈ A, in order to find (g ◦ f )(x), we apply the function f first to x, followed by the function g to
f (x)
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Example 2
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2
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2
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The first of these give the altitude or depth of points on earth, while the second one gives
colours corresponding to ranges of these altitudes and depths
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The picture
h

(x, y) 7−→

h(x, y)

s

7−→

s(h(x, y)) = (s ◦ h)(x, y)

describes this composition
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Chapter 2 : Functions

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First Year Calculus

W W L Chen, 1982, 2008

Example 2
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3
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Then
(g ◦ f )(x) = g(f (x)) = g(x2 ) = x2 − 1
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It will be a little clearer if we think of the question as
follows
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After
all, x and y are “dummy” variables which we simply use to represent arbitrary elements of R
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(2)

g(x2 ) = g(y) = y − 1 = x2 − 1
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Example 2
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4
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2
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We have
(f ◦ g)(x) = f (g(x)) = f (x − 1) = (x − 1)2
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This simple example shows that (g ◦ f )(x) = (f ◦ g)(x)
does not hold in general
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2
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Suppose that the functions f : R → R, g : R → R and h : R → R are defined by
f (x) = x2 , g(x) = x − 1 and h(x) = x3 + 3x for every x ∈ R
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Here it is convenient to think of the functions g : R → R and h : R → R as defined by g(y) = y − 1
for every y ∈ R and h(z) = z 3 + 3z for every z ∈ R
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Then
(g ◦ f )(x) = x2 − 1 as before, so that
(h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = h(x2 − 1)
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(5)

On combining (4) and (5), we obtain
(h ◦ (g ◦ f ))(x) = (x2 − 1)3 + 3(x2 − 1)
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To do so, we first study h ◦ g
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(7)

Now write z = g(y) = y − 1, so that
h(y − 1) = h(z) = z 3 + 3z = (y − 1)3 + 3(y − 1)
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(9)

((h ◦ g) ◦ f )(x) = (h ◦ g)(f (x)) = (h ◦ g)(x2 )
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In view of (9), we have
(h ◦ g)(x2 ) = (h ◦ g)(y) = (y − 1)3 + 3(y − 1) = (x2 − 1)3 + 3(x2 − 1)
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(12)

Note that the right hand sides of (6) and (12) are identical
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ASSOCIATIVE LAW
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Then h ◦ (g ◦ f ) = (h ◦ g) ◦ f
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On the other hand, we clearly have
(h ◦ (g ◦ f ))(x) = h((g ◦ f )(x)) = h(g(f (x)))
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2
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Real Valued Functions
We are primarily interested in real valued functions
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Suppose now that some defining formula is given
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We illustrate this point by a number of examples
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3
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We wish to find the largest set D√of real numbers such that f : D → R, defined by
f (x) = x for every x ∈ D, is a function
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However, as long as x ≥ 0, and as long as we specify which square root we take, then the function is
clearly defined
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Example
√ 2
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2
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Then for x2 + x to be real valued, we must make sure
that x2 + x = x(x + 1) ≥ 0; in other words, we must have x ≥ 0 or x ≤ −1
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In
this case, we can therefore take D = {x ∈ R : x ≥ 0 or x ≤ −1}
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3
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We wish to find the largest set D of real numbers such that f : D → R, defined by
f (x) = (x2 − 4)−1 for every x ∈ D, is a function
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However, as long as x2 − 4 6= 0, then the function is clearly defined
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Example 2
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4
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Then for (x2 − 4)−1 to be real valued, we must make
sure that x2 − 4 6= 0 (to ensure that we do not divide by 0) and x2 − 4 ≥ 0 (to ensure that the square
root is real)
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However, as long as x2 − 4 > 0, and
as long as we specify which square root we take, then the function is clearly defined
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Chapter 2 : Functions

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First Year Calculus

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W W L Chen, 1982, 2008

We can in fact vary the question somewhat
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3
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Consider the set N = {1, 2, 3,
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We wish to find the largest
set D of real numbers such that f : D → N, defined by f (x) = x − 1 for every x ∈ D, is a function
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However, as
long as x ≥ 2, then the function is clearly defined
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In Chapters 3 and 6–8, we shall adopt the following convention
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Furthermore, the domain D is a set of real
numbers and, unless specified, is chosen to be the largest such set so that f : D → R is a function
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4
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The choice of domain and codomain is
entirely at our disposal
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However, we need two definitions
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A function f : A → B is said to be one-to-one if x1 = x2 whenever f (x1 ) = f (x2 )
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A function f : A → B is said to be onto if for every y ∈ B, we can find x ∈ A such that
f (x) = y
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A function f : A → B is
one-to-one if no two different elements in the domain can share the same image
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Example 2
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1
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Example 2
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2
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Example 2
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3
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Example 2
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4
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Then the function f : R → S,
defined by f (x) = x2 for every x ∈ R, is onto but not one-to-one
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Let y ∈ B
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Since f is one-to-one, there cannot be more than one such x ∈ A,
for otherwise they would share the same image y
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This means that we can define a function g : B → A, with domain B and codomain A and
such that g(y) = x precisely when f (x) = y
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It is not difficult to see that g : B → A is also one-to-one and onto
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PROPOSITION 2A
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If the function f : A → B is one-to-one and
onto, then there exists a function g : B → A such that g(y) = x whenever f (x) = y
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Example 2
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5
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Clearly the inverse function g : R → R is defined by g(y) = y/2 for every y ∈ R
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4
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Consider the function f : R− → R+ , where f (x) = x2 for every x ∈ R−
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It
is not difficult to see that the function is one-to-one and onto
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Example 2
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7
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Note that f (1) = 2, f (3) = 4, f (5) = 6,
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Hence f is one-to-one
and onto
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2
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One-to-One and Onto Real Valued Functions
By Proposition 2A, a given function f : A → B has an inverse if it is one-to-one and onto
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Our task is to
find sufficient conditions for f to be one-to-one and onto, so that it has an inverse
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(1) By an open interval in R, we mean a set of the form (A, B) = {x ∈ R : A < x < B}
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Remarks
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(2) The definition is extended to A = −∞ and B = ∞, provided that the interval is open at that end
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The last one
is simply R
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(1) A function f is said to be strictly increasing in an interval I if f (x1 ) < f (x2 ) for every x1 , x2 ∈ I
satisfying x1 < x2
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Example 2
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1
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Example 2
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2
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To see this, suppose
that x1 < x2
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PROPOSITION 2B
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Suppose further that the function f : I → R
is strictly increasing or strictly decreasing
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Proof
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However, we still need to have the onto property
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We have the following result
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Suppose that I is an interval in R
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Then f : I → T is one-to-one and onto,
and there exists a function g : T → I such that g(y) = x whenever f (x) = y
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Proof
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The result now
follows from Propositions 2A and 2B
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5
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The function f (x) = x2 is strictly increasing in the interval [0, 2)
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It follows from Proposition 2C that f : [0, 2) → [0, 4) has an inverse function
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On the other hand, the
function f (x) = x2 is strictly decreasing in the interval (−2, 0]
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It follows from Proposition 2C that f : (−2, 0] → [0, 4) has an inverse function
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Finally, consider the
function f (x) = x2 in the interval (−2, 2)
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Clearly the function f : (−2, 2) → [0, 4) is neither strictly increasing nor strictly
decreasing in the interval (−2, 2), so Proposition 2C does not apply in this case
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In the statements of Propositions 2B and 2C, it is not necessary for the domain of the
function to be an interval I
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Chapter 2 : Functions

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c


First Year Calculus

W W L Chen, 1982, 2008

Problems for Chapter 2
2x+2
and h(x) = |x|
1
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Here [x] denotes the greatest integer not
exceeding x, so that, for example, [5] = 5, [4 12 ] = 4 and [−4 21 ] = −5
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h) Describe the function f ◦ g : R → R
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j) Show that h ◦ h = h
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Consider the functions f : R → R, g : R → R and h : R → R, defined by f (x) = sin x, g(x) = x − π
and h(x) = x2 + 1 for every x ∈ R
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d) What is h ◦ (g ◦ f )?
√
3
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Given f (x) = cos x and g(x) = x2 − x + 1, find each of the following composite functions:
a) f ◦ g
b) f ◦ f
c) g ◦ g
d) g ◦ f
5
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Find the largest possible domain and corresponding range for each of the following functions as a
real valued function:
√
x+1
a) f (x) = x2 − 4x + 3
b) f (x) = 4 − x2
c) f (x) =
x−2
√
1
d) f (x) = |x + 2| − 1
e) f (x) = x + 1
f) g(x) =
x
x
3
g) f (x) = e
h) g(x) = x + 1
7
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For each of the following functions f , draw a graph of the function with the given domain D,
determine whether with a suitable choice of codomain T , which you must specify, the function
f : D → T has an inverse function and, if so, find the inverse function:
a) f (x) = 1 + 2x; D = (4, 7]
b) f (x) = sin x; D = [0, π]
c) f (x) = cos x; D = [0, π]
d) f (x) = x2 − 2x + 4; D = [1, 2]
e) f (x) = √
x2 − 2x + 4; D = [0, 2]
f) f (x) = √1 − x2 ; D = (−1, 1)
g) f (x) = 1 − x2 ; D = (0, 1)

Chapter 2 : Functions

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