Functions
Functions
Functions
This is the second of three
chapters that are meant primarily as a review
Functions
ofAMath
1050.
willofmove
quickly
through thisbetween
material.two sets.
function
is We
a way
describing
a relationship
A function is a way of describing a relationship between two sets.
A
isisa away
between
two
sets.
To
have a function
we
first
need atwo
sets, so lets
suppose
that
D and T
Afunction
function
wayof
ofdescribing
describing
arelationship
relationship
between
two
sets.
To have a function we first need two sets, so lets suppose that D and T
are
sets.
Then
a function
is something
that
assigns
every
x ⇥ that
D to aand
single
To
have
aafunction
we
need
sets,
soso lets
suppose
To
have
function
wefirst
first
needtwo
twothat
sets,assigns
let’severy
suppose
andTT
are
sets.
Then
a function
is something
x ⇥that
D toDD
a single
object
in TThen
.
are
aresets.
sets.
functionisissomething
somethingthat
thatassigns
assignsevery
everyx xC2DDtotoaasingle
single
object
in Then
T . a afunction
object
D is incalled
object
inT.T . the domain of the function, and T is called the target of the
D is called the domain of the function, and T is called the target of the
function.
We the
usually
assign
names
to our functions
—
though
usually
simple
DDisiscalled
ofof
the
and
called
the
target
ofofthe
called
thedomain
domain
thefunction,
function,
andTTisis
called
the
targetsimple
the
function.
We usually
assign
names
to our functions
—
though
usually
and
generic
names
—
like
g,
for
example.
Naming
the
function
lets
us
give
function.
We
usually
totoour
usually
simple
function.
We
usually
assign
names
ourfunctions
functions
—though
though
usually
simple
and
generic
names
—assign
like g,names
for example.
Naming the
function
lets us
give
a
specific
name
to
the
object
in
the
target
that
the
function
assigns
to give
an
and
generic
names
like
g,g,for
example.
Naming
the
function
Inlets us
and
generic
names
—
like
for
example.
Naming
the
function
a specific name to the object in the target that the function assigns to an
object
in the
domain
as object
follows:
aobject
name
totothe
ininthe
aspecific
specific
name
the
thetarget
targetthat
thatthe
thefunction
functionassig~
assigns to an
in the
domain
asobject
follows:
object
in
the
domain
as
follows:
object in the domain as follows:
If x ⇥ D, then g(x) ⇥ T is the object that g assigns to x.
If x ⇥ D, then g(x) ⇥ T is the object that g assigns to x.
IfIfx xe2D,D,then
theng(x)
g(x)e2TTisisthe
theobject
objectthat
thatg gassigns
assignstotoz.x.
Writing the symbols
Writing the symbols
Writing
Writingthe
thesymbols
symbols
g:D
T
g:D
T
g:D-*T
g:D!T
is a shorthand for writing that g is a function that assigns every x ⇥ D to
is a shorthand for writing that g is a function that assigns every x ⇥ D to
a issingle
object in for
T . writingthat
isa ashorthand
shorthand
thatg gisisa afunction
functionthat
thatassigns
assignsevery
everyxxe2DDtoto
a single
object infor
T .writing
a asingle
singleobject
objectininT.T .
8
D
19
11
8
8
Polynomials. Some of the most important examples of functions in mathematics are polynomials. These are functions of the form p : R ! R
where p(x) = an xn + an 1 xn 1 + · · · + a1 x + a0 for some fixed real numbers
an , an 1 , . . . , a0 .
For example, q : R ! R where q(x) = 5x3 2x2 + x 1 is a polynomial.
The domain of a polynomial is R. That is, you can put any real number into
a polynomial function. If you do put a real number into a polynomial, then
the output is another real number. In other words, the target of a polynomial
is R.
Identity function. The identity function is the function that assigns every
object in the domain to itself. (To have an identity function, the domain and
target have to be the same set.) Identity functions are important enough that
they get to have a name that is reserved for identity functions only: id. In
other words, if D is a set, then the identity function for the set D is described
by
id : D ! D where id(x) = x.
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Some important functions
For much of this course, the functions we’ll be interested in are functions
whose domains are subsets of R. Below are 12 very important examples of
functions that should be familiar to you.
I
n
x’)C.
contvit c’tion
e.Iev1
20
21
AL
LD
A
4
(b
(b
v
0-
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Graphing functions with restricted domains
Sometimes we are interested in functions that have smaller domains than
perhaps they possibly could have. These smaller domains are sometimes
called restricted domains.
Examples.
• The function f : [1, 2) ! R where f (x) = x2 has a restricted domain,
because the domain is told to us as being the rather small interval [1, 2) even
though it makes sense to square any real number.
If someone gives us a function, and tells us that the domain of that function
is [1, 2), then that’s its domain, regardless of what we think its domain should
have been.
When graphing a function with a restricted domain, we must only draw
the graph of the function above or below the restricted domain on the xaxis. That’s because the restricted domain is the only set of inputs that are
relevant for the function. You can see this in the picture below. The graph
for f : [1, 2) ! R is the portion of the graph of x2 that lies between the
numbers 1 and 2 on the x-axis.
z
I
-
£
4 40
The number 3 is not an object of the domain [1, 2), so it doesn’t make sense
to use 3 as an input for the function f , and therefore there is no point in
the graph of f that lies above or below the number 3 on the x-axis. More
22
generally, there are no points of the graph of f to the left of the number 1
on the x-axis, or to the right of the number 2 on the x-axis.
generally,
there
no
of
of
ff to
left
11
generally,
there are
arenotice
no points
points
of the
the graph
graph
of interval
to the
the[1,
left
of the
the
numberfor
It’s important
to
the endpoints
of the
2) of
on
the number
x-axis
on
x-axis,
or
the
the
number
22 on
the
on the
the
x-axis, The
or to
tonumber
the right
right
ofan
the
number
on
the
x-axis.
this
example.
1 isof
object
in the
set
[1,x-axis.
2), and f (1) = 12 = 1,
to
the
of
[1,
x-axis
for
It’s
important
to notice
notice
the endpoints
endpoints
of the
the
[1,2)
2) on
onusthe
the
x-axis
for
soIt’s
theimportant
point (1, 1)
is
a point
in the graph
of interval
finterval
. To remind
that
this
is
22
this
example.
The
number
1
is
an
object
in
the
set
[1,
2),
and
f
(1)
=
1
=
1,
this
example.
The
number
1
is
an
object
in
the
set
[1,
2),
and
f
(1)
=
1
=
1,
actually a point in the graph, we place a giant dot on the point (1, 1).
so
point
1)
aa point
in
graph
.. To
remind
us
isis
soInthe
the
point (1,
(1,
1) isisthat
point
in the
the
graph of
ofaffgiant
To dot
remind
us that
that this
this
contrast,
notice
instead
of drawing
on the
point
(2, 4)
actually
aa point
the
we
aa giant
dot
on
point
actually
point
in
the graph,
graph,
we place
place
giant
dotThat’s
on the
thebecause
point (1,
(1,
in
the graph
of fin
, instead
we drew
a little
circle.
2 1).
is1).not an
In
notice
of
aa giant
point
In contrast,
contrast,
notice that
that
instead
of drawing
drawingwhile
giant
dot4,on
on
the
point
(2,
4)
object
in the domain
of f ,instead
[1, 2). Therefore,
22 dot
=
thethe
point
(2,(2,
4) 4)
is
in
the
graph
of
f
,
instead
we
drew
a
little
circle.
That’s
because
2
is
not
an
in
the
graph
of
f
,
instead
we
drew
a
little
circle.
That’s
because
2
is
not
an
not a point in the graph of f . We draw the little circle
to remind us that the
22
object
in
the
domain
of
f
,
[1,
2).
Therefore,
while
2
=
4,
object
in
the
domain
of
f
,
[1,
2).
Therefore,
while
2
=
4, the
the point
point (2,
(2,4)
4) isis
point (2, 4) is not a point of the graph.
not
not aa point
point in
in the
the graph
graph of
of ff.. We
We draw
draw the
the little
little circle
circle to
to remind
remind us
us that
that the
the
• The
g : {2}
!graph.
R where g(2) = 2 has a restricted domain
point
4)
not
of
point (2,
(2,
4) isisfunction
not aa point
point
of the
the
graph.
that consists of the single number 2, so its graph should consist of a single
•• The
function
:: {2}
R
23 has
domain
The
function
{2} !
!
R where
where
g(2)
=
has a2a restricted
restricted
domain
point that
lies
directlyggabove,
below,
or ong(2)
the=
number
on
the x-axis.
In
that
consists
of
the
single
number
2,
so
its
graph
should
consist
of
a
single
that
consists
of
the
single
number
2,
so
its
graph
should
consist
of
a
single
this case, the graph of g consists of the single point (2, 2), since g(2) = 2, and
point
that
lies
directly
below,
on
number
22 on
x-axis.
In
point
that
liespoint
directly
above,
below,
oruse
onathe
the
number
on
the
x-axis.
In
we
draw
that
as aabove,
giant dot.
Weor
giant
dot here
forthe
two
reasons.
this
the
gg consists
of
point
(2,
since
g(2)
and
this case,
case,
the graph
graph
of
consists
of the
the
single
point
(2,2),
3),
sinceunlike
g(2) =
=a2,
3,little
and
First,
to remind
us of
that
it actually
is single
a point
in the
graph,
we
that
as
dot.
We
aa giant
dot
we draw
drawSecond,
that point
point
as aa giant
giant
dot.big
Wesouse
use
giant
dot here
here
for
two
reasons.
circle.
we make
the dot
that
it’s clear
thatfor
thetwo
dotreasons.
isn’t a
First,
to
remind
us
that
it
actually
is
a
point
in
the
graph,
unlike
a
little
First,
to
remind
us
that
it
actually
is
a
point
in
the
graph,
unlike
a
little
stray mark that we accidentally made with our pen.
circle.
circle. Second,
Second, we
we make
make the
the dot
dot big
big so
so that
that it’s
it’s clear
clear that
that the
the dot
dot isn’t
isn’t aa
stray
stray mark
mark that
that we
we accidentally
accidentally made
made with
with our
our pen.
pen.
ft
ft
c;.
c;.
I
-t
I
-t
40
40
• The function h : (2, 1) ! R where h(x) = x 1 has a domain of
(2, 1), so 2 is not an object in the domain. Thus, the point (2, 2 1) = (2, 1)
hh :: (2,
1)
!
R
h(x)
=
aa domain
of
The function
function
(2,So
1)we
!draw
R where
where
h(x)
= xxon 1the
1 has
has
domain
of
is not •a• The
point
in the graph.
a little
circle
point
(2, 1) to
(2,
not
object
domain.
point
(2,
1)
1)
(2,1),
1), so
so 22ofisisthat.
not an
anOtherwise,
object in
in the
the
domain.
Thus,
the
point
(2,22number
1) =
= (2,
(2,
1)
remind
us
every
point Thus,
to thethe
right
of the
2 on
is
aa point
the
graph.
So
aa little
on
to
is not
not
point
inthe
thedomain
graph. of
Soh,we
wesodraw
draw
littleofcircle
circle
on the
the point
point (2,
(2,to1)
1)the
to
the
x-axis
is inin
the graph
h stretches
forever
remind
point
remind us
us of
of that.
that. Otherwise,
Otherwise, every
every 22
point to
to the
the right
right of
of the
the number
number 22 on
on
the
the x-axis
x-axis isis in
in the
the domain
domain of
of h,
h, so
so the
the graph
graph of
of hh stretches
stretches forever
forever to
to the
the
22
23
5
—
,
13
right. We put a little arrowhead on the rightmost part of the graph that we
drew to indicate that the graph continues forever to the right.
-
0
0
In each of the three examples above, we labeled the domain of the function
on the x-axis. That isn’t really part of the graph. It was only drawn in these
three examples to illustrate the process of graphing functions with restricted
domains. If someone asked you to draw the graph of a function, you don’t
have to label its domain on the x-axis.
• The function q : R {2} ! R where q(x) = x2 is another example
of a function with a restricted domain. We are used to graphing x 2 when the
domain is R. The function q will have a very similar graph, we just have to
draw the graph of x2 with a little circle on the point of the graph of x2 that
reminds us that 2 is not in the
the domain.
domain. More
More precisely,
precisely, because
because 222 is= not
in
4, we
2
the domain,
because
4, we
need
need
to drawand
a little
circle2 at=the
point
(2,to
4).draw a little circle at the point
(2, 4).
24
23
Oncewe’ve
we’vedrawn
drawnthe
thelittle
littlecircle,
circle,we
wecan
candraw
drawthe
therest
restofofthe
thegraph
graphofofxx22
Once
makingsure
surethat
thatwe
wedo
donot
notpuncture
puncturethe
thelittle
littlecircle.
circle. There
Thereshouldn’t
shouldn’tbe
beany
any
making
partofofour
ourgraph
graphthat
thatisisinside
insideofofthe
thelittle
littlecircle
circlethat
thatwe
wedrew.
drew. That’s
That’show
how
part
weknow
knowthat
that(2,
(2,4)
4)isisnot
notaapoint
pointininthe
thegraph
graphofofq.q.
we
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Implied domains
domains
Implied
Sometimeswe
wewon’t
won’tgo
gothrough
throughthe
thetrouble
troubleofofwriting
writingthe
theentire
entirename
nameofofaa
Sometimes
5
5
functionas
as“f
“f : :DD!
!TT where
whereff(x)
(x)==xx ”.”. This
Thisisissimilar
similarto
tohow
howwe
weusually
usually
function
call people
people by
by their
their first
first names,
names, omitting
omitting their
their middle
middle and
and last
last names,
names, just
just
call
becauseit’s
it’seasier.
easier.
because
we are
are introduced
introduced to
to aa function
function that
that isis given
given by
by an
an equation,
equation, and
and its
its
IfIf we
domainisisnot
notspecified,
specified,we
wewill
willassume
assumethat
thatthe
thedomain
domainfor
forthat
thatfunction
functionisis
domain
the largest
largest subset
subset ofof the
the real
real numbers
numbers possible.
possible. This
This set
set will
will be
be called
called the
the
the
implieddomain
domain ofofthe
thefunction.
function.
implied
.1)
Examples.
Examples.
Letg(x)
g(x)==log
logee(x).
(x). The
Thepositive
positivereal
realnumbers
numbersare
arethe
theonly
onlyreal
realnumnum••Let
bersthat
thatwe
wecan
cantake
takeaalogarithm
logarithmof,
of,so
sothe
theimplied
implieddomain
domainofofthe
thefunction
function
bers
(0,1).
1).
gg isis(0,
polynomial —
— for
for example
example ifif p(x)
p(x) == 2x
2x 3,3, or
or p(x)
p(x) ==
•• IfIf pp isis aa polynomial
77
22
4x 17x
17x 2,2,or
orififp(x)
p(x)==44—
—then
thenwe
wecan
canuse
useany
anyreal
realnumber
numberas
asan
aninput
input
4x
forp.p. That
Thatis,
is,the
theimplied
implieddomain
domainisisR.
R.
for
25
24
• Recall that a rational function is a function of the form r(x) = p(x)
q(x)
where p(x) and q(x) are polynomials. For any real number a, p(a) and q(a)
are each real numbers, so the only problem here might be if the denominator
of the fraction equals 0. We can never divide by zero, so the implied domain
of r(x) = p(x)
q(x) is the set of all real numbers a for which q(a) 6= 0. That is, the
implied domain is the set of all real numbers except for the roots of q(x).
2
For example, the implied domain of 3x
{1}, because 1 is the only
x 1 is R
3
2
+4x 2
root of x 1. The implied domain of x 7x
is R { 2, 2}, because 2
x2 4
5
and 2 are the roots of x2 4. The implied domain of 4xx2 +117 is R, because
x2 + 1 is never equal to 0.
•p
If n 2 N is an even number, such as 2, 4, 6, 8, etc., then the function
h(x) = n x has an implied domain of [0, 1). We can’t take an even root of
a negative number.
On the other hand, we can
take an odd
root of any real number, so the
p
p
3
5
implied domain of f (x) = x or g(x) = x is R.
Examples to be careful with.
• The implied domain of the constant function g(x) = 1 is R.
Now suppose that f (x) = 1 + loge (x) loge (x). Because the function f
asks us to take a logarithm, and because we can only take the logarithm of a
positive number, the implied domain of f is (0, 1).
26
You must find the implied domain of a given function before you simplify it.
The function f above can be simplified as f (x) = 1, but the implied domain
is still (0, 1).
Because they have di↵erent implied domains, g and f are not the same
function.
• The implied domain of the polynomial p(x) = 2x is R.
Lf
/
2
Let r be the rational function r(x) = 2xx . We must never divide by 0, so
the implied domain of r is R {0}.
2
Now that we know that r(x) = 2xx has a domain of R {0}, we are free
to divide by x, because x can’t be 0 if x 2 R {0}. Then the function r
27
simplifies as r(x) = 2x. This doesn’t change the implied domain though. We
already said that the implied domain is R {0}, and it has to stay that way.
r
In particular, p is not the same function as r. They have di↵erent domains.
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Composition
Suppose that f : A ! B and g : B ! C are functions.
If a 2 A, then a is in the domain of f and f (a) 2 B. Since f (a) 2 B, we
have that f (a) is in the domain of g, so g(f (a)) is an object in C.
This process defines a third function, named g f : A ! C that is defined
by
g f (a) = g(f (a))
28
The function g f is pronounced “g composed with f ”.
Examples.
• Suppose g : R ! R and h : R ! R are functions and that g(1) = 3
and h(3) = 7. Then
h g(1) = h(g(1)) = h(3) = 7
• Suppose f : R ! R and g : R ! R are functions that are defined by
f (x) = x2 and g(x) = x 1.
Then g f : R ! R is the function defined by
g f (x) = g(f (x)) = g(x2 ) = x2
g : R ! R is the function defined by
And f
f
g(x) = f (g(x)) = f (x
1) = (x
1
1)2
Important: Notice in the previous example that g f (2) = 3 and f g(2) = 1.
That means that g f is not the same function as f g. In other words,
g f 6= f g.
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Piecewise defined functions
A piecewise defined function is a single function that is described using
multiple functions that are more common to us.
Examples.
• The function p : [1, 1) ! R where
8
2
>
<x
p(x) = 3
>
:x 1
if x 2 [1, 2);
if x = 2; and
if x 2 (2, 1).
Here’s how the function p works. If you are given a number in the domain
of p, [1, 1), and you want to see which number p assigns it to in the target,
first check which piece of the domain the number is in. Is the number in the
interval [1, 2), or does it equal 2, or is the number in the interval (2, 1)? If
29
—
—
p,
—
p,
p,
—
p,
the number is in [1, 2), then use the function p(x) = x2 ; if the number equals
2, then use the function p(x) = 2; and if the number
is in (2, 1), then use
the
thenumber
numberisisinin[1,
[1,2),
2),then
thenuse
usethe
thefunction
functionp(x)
p(x)==xx2 ;2 ;ififthe
thenumber
numberequals
equals
the function p(x) = x 1.
function
p(x)
==function
3;2;and
number
isisinnumber
2,then
thenuse
use
the
function
p(x)
andififthe
the=
number
in(2,
(2,1),
1),then
thenuse
use
the 2,
number
inthe
[1,
then
use
the
x2 ; if of
the
Let’s isgive
it a2),try.
The
number
4 is in p(x)
the domain
p, that is, equals
4 2 [1, 1).
the
p(x)
thefunction
functionfunction
p(x)==xx 1.1.= 2; and if the number is in (2, 1), then use
2, then
To use
findthe
p(4), noticep(x)
that 4 is in the third piece2 of p because 4 2 (2, 1).
the number
isgive
init[1,
2),
then
use
the function
p(x)
= x ; ifofof
the
Let’s
give
a
try.
The
number
44isisininthe
p,p,number
that
Let’s
it
a
try.
The
number
thedomain
domain
thatis,
is,4equals
422[1,
[1,1).
1).
the function
p(x)p(4)
= x = 1.
Therefore,
4 1 = 3. To check your understanding, verify that
2, then
use
function
p(x)
and
ifdomain
the
number
inp (2,
1), then
use
To
p(4),
notice
that
thethe
of
theisof
third
piece
of4 p2because
Tofind
findthe
notice
that4=4is2;is
in
third
piece
because
(2, 1).
Let’s
give
itp(4),
ap(2)
try.
number
4inis
p(1)
= 1,
=The
2, and
p(3) =
2. in the domain of p, that is, 4 2 [1, 1).
the function
p(x)
=
x
1.
4Therefore,
2 (2,
1).notice
Therefore,
= 3.
To
check
understanding,
verify
p(4) that
= 4 4p(4)
1 in
= 4the
3. 1To
check
your your
understanding,
verify
that
To find
p(4),
is
piece
because
4 we
2 (2,
The
function p(x)
comes
in threethird
pieces,
andoftopgraph
p(x),
just1).
have to
Let’s
give
it=ap(2)
try.p(2)
The
number
4 is =
in 2.
the domain of p, that is, 4 2 [1, 1).
that
p(1)
1,
=and
p(1)
= p(4)
1,
=
p(3)
=
Therefore,
= the
4 2,three
1 3,=and
3. p(3)
To2.
check
understanding,
verify
graph each
of
pieces.
That
is, your
we have
to graph p : [1,
2) !that
R where
To find
p(4),
notice
that
4
is
in
the
third
piece
of
p
because
4
2
(2,
1).
The
function
p(x)
comes
in
three
pieces,
and
to
graph
p(x),
we
just
have
The
function
p(x)
comes
in
three
pieces,
and
to
graph
p(x),
we
just
havetoto
2
p(1) =
1, p(2)
2, well
and p(3)
2. ! R where p(x) = 2, and also p : (2, 1)
p(x)
= x=
, as
as p =
: {2}
!R
Therefore,
p(4)ofof
=the
4 three
1 =pieces.
3. To
check
your
understanding,
verify
that
graph
each
That
is,
we
have
to
graph
p
:
[1,
2)
!
R
where
graph
each
the
three
pieces.
That
is,
we
have
to
graph
p
:
[1,
2)
!
R
where
Thewhere
function
p(x)
pieces,
and toeach
graph
justpieces
have to
p(x)
= xcomes
1. in
Wethree
already
graphed
of p(x),
thesewe
three
earlier
2 2= 2,
p(1)p(x)
=
1, =
p(2)
andas
p(3)
=
2. !
x
,
as
well
p
:
{2}
R
where
p(x)
=
3,
and
also
p
:
(2,
1)
p(x)
=
x
,
as
well
as
p
:
{2}
!
R
where
p(x)
=
2,
and
also
p
:
(2,
1)
!RR
graphineach
the three(Earlier,
pieces. That
is, we
have
to graph
: [1,h,
2) respectively.)
! R where!
thisofchapter.
they had
the
names
f , g, pand
So
Thewhere
function
p(x)
in
pieces,
and to
graph
p(x),
we
justpieces
have earlier
to
==xas
1.1.
We
already
graphed
each
ofofthese
three
p(x)
xcomes
Wethree
already
graphed
each
these
three
pieces
earlier
2 p(x)
p(x)where
=
x
,
as
well
p
:
{2}
!
R
where
p(x)
=
2,
and
also
p
:
(2,
1)
!
R
now we’ll just draw all three pieces together, and the result is the graph of
graph
each
the three
pieces. they
That
is, we
have
to graph
pand
: [1,
2)respectively.)
! R where So
in
chapter.
had
the
names
h,h,
in
thisof
chapter.
(Earlier,
theygraphed
had
the
names
f, ,g,g,and
respectively.)
wherep this
p(x)
=
x! R.
1.(Earlier,
We already
each
of fthese
three
pieces
earlier So
: [1,
2 1)
p(x)now
=
x we’ll
,we’ll
as well
as
p : all
{2}
! Rpieces
wheretogether,
p(x) = 2,and
andthe
alsoresult
p : (2,
R
just
draw
three
isis1)
the
graph
now
just
draw
all
three
together,
result
the!
graph
in this
chapter.
(Earlier,
they
hadpieces
the names
f , g,and
andthe
h, respectively.)
So ofof
where
=!
x!R.
1. We already graphed each of these three pieces earlier
: p(x)
:[1,
[1,1)
1)
R.
nowppwe’ll
just
draw
all three pieces together, and the result is the graph of
in this chapter. (Earlier, they had the names f , g, and h, respectively.) So
p : [1, 1) ! R.
now we’ll just draw all three pieces together, and the result is the graph of
p : [1, 1) ! R.
Sometimes, depending on whose writing it, the function p : [1, 1) ! R
above might also be written as
Sometimes,
is writing
Sometimes,depending
dependingon
onwho
whose
writing it,
it, the
the function
function pp : : [1,
[1,1)
1) !
! RR
above
also
asas8writing it, the function p : [1, 1) ! R
abovemight
might
alsobe
bewritten
written
Sometimes,
depending
on
whose
2
>
if 1  x < 2;
above might also be written as
<x
Sometimes, depending on whose88writing it, the function p : [1, 1) ! R
p(x) = >xx22 2
and
ifif1x=xx2;<<2;
2;
<<
>
above might also be written as8 >
:x 1 ifif 1x 
>
2.
2 32
p(x)
p(x)
=2;
2; and
and
>
if 1 if
ifxxx=
<
2;
<=x=>
>
::x 1 if x > 2.
8
x if1x =if 2;
x >and
2.
p(x)which
= >x2of
2 these
It doesn’t matter
two
ways
that you’d prefer to write p.
if
1

x
<
2;
>
<
:x 1 if x > 2.
Both are correct. p(x)
You=should
be used
reading
both though, because both
2of these
x to
=ways
2;
and
ItIt doesn’t
of
that
doesn’t matter
matter which
which
theseiftwo
two
ways
that you’d
you’d prefer
prefer toto write
write p.p.
>
are common ways of writing
defined functions.
:x piecewise
1 used
if xto
>reading
2.
correct.
You
be
both
though,
Bothare
arematter
correct.which
Youshould
should
be
used
to
reading
bothprefer
though,
because
both
29
It Both
doesn’t
of
these
two
ways
that you’d
tobecause
write p.both
defined
functions.
are
common
ways
ofwriting
writing
piecewise
defined
functions.
Bothare
arecommon
correct.ways
You of
should
be piecewise
used
to reading
both
though, because both
3029
It doesn’t matter which of these two ways
that you’d prefer to write p.
are common ways of writing piecewise defined functions.
Both are correct. You should be used29to reading both though, because both
are common ways of writing piecewise defined functions.
29
• Let’s look at the piecewise defined function q : R ! R where
q(x) =
(
x2
1
if x 6= 2; and
if x = 2.
There are two pieces here, q : R {2} ! R where q(x) = x2 , and q : {2} !
R where q(2) = 1. So q( 1) = ( 1)2 = 1, q(0) = 02 = 0, q(1) = 12 = 1,
q(3) = 32 = 9, and q(2) = 1.
We drew the graph of the first piece earlier in this chapter. The second
piece has as its graph a single giant dot on the point (2, 1), since q(2) = 1,
and the two pieces drawn together make up the graph for q : R ! R.
• The single most important piecewise defined function is the absolute
value function. It is the function defined by
|x| =
(
x
x
if x 0; and
if x < 0.
The domain of the absolute value function is R.
If x is a number that’s positive or zero, then |x| = x. This first piece of the
absolute value function is the identity function, with a domain of [0, 1). If
you take the absolute value of a positive number, or of 0, nothing happens.
You are returned the same number that you p
put in p
to the absolute value
function. For example, |0| = 0, |5| = 5, and | 2| = 2. The graph of the
31
1
-
first piece of the absolute value function is below. Notice that there is a giant
dot on the point (0, 0) since 0 is an object in the domain of this first piece,
[0, 1).
The domain of the second piece of the absolute value function is the set of
negative numbers, ( 1, 0). If x is a negative number, then |x| = x. For
example, | 3| = ( 3) = 3 and | ⇡| = ( ⇡) = ⇡. Notice that for this
second piece of the absolute value function, the e↵ect of the function is to
erase the minus sign that is in front of the negative number. Therefore, the
absolute value of a negative number is always positive. The graph of this
second piece is drawn below. Notice that there’s a little circle drawn at the
point (0, 0). That because 0 is not negative, and therefore it is not a number
in the domain of this second piece, ( 1, 0).
To draw the graph of the absolute value function, just graph both of its
pieces at once. Notice that the giant dot of the first piece at the point (0, 0)
32
fills up the little circle of the second piece at the point (0, 0). The result is
that there is no little circle in the graph of the absolute value function
33
Exercises
For #1-18, match the numbered functions on the left to the lettered implied
domains on the right.
1.) 3x2
2.) ex
11.)
2x+3
x7
3.)
p
2
x
13.)
x2
5
15.)
7.) ex
2
8.)
p
2x
p
2x
9.)
p
9
x3
p
2
B.) (0, 1)
x
C.) R
{0}
4x
x2 9
D.) R
{ 3, 3}
14.) e3x
5.) 11
6.)
2x3 4x+2
x+5
12.) e
4.) loge (x)
A.) R
10.) loge ( ex )
7x + 4
7
5
E.) [0, 1)
loge (x)
x+5
F.) ;
G.) R
16.) x + 5
17.)
13x2 + 2x
7
q
2
{ 5}
1
x
18.) loge ( |x| )
For #19-27, match the numbered functions below to their lettered graphs
on the next page.
19.)
p
2
: (1, 4) ! R
22.)
p
2
: [1, 4] ! R
25.)
p
2
20.)
p
2
: (1, 1) ! R
23.)
p
2
: {1} ! R
26.)
p
2
21.)
p
2
: [1, 4) ! R
24.)
p
2
: {4} ! R
27.)
p
2
34
: [1, 1) ! R
: (1, 1)
{4} ! R
: (1, 4] ! R
___
___
___
V.
H.)
4-
C.)
F:)
I.)
35
34
Li
‘43-
___
Let f : R ! R, g : R ! R, and h : R ! R be the funtions f (x) = x3 ,
g(x) = x2 , and h(x) = (x 1). For #28-33, match the numbered functions
on the left to the lettered functions on the right that they equal.
g(x)
A.) x3
29.) g f (x)
B.) x6
30.) f
h(x)
C.) x2
2x + 1
31.) h g(x)
D.) x2
1
32.) g h(x)
E.) x3
3x2 + 3x
28.) f
1
1
33.) h f (x)
For #34-38, identify the given absolute value.
34.) |5|
35.) |0|
36.) |
7|
37.) | 38 |
38.) |
2
3|
Match the numbered piecewise defined functions below with their lettered
graphs on the next page.
39.) f (x) =
(
40.) g(x) =
(
ex
1
if x 6= 1; and
if x = 1.
loge (x)
1
8
>
if
<5
41.) h(x) = |x| if
>
:3
if
8
1
>
<x
42.) p(x) = 2
>
:5 x
if x 2 (0, e); and
if x 2 [e, 1).
x < 3;
3  x  4; and
x > 4.
if x 2 (0, 1);
if x = 1; and
if x 2 (1, 4].
36
37
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