tffi
iiiiiriiii1:i:iilii:i:liliiiiiiriiti:iii:iiiii
iliiffi
INrquALlnES
FuNcrroNS
Tnr LINE
ln this chapter we develop the topics in algebra and
Beometry needed for learning calculus. The subjects we
emphasize are algebraic inequalities, the definition of
function, and the basic properties of straight lines in the
plane. These are essential tools for mastering the main ideas
of calculus.
INEQUAUTTES
In elementary algebra and plane geometry we study equalities almost
exclusively. The solution oflinear and quadratic equations, the congruence of
triangles, and relationships among trigonometric functions arc topics concerned with equality. As we progress in the development of rnathematical
ideas, we shall see that the study of inequalities is both important and useful.
An inequality arises when we are more concerned with the approximate size
of ,a quantity than we are with its true value. Practically all laboratory
experiments in science deal with such approximations. AIso, since the proofs
of the most important theorems in calculus depend on approximations, it is
essential that we develop a facility for working with inequalities.
We assume the reader is familiar not only with ordinary numbers which
we call the reel number system but also with the laws of elementary algebra.
In this section we shall be conceroed with inequalities among real numbcrs,
and we begin by recalling some familiar relationships. Given that a and D are
any two real numbers, the sYmbol
a<b
CHAPTER
lNrqumrtrs. FuNCrloNs.
1
THE
tlNt
means that a is less than b.
we may also write the inequaiity in the opposite
direction.
b>a.
which is read b is greater than a.
The rules for handling inequalities are only slightly more cornplicated
However, the
than the ones we learned in algebra for manipulating equalitie.s.
about
Theorems
four
as
differences are so important that we state them
Inequalities and they should be-studied carefully'
THEoREM
1 If a <b and b < c, then i<c.
lnwords: if a is
less
thanb and b is
less
thqnc'
then a is less than c'
flffOAfnf
and
2 lf c is any number and a<b, then it is also true thatora+c<b+c
subtrscted
to
is
added
I
number
from
same
if
the
o - , . b c. ln worils:
each
THEOREM
side
of an inequalityr, the resuk is an inequality in the satne direction.
a o"a, <ithen
may be added
3 Ii r.
a* c <b + d. That is,inequalitiesinthe
same
directio,
is important to note that in general inequalities rnay not be subtracted'
For exampie,2 <5 and I < 7. We can say, by addition, that 3 < 12, but note
It
that subtiaction would state the absurdity that
ffffOnff"f
i
is less than
;2'
a If a < b and c is any positiue number, then
ac
<bc,
while if c is a negatiue nuwtber, then
ac
In
>
hc.
a! both stdes af an ineqwality by the same positiue
the diriction, while multiplicatian hy a negatitse number
words: multiplication
number
preserl)es
reuerses the directian af the inequality.
Since dividing an inequaiity by a number d is the-same as multiplying
by lld, we see-that
#
-3-2-l
\
TICURE
1
flGUff
2
0 1 2
3
Tireorem
4
it
applies for division as well as for
multiplication.
From the geometric point of view we associate a horizontal axis with the
totality of real numb*.r. Th" origin rnay be selec,ted at any eonvenient poiir!
with pos:tive nurmbers to the right and negative numbers to the left {Fig. 1).
For wery real numtrer there will be. a corresponding point on the line aod'
every point will represeni a real nurnber. Then the inequality
"onr.rr.iy,
a<b may be read: a is to the left of b. Tliis geornetric way cf looking at
inequalities is frequently of help in solving problerns. [t is efuo helpfitl to
introduce the notion of an interual o! nuwbers,or poie$s" {f a and b are
nurnbers {as shown in Fig.2}, then the open imterval froni a to b ii tbe
collection of all numbett itti"t are both iarger than e and smaller than b'
sEcTtoN
1.1
lNreu uflcs
That is, an open interval consists of all numbe rs between a and b,A number
x
is between a and b il both inequalities a < x and x < b are
true. A *.;;;;;
of writing'this is
a<x<b.
rh9 qlosed intervsl from a to D consists of all the points between a and b,
incluiling a and, (Fig. 3). suppose a number x is either equal to c o. targei
_b
than a, but we don't know which. we write this convenienily
, > o, *hi-"h
is read: x is greater than or equil to a. Similarly, x ( b is ,.ud, ,", is
iss than or
equal to b, and means that x may be either smaller than b or may be b itsclf.
A
compact yay of designating a closed interval from a to b is to ,t"t. that it
consists of all points x such that
a<
x Sb.
An interval which contains the endpoint b but not a is said to be helf-open
on
the left. That is, it consists of all points x such that
a<x<b.
similarly, an interval containing a but not b is called helf-open on the right,
and we write
aSx<b.
Parentheses and brackets are used as symbols for intervals
way:
in the following
(a,b) for the open intervat a<x<b,
la,b)for the closed interval: aSxlb,
{a, b) for the interval half-open on the left a < x < D,
lqb) for the interval half_open on the right c < x i D.
we can extend the idea of. an interval of points to cover some unusual
to consider ail numbers rarger than ?. Tlri. i";;
thought of as an interval extending te infnily on tf,e right. (Sec
Fig. 4.) Of
course' infinity is not a number, but we use the symbol (2, o) io
reprcscnt alr
numbers larger than 7. we could also write: all numbers x'such
that
cases. supposc we wish
7<x<o.
In a similar way, the symbol
The double inequality
(
-
co, 12) will stand for all numbers lcss than 12.
-o<'x<12
is arr Equivalent way of representing all numbers x less than
tr2.
The first-degree equation 3x *
19 has a unique solutioa
7:
x = 4. The
quadratic equation x2
has
two solutions, x= _f aoi i=2,_The
-x-2:0
trigonometric equation sin x : l has an infinite number of
solutions: x = 30o,
150o, 390o, 510", .... The sorution of an equation involviaj
a ,ioE, 'ortno*o,
say x, is the colbction af afi rumbers whtchmake ttn
,qr;lrliton oiiirorr rn .
This is. called the solution set of the equation. similarly, tt
a .oiutioo or *
incewlitr involving a single unknown, say x, is the couectioi oi"ri'ou.*o
which make the incquality a true statemint. For exa.pte, tt
rioeqLuty
3x-7<8
CHAPTER
INEQUALmES.
1
FuNcnoNs' Txt t'ttlr
this we'argue in
has as its solution oll numbers less than 5. To demonstrate
inequality we
above
the
satisfies
which
If x is a number
the following way.
and obtain a true
can, by Theirern 2, add 7 to both sides of the inequality
statement. That is, we have
3x-7 +7 <8+7
or
3x
<
15'
Now, dividing both sides by 3 (Theorem 4), we obtain
-x<5
speaking,
and observe that dx is asolutior.,then it is less than 5. Strictly
5 is a
than
is
less
which
number
every
that
proued
however, we have not
any
is
x
that
supposing
by
begin
proof
would
we
solution. In an actual
number less than 5; that is,
x<5.
Wemultiplybothsidesby3(Theorem4)andthensubtractT(Theorem2)to
get
3x-7<8,
the
the original inequality. Since the condition that x is less than 5 implies
to
noJice
originai inequaliiy, *i hur. proved the result. The important thing
is that the iroof consisted of reuersing the steps of the briginal argument
which led to the solution x < 5 in the first place. So long as each of the steps
so fa11s
we take is reoersible,the above procedure is completely satisfactory
< l5 is
<
8
to
3x
The step going from 3x -'7
obtaining solutions i,
"on."tn"d.
inequali'
the
reversible, since these two inequalities ari equivalent. Similarly,
set
ties 3:t < 15 and x < 5 are equivalent. Finally, note that the solution
consists of all numbers in the interval ( - m' 5)'
.
Methods for the solution of various types of simple algebraic inequalities
carefully.
are shown in the following examples, which should be studied
EXAMPTE
1
Solve for x:
-7-3x<5x*29.
Solution Subtract 5x from both
sides, getting
-7 -8x<29'
Multiply both
sides
by
-
1,
reversing the direction of the inequality, to obtain
7+8x>-2g.
Subtracting 7 from both sides yields 8x
solution
> -36, and dividing by
*> -2,
or, stated in interval form: all x in the interval
(
-3, o)'
8 gives the
fi
To verify the correctness ofthe result, it is necessary to perform the above
is
steps'in ,.ror" order. However, the observation ihat each individual step
revlrsible is sufficient to check the validity of the answer'
$ECriON
EXAMru
2
Solve for x(x
*0):
3-<).
x
Solution
we have an immediate inclination to murtiply both sidei by x. However,
since we don't know in advance whether x ii positive o, n"gutiu., we must
proceed cautiously. we do this by considering two cases: (l) iis positivq
and
(2) x is negative.
case I. suppose x > 0. Then multipllng by x preserves the direction of the
inequality (Theorem 4), and we get
3<5x.
Dividing by 5, we find that x > t. This me4ns that we must find all numbers
which satisfy both of the inequalities
x>0
and x>$.
clearly, any number greater than $is also positive, and the solution in case r
consists of all x in the interval (t, o).
case
2.
x < 0. Multiplying by
x
reuersesthe direction o[ the inequality.
have
we
3>5x,
and therefore $ > x. we seek all numbers x, such
x<0
thrt bothof the inequalities
and xct
hold. The solution in case 2 is the collection of all x in the interval ( o,0).
A way of combining the answers in the two qases is to state that the solution
set consists of all numberc x not in the closed interval
Fig.5.)
t0il.(see
D
EXAMPTE 3
Solve for
x(x*
-2):
2x*3.1
x+2 - 3'
Solution
As in Example 2,we must consider two cases, according to whether x
positive or negative.
Case
l. x*2>0.
We rnultiply by
*
2 is
3(x+2), which is positive, getting
6x-9<x+2.
Adding
9
-
x to both sides, we have
5x<
ll,
from
which x<$.
Since we}ave already assumed that x + 2 > e and since we must have,
< +,
we see that x must be larger than
and smaller than $. That is, tle
-2
solution set consists of all x in the interval (-2,$).
cHAPrEt
lNEeu[mEs. fuxcnons. Trc um
1
Case2.
x*2<0.
Again multiplying
by
3(x
* 2) and reaersing the in-
equality, we obtain
6x-9) x *2,
$. In this case, x must be less than -2 and greater than
irnpossible.
Combining the cases, we get as the solution set all
which
is
$,
numbers in (-2,$). See Fig.6.
or 5x > l1 and x >
1
In Problems
I
PROBLEMS
through 18, solve for x.
l2x-3<1
22x*4<x-5
3 5-3x<14
4
2-5x<3+4x
,i)tlr-6)<4-(2+5x) G)*-+.?.?
2l
-<3x
t2
9_<_
5 l-x
43
ll -<:
xj
7
,r'+l.s
rsfi.-z
<z
n =IJ-x
42
8:<-3x
32
lo
4< x1
:? <2 and (x- l)(x+4)<0
x
*27
28 Show that Theorem
I
Theorems
29 Given that a,
a<b
arrd
3l
x-l
12-<4
x+l>2.
x
)t
t4=-3<-+l
xx
2-x
16
r+l <,
x-2 x* I
x+J x
32 a)
lf
x and y are positive nurnbers, show that
lr
2tT-i.t
3-6x<2(x+5) and 7(2- x)<3x*8
2_3
23;:1<4 and x_2<7
3-x
*24 x-2 <3 and jI.
-'x* I x-2 s
'251<5 and x+2<7-3x
<i and (x-2)(x+3)<0
r\
I-+-l(x+},)>4.
\' Y/'
3
192x-7<5-x and 3-4x<{
3x-8<5(2-x) and 2(3x+4)-4x+7<5+x
2x*6 x - and t5-3x<4+2x
'25
c, and dare all positive numbers, and that
IFx is a positive number, proie that
b) If x, y, alrrd z are positive
numbers, show that
(t*t*1\t,
z/
\x Y
+y+z)>e.
c) If x, y, z, and w are positive
numbers, show that
it: + :I + II + a1\
|1x+r+i+w)
\x y z w/
I
20
1
for inequalities may be derived from
*30 a) State the most general circumstances in which the
hypotheses a < b and c < d imply that ac < bd.
b) Given that a < D and c < d, when is it true that ac> bd't
both inequalities hold.
--
D,
3
2.
c<d; show thal ac<bd.
In Problems 19 through 27,find the values of x, if any, for which
22
and
*d)
*33
Ifx
> t6.
Generalize the above results to n numbcrs 11; x2, ... r x6.
and y are any numbers different from zero, show that
\.Y{+z+>E+Q.
y'x-yx
*34 l-et x and y be positive numbers with x 2 y. Show that
1+:r=4*r.
yxx'
Show that the inequality is revorsed if y 2 x.
*Starred problerns are those that are unusually difficult.
sEcTtoN
Assoturr valur
1.2
_
)
ABSoI-UTE VALUE
If c is any positive number, the absolute value of c is defined to be a itself. If o
is negative, the absolute value of a is defined to be -a. The absolute value of
zero is zero. The symbol for the absolute value of a is lal. In other words, we
havc
DEFINITION
I r'rr t:xlimpl,-..
i1','=J. i l.]i:13,
t2-
5t:l-3i:3.
This rathcr sirnpic idca has irnportant consequences but, before we can
consider thcrr. we nrust discuss methods of solving equations and inequalities
involving absr:lute values.
EXAMPLE
1
Solution
EXAMPLT 2
Solution
Solve for x: lx
-
7l
- l.
This equation, accorr.ling tt; rhe definition of absolute value, expresses the
fact that x - 7 rnust bc J or .\, since in either case the absolutJvalue is 3.
If x- 7:3, we have.r;: l0; and if .x- ?= *3, then x:4. Wesee that there
are two values of
x which solvr: the equation: x :4,
Solve for x: l2x
6l
-
:
la
-
10.
fl
5xl.
The trvo possibilities are
2x*6:4-5x
and 2x-6:-(4-5x).
Solving each of these for x, we ohtain the two solutions
x:+' -3
d
z---=_\
____t__-.1___1__
llbil
ncunr
7
t
1
If c and
b are any two numbers, we may represent them as points on the
line as shown in Fig. 7. Then the distance beiween a and b, denotea by a, is the
length of the line segment from b to a. This distance is always positive (or zcro
if a: b). From the definition of absolute value we see that
-.
u'tlli*ll+t'ritiffi
If x
- {-----+}0
-4
RGt St 8
4
is any number, then
l"x I represents the distance d of x from the origin, i.e.,
The
statement
lxl.
lxl < 4 is equivalent to the condition that x is
any number in the interval extending from
to +4 (Fig. g). or, in terms of
tl:lx - 0l :
the symbol for intervals,
-4
x must lie'in the interval (-q,q\.
Sometimes a
cnAPrEn
tNEQUAtlrlEs. FuNcrloNs.
1
Tttt
t-tNt
interval
statement such as lxl <4 is used to denote the
statemen
the
inequalities without absolute value signs,
(-4'4)' In terms of
-4<x<4
the inequality i' - 3l <
is equivalent to lxl < 4. In a similar way-'
could also write
We
(
5)'
5,
lie in ihe interval ,
5 means
that
-'i**tt
-5<'x-3<5'
must satisfy' If we add 3 to
This consists of two inequalities, both of which x
then
each member of the double inequality above'
Therefore
gAMPI.E 3
Solution
-2<'x<8'
x must lie in the interval ( - 2, 8) See Fig' 9'
Solve for x: | 3x
-
4l <
't
'
form
n z1-A/
-7 <3x-4<7'
We rvrite the inequality in the equivalent
'
l
Now we add 4 to each term:
-3<3't<11'
and divide each term bY
3:
-1<x<#.
-I
flGUnt
il; i6-
Thesolutionsetconsistsofallnumbersxintheclosedinterval[_1,+].ST D
()
10
EXAMPLE 4
Solve for
x:12- 5xl<
3'
Solution
5<-5x<1'
Dividing by
- 5 wili
get
reverse each of the inequalitie5' and we
l>x>-t.
interval
The solution set consists of those x in the ofen
TTGURE 11
TXAMPIE 5
Solve for x:
It-ll . r
lr-61
Proceeding as before, we see that
2x-5
-'- x-6;<3.
-3<-"
(-. 1, 1)' See Olt
',J
sEcrl0N
AssorurE vlr.ug
1.2
We would like to multiply by x - 6, and in order to do this we must
distinguish two cases, depending on whether x - 6 is positive or negative.
l. ,x - 6 > 0. In this case multiplication by x - 6 preserves the direction of the inequalities, and we have
Case
oi < :,r
-3(x -
-
5
< 3(x
-
6).
Now the left inequality states that
-Jx*18<2x-5,
or lh;rl
f.r.
The right inequality states that
2x-5<3x-18,
or that
13
<
"t.
In Case I we must have x:6 > 0 and! < x and 13 < x. If the third inequality
holds, then the other two hold as a consequence. Hence the solution in Case I
consists of all x in the interval (13, m).
Case2.
x-6<0.
The inequalities ret)erse when we multiply by
x-6. \\e
then get
-3(x- 6)>2x-5>3(x-6).
The two inequalities now state that
+
ftcutt
0235
12
2
ln Problems I through 16, solve for
lxl
5
PROBTEMS
x.
t3
2;x-l;:t
4l7-5xl:8
6l-2x-41:g
8
9ix-61:13-2xl ,,
ll 1:?
> x and l3 > x. The three inequalities
x-6<0
and zfr,
and 13>x
all hold if x < f . In case 2, the solution consists of all nurnbers in the interval
(
- m, ?). we could also describe the solution set by saying that it consists of
all numbers not in the closed interval [?, 13]. See Fig. 12.
n
13
I l2x+11:4
3l4-2xl:3
s lx1l1:s
7 l3x - 5l-l7x * 2l
$
n
l2x
+ tl
:
l#l:,
4:
lx-U
l3xl
I
12-xl
3:
lxl
s
lx
-
3l
"l*:l:'
,nlx-ll_l
14-xl
3
rolffil=o
In each of Problems 17 through 27 , frnd. the values of x, if any,
for which the tbllowing inequalities hold. State answers in terme
of intervals.
17 lx+ll<4
19 [2x+51<3
It lx-31<2
m p-?xl<6
10
cHAPrtn
lNEeuAUnEs. Fuxcnoxs. THE
1
zzffi.s
2t lt?+ 5xl3l
23
13
-
2xl
llx
lx(x+l)l<lx+41
28
A lx+31<l2x-61
"l#l.-,
+ 4l
LINE
Find all solutions of the simultaneous equations
:
2, l2x
+ yl:
lx-vl
4.
29 Find all solutions of the.simultaneous equations
:l,l2x-3yl:8.
lx+vl
30 Find all solutions of the simultaneous equationslx2 -2y21
"l#l=,
:
t,l2x
- yl:4.
,
J,
ABSoLUTE vALUE AND lNEQUALlrlEs
THEOREM
s
If a
Proof The statement of the theorem is equivalent to the double inequality
-(lal + lbD < a+ b< lal + lbl.
However, we know from the definition of absolute value that lcl is c ot -a
and lbl is b or -b. Therefore, we may write -lal < a<laland -lbl <b< lbl
since one ol the equal signs must aiways hold. Now adding these last two sets
of inequalities, we get
-
lal
- lbl < a+ b < lal + lbl,
n
which is the desired result.
COROLLARY
Il
a and b are any numbers, then
iinl$fi$-1tr1iffi
Proof We write a-b
as
4+(-b),
and apply Theorem 5 to obtain
la
- bl:
la +
(-b)l
< Icl + l_bl
:
lal
+ lbl.
The final equality holds since, from the definition of absolute
value,
always true that l- bl lbl.
:
In calculus we frequently
qua,tity although
need
it
to know the approximate value of
we cannot find iis exact varue. Theorem 5 and its
is
f
a
corollary
are invaluable if we wish to make such approximations.
we show how this is
done by working several examples.
EXAMPI.E
1
Estimate how large thc expression x3
interval
Solution
l-4,41.
-
2 can become
if x is restricted to the
From the Corollary, we have
lr.-zl<lx3l+121.
since the absolute varue of a product is-the product of the absolute
varues, we
have lxsl lx .x. xl lxl . lxl . lxl lxl3. Tirerefore we get
:
:
:
lxt -21<lxl3 +2.
By hypothesis we stated that lxl is arways less than 4,
and we concrude that
l*'-21<43+2:66
if x is any number in
EXAMPTE 2
[-4,
4].
fl
Find a positive number M such that
lxt-2xz *3x- 4l<M
for all values of x in the interval
L_3,2).
Solution From Theorem 5 and the Corollary we can write
* 3x - 4l < lx3l + l2x2l+ l3xl + l4l,
and from our knowledge of the absolute value
of products, we get
lx3
-
2x2
lxsl + l2x2l+ l3xl + l4l
Since lxl can never be larger than 3,
:
lxl3
+ 2lxlz+ 3fxl +4.
it follows that
lxl3 + 2lxl2 + 3lxl + 4 <27 + 2.g
The positive number
M we serk is 5g.
+3.
3
+4:
58.
n
THEOREM 6
Proof
sine c is less than or equar to d the reciprocal of.q the quantity
r/c, is rargcr
than or equal to the reciprocal of d. This forows
from Theorem 4 for
inequalities, since dividing both sides of th;l*q;;iit
i'" i'oly"t"iJ'poritir.
quantity cd gives
CHAPTTR
lNEeuAUTtEs. FuNcrloNs. THt tlNE
1
l r1
d
c-
> b' we have
Again from Theorem 4, and from the fact that a
,(:) =,(:) =,(;)
The first and last terms yield the result'
l-l
Theorem6maybecombinedwithTheorem5toyieldestimatesformany
estimates are only
algebraic expressions.'-W" ob""e that since these
will give better results than
approximate, some techniques for making them
oit.... We exhibit this fact later in Example 5'
EXA {PLE
3
Find a number M such that
.,
Ir2l l*-21
when x is restricted to the interv al l+,*7'
Solution
We.
know that
lx+21= lr+21
l,*--l t.-zt
Ifwecanestimatethesmallestpossiblevalueofthedenominatorandthe
Iargestpossiblevalueofthenumeratorthen,byTheorem6,wewillhavethe
largest possible value of the entire expression. For
"rtirnrt"J-ii,.
numerator we have
lx +
2l< lxl+
l2l<|+Z:l'
Forthedenominatorwenotethatthesmallestvalueoccurswhenxisasclose
32'
ut possible to 2' This occurs when x - and so
lx- 21>li-21:+,
if x is in the interval
[]'*l'
W" conclude finally that
lx + zl
.'.:r.
l'-zl-i
ExAM?n
e
Find an estimate ror the largest
u
n;;,T;iru. "t
l.x+3
when -r is restricted to the interval
i
l-4'47'
since
Solution The numerator is simple to estimatc'
42 + 2 : l8'
1x2 + 2l< lxl' + 2 <
+ 3l if x is in l- 4' 41'.We see
However, we have to find a smaliest value ior ix
x = - 3' since' then the
first of all that the expression is not defined for
'zero
is always excluded"
by
rjivision
denominator would b" ,,,o, and
stcTroN
Furthermore, if x is a number near 3, then the denominator
is near zero. the
numerator has a value near ll,.and the quotient will
be a..large,,;;;l;;.
Indeed, by taking x sufficientry close to i *"
can make the denominator as
small as we like and hence the fraction can be made
as large as we like. In this
problem there is no largest value of the given
expression in [-4,+]. sucrr
questions are discussed further in Chaptei 2.
n
EXAMPLE
s
Find a number M such that
lx+2 5i'M
l-Y
I
I
when -x is restricted to the interval ( l, 4).
Solution Method
l.
Since, from elementary algebra,
x+2 ,__4x*2
.
xy)
we can write
lx
+z
lx
l.
We
lxl
I
Since lxl < 4, we have 4jxl
smaller than
_ sl : r_o* * rt <
+ 2 < lg, while the denominator, lxl, can never
be
:
I
'
2.
lrl
obtain
lx+2
I -sl
Method
t4lltfl?1
<18'
From the Corollary ro Theorem 5 it follows
that
lx+2 _l
t+-,1 =l+l +r5r:ff*,
Now lxl +
2
lx+21
< 6 and ixl > I. Hence we obtain
Ix+2 _l
|'
6
-'l'i*5:ll'
This pxample shows that some algebraic
manipulations lead
estimates than do others.
3
PROBI.EMS
In'Problems I through 12, find a positive number
M, if there is
one, such that the absolute value of the given
not exceed M when x is in the intervat given.
I
x2
2
x2 +
-
4: x in l-2,21
l:
- x in [*2,4]
Jx +
4r
.*prr.ri"i
Ao".
3 x3+ 2x2 -3x-6; x in [_2,5]
4 xa -Zx1 + x2-3x_5; x in [_3, _l]
x+2
5
x in [5,8]
,=;
n
to
better
14
CHAPTER
o
lNteuAUTtEs.
1
ffi;
fuxcnoNs. Tm uNr
[#.*l'
l6
l#."-'l;
x2-6x+2
, #;
x3-6x+5
,t
.\
xin(-4+,41
xin(-2,3)
t7
9
' x2+4x+4'
=*!7 ,; xin(-1,3)
l8
s
ro
ffi;
x3-3x+5
ffi;
tt ii;;$;
n
2x1-3x+l
'-:t
_;-;
19
20
xin[0,4]
2x*l
x in
l5
xin(-1,4)
r in (-3'
2l
r)
r-2,
-r]
xinro'21
l*-*l'xinro'31
lx
ft
2xl
- ;;l'
x in
r-2'2f
- lbl < la - bl'
a and b that llcl - lbll < la - bl'
Prove for any numbeis a and b that lal
Prove for any numbers
Given that a and b are positive, c and d are negative, and
a> b, c > d, show that
ab
cd
x in (0.7,0.9)
22 lf a r, a 2, &x a"tlany numbers, show that la, * a2+ a3 | S lor
+ larl+ larl.
23 lf ar, a2, a3 are any nurnbers and la, * azl> la.l, show that
I
In each of Problems
13
through
18,
find in two ways how large
the
the given expression can become in the given interval' Use
methods in ExamPle 5.
,,
,,
lfr.rl'
ot
xinrl,3l
ar+a2
Is the same result true
if
lo,
al.
* azl>lazl is replaced by
A, + a2> ar?
l#-31'xin12,3l
SETS. SET NOTATION. GRAPHS
R6URT 13
(numbers,
A set is a collection of objects. The objects may have any character
given
set and
in
a
points, lines, etc.) so long as we know which objects are
that
say
and
P
e
S
*hi"h u.. not. If S is a sei and P is an object in it, we write
their
sets,
two
p belongs to s or that P is an element of s. If s, and s, are
is in at least
union, dinoted by S,uS2, consists of all objects each of which
n
on. oi tt . two seis. ihe intersection of 51 and 52, denoted by 51 52, consists
of all objects each of which is in both sets. Schematically, if sr is the
S,
horizontally shaded set ( Fig. 1 3) and S, the vertically shaded set, then S t u
shaded
the
doubly
consists of ihe entire shaded area and S, nS, eonsists of
of
atea. Similarly, we may'fqrm the union and interEection of any number
.
..
s2,..
u s7 for the union of the seven sets s1,
sets. when we write s, u s, u
" sl,
of
the :7
bne
loast
in
at
is
which
of
each
of
alt'Otements
consists
union
this
may
sets. The intersection of these 7 sets is written SrnSrn
case
such
a
In
happen ihat two sets S, and 52 have no elements in common.
the
for
set
ernpty
J,"V that theirintersection is empty, and we use the tetm
""^Sr'It
"r
set which
is devoid of mernbers.
sEcTtoN
1.4
SErs.
Srr Nor noN.
Gr
15
pHs
Most often we will deal with sets each of which is specified by some
property or properties of its elements. For example, we may ipeak
of the
set
{x:
of
x: Zn and n is an integer}
to represent the set ofall even integers. In this notation the letter
x stands for
a generic element of the set, and the properties which determine
membership
in the set are listed to the right of thi coion. we recall that a rational
number
y gne $1h cen be expressed as a/b where a and D are integers rtil*u,,pt.,
517, -3l2,and 1713 are rational numbers. The symbol
{x: x e (e
l) and x is rational}
represents the rational numbers in the open interval (0,
1). If a set has only a
few elements, we may specify it by listini its members
between bruor. thu,
the.symaol {-2,0,1} denotes the set whose elements are the
- i: 0,
and l. Other examples are
ou*b;
(0,2):{x:0< x<2},
12,14):{t:2<r<14}.
The last set is read as "the set of all numbers r such that
r is greater than or
equal to two and less than fourteen.',
To illustrate the use of the symbols for set union and set
intersection, we
observe that
U,3l: ll,2llvl2,3l,
(0,
Notation. The expression "if and only if,"
l)
:
(0, co) n(
-
m,
1).
a technical one used frequentry in
mathematics, requires an explanation. Suppose A and, g
stand for propositions which may be true or false. To say trrut ,l is
trueif Bis rnre means that
the truth of B impries the truth of ,{. fue often shorte,
tiri.
saying that
I
implies A, and we denote this by
B+
"-pi"ssion
by
A.
The statement A is true onry
if B is rrue means that the truth of z{ impries the
truth of 8. we shorten this expression by saying that-,4 i,mptiir-n,
and we
denote this by
A+ B.
"/
The shorthand statement
is true if and only if g is true" is equivalent to the
double implication:the truth of ,4 impties and is impliea
uv ttreiiuir,-or r. a,,
further shorthand notation we use the symbol.* io,"p*"n; ;if
onry if,,,"
il
and we write
A+B
for the two implications above. The term rcces*rry
and sufficient is sometimes
used as a synonym for ..if and only if."
In Sections 1-3
we sawr that the study of inequalities is greatry
illuminated when we represent real numbers by
means of points on a Etraight
Iine. In the study of problems invorving t*o uoto*;;;"ll
sL tnat a
corresponding geometric interpretation-is both usefur
and i,oilrt*,.
CHAPTTR
lovrQurunrr FuNcnoNs' THt
1
tlNE
equallions and inequalities in two or
The determination of solution sets of
tn mathematics' especially from the
more upknowns is an important toplc
of a soiution set"
of applications. The geometric representation
geom€trv'
analytic
of
in the studv
A. graph, is particularly heipful
the
throughout
used
be
witl
iology which
We now introduce
terms as quickly
these
with
text. It is essenrial ttrat tilr'eaoe;;;#ifamiliar
real numbers
numbers is denoted by R1' Any two
as possible. The set of
When the
numbers'
of
pair
a
a artd bform a p"ir. f"i"-"mple' 2 and 5 arewe hu'e an ordered pair' The
that
order ofthe pair ptttttiUta,
'uythe ordered pair 5 and 2' We usually
from
different
is
5
arrd
2
pait
ordered
with a comma separating the first
designate an ordered ;;'i;;;t;'theses
ordered pair consisting
an rl used for the
and second
tt . norUers a (first) and b (second)'
;;i;; il;;
;;Iil
'"t" "t*f
,iii*f
*'
i'
elements;;#;'
l
oi
called rhe number plane and is
is
DEflNtTtoNs The set o{ all ordered w1r, ?f7*t ""*irs
pait
is called
o'aie'ed
Each indi'uidu'l
denotedi,
uztlvt's
f:i:-i!e
its-'coordinates'
:
^;
plane. The two eiements in a
It
numfu
number
cailed
ThenumberplanecanberepresentedonageometrrcorEuclideanplane.
plane' R2'
;e k;ep seiarate-'t'"."on"I"p' of theaumber
is important that
pairi from the..concept of the
which is an abstra;; ;#* ;i;rdered
object we studied in Euclidean
geometric pirn., *hi-.h iJiiri*"-ai*ensional
,
H6URE 14
o'*;.*;";T$',o"an
line, denote
plane we draw a horizontal and a verricar
O
intersection
of
point
their
lt:t
them as the r and , I'*tt, tttp"ctiveiy' 1"9
of
unit
convenient
a
select
prt{,; it tuiltd tht-origin' we
F;;. iili.
zeto' mark off a number scale on the
length and, starting from the origin as
negative to the left' Similarly' we
horizontal u*ir, po,iiiitt'o ittt rlght and
upward
axii' with po'ii" numbers extending
insert a scale along ti"
the
along
units
"ttitut
the
that
necessary
and negative ones downward' It is not
axis'
vertical
the
along
units
as the
horizontal u*i, t'uit tt" rame length
equal'
as
them
take
usually
will
we
between the points of the
"f,-fr""gf.r
We now set up a one-to-one correspondence
thi Euclidean (geometric) plane. For
number plane, R2, *a ii,, points of
P to
w-e construct perpendiculars from
each point P in the Euclidean plane
is at
axis
x
the
with
intersection
in rig. 15. The
;;*;
the coordinat. a*es, ;,
the y axis is at the point R" The distance
the point Q,andtt't int""""tion with
of o and negative if
Q is to the right
from the origin to
uv.o' Similarlv' the distance oR is denoted
Q is to the left
is (a' b)'
;;;;t;;positive-if
6i';;;;'"tti
"f
in it'" number plane Lorresponding to P
point
trre
by b. Then
corresponds exactly one
there
conversely, to .u.r., point in the numbeiplane
correspondence
the
of
point in the Euclidean plane' Because
-one-to-one
frequently
geometry,we
the plane of Euclidean
between the number plane and
find it conveni"nt tlo
;;;;;;;;"
t".*,
for the number plane. For example, a
>.t, .,line,,inrnenumbe"rpii*u"*4rn r*setofpbintscorrespondingtoalinein
the geometric Plane'
ThedescriptionaboveSuggeststhemet}'odlobeusedinplotting(thatis,
of the.number plane' When plotting
HGURT 1s
representing
g**"t"*ffyl
i"oints
sEcTtoN
1.4
Ssrs. Sgr NorATtoN.
17
Gn*xs
points, we enclose the coordinates in parentheses
adjacent to the point,
illustrated in Fig. 16.
DEFINITIONS
Tlre.solution xt of an equation in Mo ilnknowns
consists oy rtt puirii tln
number plane, R2, whose. coordinates satisfy
th, ,quotiu,r.'i""g"o.rrrr,
repre.sentation of the sorution sd
lthat is, the actuar drawingl
isiaaed
graph of the equation.
',
as
The solution set of an equation in two unknowns
is, of course, a set
is a simple matter to extend the set
notation already in,roar".ai;
sets in the number prane If s denotes the
2
It
the
in Rz.
#;.
properties, which, for example, we ca1
S
:
set of ail (x, y) in R having certain
and B, we describe s as follows:
A
{(r, y): (x, y) has properties
.4 and B }.
As an illustration, if S is the solution set of the
equation
2x2
FIGURE 16
-
3yz
:6,
we write
S
= {( ., y): 2.+7
-
Jyz
:
6,r.
To construct the exact graph of the sorution
set.of some equation is
generaily impossible, since it would rbquire
the protting of infinitery many
points' Usually, in drawing a graph, we
select
p"i"rr'i. exhibit the
general nature of the, graph, ploi these
points,"rroughrmn
ttre
remaining points by drawing i "smooth
"ni tt""ppr-oximate
through
poirt,
already
plotted.
"u.u""
It is usefur to have a systematic method of choosing
points on the graph.
we can do this for an equation in two unknowns
when we can solve for one of
the unknowns in terms
of the other. d;;,'iro,, the formura
outuin"a]'#'l#
assign varues to the unknown in the
iormula and obtain values of the
unknown for which we.sorved. The corresponding
varues are then taburated
as shown in the examples below.
EXAMPLE
1
Sketch a graph of the solution set of
the equation 2x _t 3y: 5. Use
notation to describe this set.
set
The solution set S in set notation is
S
= {(x, y):2x
* 3y :
5}.
To sketch a graph, we first solve for one
of the unknowns, say x. we find
,: * - ty. Assigning values to 1 we
the following table.
"U"i,
EGIJIT
17
Plotting these points, we see that they appear
!v .v
lie vu
J -rr--' to
on the
rus Dlralt'fl[
straight lln
rine shown
in Fig. 17.
tr
t8
cHAPrtn
lNreu^l'mEs. fuNc-tloN' Ttlt uM
1
EXAI\,IPLE
2
equation
Sketch a graph of the solution set of the
***:r.
25'16
set S'
Use set nbtation to describe the solution
Solution S = {(x,
y\: (xz
we obtain
+ (y2ltOr: 1}' Solving for y in terms of x'
125\
y: xlJx-7.
Since x enters into the equation only
abbreviate the table as indicated below'
in the term involving x2' we may
+2
0
t1
v
+4
tt'tfr *Jn
y (apProx
+4
!3.92
:
t3.67
+3
!4
rf
r*
0
L3.2
!2.4
0
+5
to lie on the oval curve of
Plotting these points, we see that they appear
iu.htrt and last really represents four
i;;. ifli;i;
tr,at eact cotu*o"*.rpt
points. What happens if we choose x > 5 or x
FIGURE 18
4
;r*e;;
svmbol' {Example: The set
3) can bc describcd as s = [0' 3))
;;ion or intersection
sl-[0, il rtr,
I S= [-1, 2)u[], 3]u(2, 7l
2 s:[-a,6)n[3,8]
3 S: {2, 6, 8}u{1,6, 8}u{Q
a S: [0, 2)ala,7)
5,9}
s S = {{2, 5,e}u{4, 6, 10}}n{3,6,
6 S: [-5, l]n[-3, 8]n{O 2l
1l}
14 3x+2Y:6
x*y:Q
16 x=1Y2
y=|x2
lE x:Y2-l
y-x2-4
17
20 Y=x2-2x+3
19 y:x2-2x-3
2l xz+y2=25 22 x2+Y2:9
?4 x2 - Y2 :!
23 4xz * y2:36
26 x7 -y2 *9:0
23 Zxz *y2:17
27 y:-x2+2x+3 28 Y=$x3
30 Y=x3-3r*?
29 y:x3-4x
:0
(fllnr: lHrite as a quadlatic il
31 2x2 + 3xy + y2
13
15
each of the following
Sketch a graph of the solution sct S of
32 x2+xy+Y3=5
equations"
33
7 x-Y:$
9 y-2x=4
11
x:4
E x+3Y:l
10 3x - 2Y:6
17 y= -l
n
- 5?
PROBLEMS
given-as the union or
In each of Problems I through 6, a set S is
the sets without
Describe
of sets of rcal n-umbers'
i"itit*,i",
<
2x2
+-3xi+
y?
+ 2x + 3Y= i6
34 x7-5x*6-0
36
y:lr-11+t
35 y=fxl
37 Y=l/(lxl+f)
y')
srcrroN
19
fuxcnors Rncnol^I' norrnon
1.5
t.
!
5
FUNCTIONS. FUNCTTONAL NOTATION
In
rnathcmar.ics and many
repcatedly. For cxample, if
of the physical scienccs, sirnplc ['rmulus.ccur
r is the radius of a circle an<I I is its area. then
.
A:nr2.
If heat is added to an ideal gas in a container of fixed volume, the pressure p
and the tcmperature 7'satisfy the relation
P:q+cT
tvhere a and c are fixed numbers with values depending on the properties .f
the gas, the units used, and so forth.
The relarionships expressed by these formulas are simple examples of the
concep'r of lunction, to be defined precisely later. However, it is noi essential
tha-i a I'unction be associated with a particular formula. As an example,
consider the cost c in cents of mailing a package which weighs x grams.
Suppose postal regulations in some country state that the cost is'.61 per gram
or fraction thereof." We can construct the following table:
Weight x
in grams
0<.x<l
l<x<2
2<x<3
t2
l8
3<x<4 4<x<5
Cost C
in cents
24
30
This table could be continued until x: the maximum weight permitted by
postal regulations. To each value of x between 0 and the *u*irnu. weight
there corresponds a precise cost C. we have here an example of a function
rclating x and C.
_ It frequently happens that an experimenter finds by measurement that
the numerical value y of some quantity depends in a unique way on the
measured value -x of some other quantity. It is usually the casi that no known
formula expresses the relationship between x and y. All we have is the set of
ordered pairs (x,.y). In such circumstances, the entire interconnection
between x and y is determined by the ordered pairs. often this correspondence is denoted by a letter such as
/, which indicates that each value of
yis obtained from a particular value of x. we write y:f(x) to show the
relationship. If D denotes the set of alr values of x which octur, and E the set
of all values of y which occur, then / indicates that each value in D gives rise
to precisely one value in E. The value y in E is denoted
/(x). (see Fig. r9.)
lt = IQ')
I)orrrairr
Ft6utr
19
20
CHAPIER
tNreuAlmts. FuNcjrtoNs. Txr uxe
1
a set of numbers
set of numbeis D
-too
O
i7
*
numb"
to
each
correspontlence thatassigns
""i!,y:,".1-:O:::^!\:rY',
";:,:,7;;i':\"
rhe sit D is catted the domain or the
from
i""rrin, i *,t
!-i:-!
a
i;';;;;;;;;;;p;;;,*'
ihe ,et E
it
'41'd 'h' '""g" "f
th'
f"
the formula for the area y of a
An example of a function is given b1
i'
f {1) 1x2' and we
,qrui. *t,"re side has lt,;i; ''; i'J, y:f wt may write
a number which is the area of a
see tha-t for each ,alue of-* > 0 we ;btain
Squareofsidex.ThedomainDofthisfunction,i.e.,thesetofallpossible
(0, .o), and the range E, i'e., the set of
values of r, is the haltinfinite interval
(0' oc')'
values for the area y, is the half-infinite interval
the temperature T and
"fip"tSUf.
The formula P:3 +27, a particular case relating
thepressurep,isanexampleofafunction.HeretheletterTreplacestheletter
we have a function since to
x and the letter p repluce, the letter.v' However'
value of p. we write p :l(Tl
each value of T there corr"rponds eiactly one
the range of p depend on
for the iunction e :3 ; ir.ln" domain of T and
so forth'
ii"
' p".il.trrr
conditions of the gas, the container of the gas' and
a function'
example oi poriat ,utt'liutn in Table I also represents
mailing c
.f
expresses the cost
although no simple ror*rru is aviilable which
(x) and use-the^table to find
in terms of the weight x. We may still write C
grams, then C 24 cents' We
a value of C for .ach ,. For ex#ple, if x 3|
call this function the postage function'
as a set of
a geometriJ;;in"t of ,ie* a function can be considered
Th.
:f
:
:
From
interpretation
is
This
ordered pairs (x, y;, *f,"t. y is identified with l(x)'
we make the
Consequently'
function'
a
gtuptt
of
ii"
useful in constructing
following alternate definitlon o[ a function'
,4 function is
DETINITIONS
a
set
two pairs
of ordered pairs (x, y) of real numbers in which no
Ii
x (the
member
frst
other lpords, rc each ualue
haue the same rst element.
member
(the
second
) ' The
y
of
uqlue
of the pair 1 there corresponds exactly..one
the function' and the set
the
srt ,,,f all ualues of r which occur is called
f
u1ait y which occur is called
of
th"'
lomain:tf
WhenwewishtoConst.ructthegraphofafunction,thedefinitionintermsof
implied by the term function
ordered pairs is most uselui. The speciil property
the graph no more than
intersects
y
axis
the
ti
parallel
assures us ihat every line
graph of a lunction
on.. tfig.20). Thevertical lines which pass through the
form a set called
axis
x
the
with
intersiction
of
points
intersect the x axis. These
theprojectiononthexaxis.Thisprojectionisthedomainofthefunction
furnction intersect the y axis
rrig. zot. Horizontal lines through t'he graph of the
range o[ the
as the
and these intersections form alet *hi.h *t recognize
of the graph of the
y
axis
the
o4
projection
is
the
function. That is, the range
graph of a function many
function. Note that horizontal lines may intersect the
,
FICURE 20
times.
It is important to be able to discuss
functions and their propertles
withoutactuallyspecifyingtheparticularoneswehaveinmind.Forthis
stand for
purpose we use , ,yrntof uruuily a letter of the alphabet' to
if
Sometimes,
g,
*,
G,
are
used
o,a.
letters most often
{,
a
a
function. The
so that,
problem concerns many different functions, subscripts are employed,
functions'
three
different
for
would
stand
,'for example , f ,, fr,
6
"rd
SECTION
21
1.5
R NcnoNs FuNcnoMr NorAnoN
Several symbols for function are now in common use. one such is
y, where x is a generic element of the domain of and is the element
:
xr*
f
/
1,
of the range which is the image of x. Another notation for function is
f:Dr--D,r,where D, is the set forming the domain of / and D, is the set
forming the range.
When specifying a function /, we must give its domain and a precise rule
for determinin-g the value 9f /(x) for each .x in the domain. For the most part,
we shall give functions by means of formulas such as
'f(')=x2-x+2'
Such a formula, by itselt does not give the domain of x. Both in this case and
in general, we shall take it for granted that if the domain is not specified,, then
any ualue of x may be inserted in theformula so long as the result ma'kes sense.
The domain shall consist of the set of all such ualues x.
In prescribing a function by means of a formula, the particular letter useil
is usually of no importancei:. The function F determined by the formula
f(x):
x3-2x2+5
is identical with the function determined by
r(r)-t3_2t2+5.
The difference is one of notation only.
DUIMPLE
1
Suppose that
f
is the function defined by the equation
f(x):x2-)x-3'
Find /(0), f ( t), f ( * 2), f {2),
the portion of the domain in
-
f
(3),
-2
Solution
f (t),
<x<
and f
(
f (x)),plot a graph
3.
of
/ for
We have
02
"f(0) =
:
f(-r):(-l)2 -Zi_l)-3:0,
-2-(-2)-3:5, Ie) :22 -2.2-3: -3,
-2.0-3= -3,
f(-2) :(-2)2
2'
:0,
- 3- 3
f(t)=t2 -2t-3.
"f(3)
The difficult part is finding f (f (x)), and here a clear understanding of
the meaning of the symbolism is needed. The formula defining means that
/
whatever is in the parentheses in
) is substituted in the right'side. That is,
32
"f(
f U$)) = (f (x)12 _ 2.(f
However, the right side again has
(x,t\ _
3.
/(x) in it, and we can substitute to get
f(f(x)) : (x2 _ 2x _ 3)2 _ 2(x' _ 2x _ 3) -:3
:x4-4x3-4x2+l6x*12.
To plot the graph we compute
,f(l):12-2'l-3:-4
and assemble all the results above to obtain the following table.
CHAPTER
lr{Equnrnrs. Fuxcrmr'rs
1
Tff
uNE
f(*) = v
The graPh is Plotted in Fig'
tr
2l'
become unwieldy, and we may use
sometimes successions of parentheses
the same as /(/(x))'
brackets or braces *i,i,t"1"''ie meaning: f Lf$))is
If we write
g(x): xt +2x-6' -2<x<3'
ihismeansthatthedomainofthefunctiongistheintervall._2,3).Ilinthe
that g
n -2 <x ( 3 were omitted' we would assume
This
o)'
(-o'
domain would be
is defined by that fo;;iu i' atl; the
by the
same formula, the portio
we may define a function F
opens up many possibilities' For example'
F'dehned in this wav has the interval
following conditions iii,.
[- 1, 5] for its domain):
F(x)
flGURt
f.,i.il",
:
- rr==r.1r,
{r.rr*_r;;r;,
Formulaswhichdefinefunctionsmayhaveobviousimpossibilities'If
in a position to find
not.*pri"iuv"ioioltJr$;, ih" rr"d", should be
them. For examPle,
these are
21
1
fr(x):; -1
except x
is a function defined for all values of x
always excluded. If
:
3, since division by zero is
Ir{x) =
thenitisclearthatthedomainof/,cannotexceedtheintervalfrom
_2to+2'sinceimaginarynumbersa,reexcluded.Anyvalueofxlarger',.
than 2 or less than - 2 is impossible'
gAlfPt"E
2
function
Discuss the distinction between the
F(x)
Sotution
:
x2
-4
and'
G(x)
:
75
L' 2'
I
it
t: - =4 is $ctorable into (x - 2)(x + 2\' at first glance
of
all
is
of
G
dornain
the
However,
appears ,iuilt. functions urq tt e same.
the function F, however, there is
For
r"lr..
;il;;"ri
rnut'il,
Since the expression
"pr.
I
for x, both numerator and
difficulty u"i ;;. ii ,iir--',]"fr- is inserted
denominatorareZ,eto.TherpforeFandGareidenticalforallvaluesofx
x:2 the function F is nor
except'=; ;;t;:z t"t ['une G(2\:4' Fol
other quantity, at our pleasure'
tlefnetl.we couta define F(2) to be 4 or any
write
But we would have to specily that fact' If we
xz -4
F(x): x_z
and
F(2\
:4,
'
23
sEcTtoN'r.s
FuNCfloNs. FuNcnoi{ L NorAnoN
then this function is identical with G. This may seem to be a minor point, but
we shall see later that it plays an important part in portions of the catculus.
LJ
DUIMPLE 3
Given that
f (x) :
x2, show
f (', +
y2)
thit
: f lf G)) + f lf ( y)) + 2f (x) f ( yt.
f(x'+ y'):1x2 + y2)'-= xa +2x2y2 + ya,
f lf txll:f lx\: (*')': -r4,
I Lf $1:f lyz): (y')' : yn,
2f
(x)|iy):2x'Y'.
Adding the last three lines, we obtain
lf (y)l + 2f (x) f (y):
which we have seen is just /(x2 + y2).
llf
EXAMPTE 4
Solutiori
Given
f(x):
G)l +f
x4
*
1'a
+
2x2 y2,
tr
lxl, plot the graph.
From the definition of absolute value, we have
x for x>0
JG):{I
[_x for x<0.
^
We set up the table of values:
0ll
x
y:f
(x)
The graph is shown in Fig. 22. Note that the graph has a corner at the origin.
n
FIGURE 22
5
PROBTEMS
Civen that /(x):.x2+1, find f(-4'1,
"f(-3), f(-Z\,
f (- t), _f $), f (l), f (2!,, f (3\. Plot the graph of the equarion
y:f (x\ for -4 < x < 3.
Given that f(xl:x2 *x+ l, find /(
-21, .f(-t), .f(0),
f (l), f (2), /(3). Plot the graph of the equation : / (x) for
-2
<x<
-y
3.
Given that f(xl :
x2*
/( -3),,f( -2), /(- l),
./(0), /(l), f(21, l9l, and f(a+2). plot a graph of the
equation
y=IGl
3x
2, find
ficr
-3<x<3.
Given that /{x) = {x3 rx * 3, find /(
-31,
Ie2l,.f( -
l),
5
f (0), f (t), f (2\,,f(3), and .f (a- tl. ptot a graph of
equation f :"ft_t) for -3 <.x < 3.
rhe
Given that
rr'r =
lrlj'
rind /( *4), /r -:), fl-2t,./( l], .f tol. f tl), t( 1000).
f(1000). and l'[/(x)]. Is
in the domain of /'? ptot I
graph of / for .r on [-4. ll using the values above anil
additional values niar - i.
-l
cHA'PTtR
6
lxiquluns.
1
FuNcnoNs' THE tlNE
17
Civen the function
what is the domain of /?
Given /(x) :1tr-r--,
Plot a graph
(ffinr, iornpf.te the square under the radical')
ot
lx- I
^''):2:=' l'
*.1' O' l' 2' 3' 4'
find /(-x) for x: - 1000, -4, - -2'
<iomain of /? Plot a
1000. Also n.a /i/(xil' Is J in the
values above and
graph of / for x t" i]+'+j using the
uOaitionut values near j'
7 Given that /(x) :x2 -'2, frnd /[/{;tx)}]'
8 Given that /(x) :a3 - 2x+ 1, find /{/t/(.)l}
9
find the domain of
;;;;ii ;
/(x) for x : - 1000, -3'
,fur'fi-rl : -/(x)
of/ior
x on
19 Same as
Problem 16 for the function
f{x+hJ-f$).
1000' Show
-
h
[-3.3]'
and sirnPlifY, assuming rhal h +
: Il-.\+',
za
/(x) -/(Y) : x- Y
1+xY
l+/(x)/(/)
26
frnd
I
ftxt:!,
/(-1), /(0]' f(rl' tQ)' f
(x):r/i,
(3)'
f(i:yG,
x>o
show that
[i9-rot :-
,-b
Plot its graph'
l?
25
3
f txt: V;: * 3
27 lf f
whenever both sides are defined'
What is ihe domain of
A'
: x2 21 f(x) :2x2 + x Zl 16:!
712 f 1x\: x3
20 f(x)
show that
f(x\:lG+t'
.
of
In Problems 20 through 26,find the value
:2' l'o: Rr' Plot a graph
for all values of x in''ll3'
- fT*t
civen
defined
/' Give, by formula' a function
/ wherever / is defined'
('+ 2)(-'-L 6*:16)(1j.
Jtxt:- $_z\@_4x_12\
Civen that
rr
-2x -3
in R rvhich coincides with
1
Given the function
find
x2
' f(x):
f$l:;+,
l0
f.
lE Given the function
1-
tt'l+ttU
ll, what is the domain of /? Find /(x) 2E A functron is.linear if it has the form flx):ax*D for
/
x on
are linear functions' show
fo, *: -2, -1,0, l,2,3,4andplotagraphof/for
some numbers a and b'lf f andg
is a linear
t-2,41.
that /+g is a linear function' Show that l[g]
linear?
x:
for
is
-2'
find
-3'
/'g
-4'
/(x)
13 Given f (x):3-lx+ 1l'
function."Under what circumstances
2]'
on
- f, O, i, 2 and plot a graph of / for x [-4' of Plot 29 Given f {x):2x- I' g(x) ; xz +2' h(x): l1+ 1' find
14 Given ,f(r): lxl +lx-21, what is the domain /?
12 Given /(x)
:
lx
-
the graPh for x on
l5
/lgttr(x)]l'
l-3,2)'
formula
30 The volume of a sphere of radius r is given by the
Given
4f(rl : 1nt" '
3
/(r)=lr_rl'
what is the domain of
and x on (1,31.
/?
Plot the graph for x on
l-2'
1)
of /?
16 Given 1111:o{Fr)G-3), what is the domain
t."ptt o"f ; fot'x
" close to 1 and 3'
values
Pl"i
between
-2
and 6' using several
volume is multiplied
Show that if the radius is doubled' the
8.
by a factor of
3lFindafunctionihatgivesthevolumeofasphereifthe
area of a circle
radius of the sphere is t-he square root of thc
of radius r.
sEcTtoN
1.5
RErATtoNs.
25
lxrrnceprs. AsvmrrorEs
RELATIONS. INTERCEPTS. ASYMPTOTES
DEFINITIONS
The solution set of qn equation in two unknowns,
say x and y,
relation.. TIre domain of a r eration is the set
of alr number, *o"r"n that the
uertical line x xo intersects the graph of.the reiatior.
rh" rangl
irffi
,t
W
^l
:
t, ;rth ,r*
Figure 23 shows
the_
tfu
6ai"tit
line
y:
yo
ofi'retotion is
intersects the graph.
graph of a typical relation. we see that
the
line
x : xo.intersects rhe ggaph in three poinis while
the tine y : yo iniirre"t, tt e
graph in two points. unlike a functlon, it
is not necessary that in a reration
vertical lines such as x:xo intersect the graph
in only
p"lri.if we are
given a relation in x and y, that is, an equatlon
connecting"r"
the variabres x and
y, then we can determine its domain
by performing ,n. foifo*lngi;;;;i,
i;
we solue for y in terms o/x. (we caution that this
step i" rrt"r"*t possible.)
(ii) In the resulting expression or expressions
in x, we determine thosi varues of
for which at least one of the expressions has meaning. The domain is
the
totality of such values of x. To find the range,
we perform steps (i) and (ii)
with-the roles of -x and y interchanged. Severar
examples illustrate the
method.
x
domain
A relation in E2
FIGURC 23
EXAMPI.E
1
Find the domain and the range of the
relation defined by
12:x2-4xi-3.
Solution
To find the domain
f
3, we see that
*:.fi:rt solve for y in terms of x. Setting (x) :
f
y2:f (x) and
Y: + Jf u),
y: _ r[f6.
xz _ 4x
26
CHAPTTR
ffir*t
,rt. FuNcrloNs' THE UNE
1
when /(x) is
Sincethereisnorealnumberwhichis-thesquarerootofanegativequantity'
-t'itt' /t'i]o' it att"t*ine
the domain consists "i"ii''r"t
ir'" t-pi"ttit i' - 4x I 3'we obtain
' nonnegativ.,
and
f
*t
f"tto'
f (x\ :
'
4x* 3 = (x - l)(x
-
xz
-
3)'
are
both factors
factors are positive or
is positive whene'rer borh
x > 3 and both
for
;: ;aij ; 3 are uoin'fotiti'"
nesative. The factors
f";; . i
ffi;il
is the set
The domain of the relation
u [3, + oo)'
the relation
for x in terms of y' From
To find the range we solve
(
-
oo, 1l
{.2_
we obtain
4x*3-Y2:$'
-2t$ +7.
A graPh of the
the range is (-m' + m).
all
for
),,
is
+
Positive
1'2
Since
24'
,"iu,ion is sketched in Fig'
1
flGURt 24
EXAMPTE 2
or not the solution
+ x'v - 4 :0' determine whether
how it can
s.1t^'ot a function' show
pro'lt'i
e'upt'
function'
a
and range'
set S is
"i's.'.rf
functions' Find the domain
;;t;;";i:;';ral
Given the equatioa
'x2
be represented as the
Solution
for y' we find
Solving the above equation
+
x2+x)'-4:0
4-*'
!=-i-- -4* _*
y:{4lx)- x. Since there is only
set of the equation
function /
The set S is the solution
s'it a fun'ction''ln fact' s is the
one value of y for t";';;l;;;r
''
defined bY
f(x):
4-x2
all x in R1 except x
The domain consists.of
x in terms of Y, getttng
*:lrt-y!JY2
:
solve for
0' To find the range' we
r_;_+16\'
the above
side of
be inserted in the right
all possible values of -v may
io piot the graph we make the following
t,
expression, rhe range
"ii "r ^;.
Since
table:
-4
"f
(x)
:
-3
-2
-3
Y.
5
I
2
-t
-3
-3
table. since / is
the points plotted from the
through
curve
smooth
or the
we draw a
uni Ttll to set an indication
i,
not defined for x = o,
t"i"riot of near x:0' (See Fig' 25')
;;;;;;
EGUTf,
-t
/
;i-
srcTloN
1.6
27
.
EXIIMPLE
ReunoNs. lrrErcErrs. AsyMprorrs
3
Let S be the solution set of the equation xa 4x2 * y, :0. plot the graph of
S. IfS is not a function, show how it can be represented as the union ofseveral
functions.
Solution Solving for y. we note that
xn-4x2 *y':A
4-x2
!:*x
Thus s is a relation which is the union of two functions
fJx) : *r/4 - *'
and
andl,
defined by
fr(x): - xuT - *r. We obtain the following table:
x
+2
v
0
y epprox.
0
+l
,l
-r-
0
t*.,f;
\E
ti$,
0
+
+
+0.97
0
,-l
1.98
The graph is shown in Fig-. 26. Note that
thr equation
f,
*
: O.
fi
1.73
and
f2 are defined implicitty by
tr
we now discuss several facts about graphs which are easily obtained
from the equation and which are substantial aids in construciing quick,
ftcuRt
x4
4x2
yz
accurate graphs. Later we shall see how the methods of calculus can be used
to get additional information about alraph before it is drawn.
It is useful to know where a graph crosses the x and y axes.
25
DEFlNlTloNs A point at which a graph cros.se.t the x axis is called an x intercept; a point
where it crosses the y axis is called a y intercept.
EXAMPLE 4
Find the,x and y intercepts of the graph of
x2-3y2*6x-3y:7.
Sotution
Setting
-y:
0, we get
x2
+ 6x *
7
:0;
thus x: -7
The x intercepts are ( - 7, 0) and
Setting x :0, we get
x:
I
( 1, 0).
3y2+3y*7:0; thus y: -3 t
!t-B
since the solutions for y are complex numbers, there are no y intercepts
and
the graph does not intersect the y
axis.
tr
:! Li
CHAPTTR
lNEeuAUnEs. FuNcrloNs. Tnr
12
EXAMPLE
5
uxr
Given the curve with equation
l(xz
- l\:2'
find the intercepts and sketch the graph'
which is the y intercept; y: 0 Yields the
2' and the curve has no x intercePt. We make uP a
Solutioir Setting x:0, we get y impossible statement
0:
-2,
table of values:
Weseethatasxincreases,the'valuesofygetcloserandclosertozero.onthe
:
1' In order to
other hand, there is no value of y correspondirtg to x 1 or an auxiliary
get a closer'look at what happens when x.is near 1, we construct
values:
of
set
following
the
obtain
We
1.
x
near
t-able of values for
1 from the left, the corresponding values of y become
the values of y
large negative numbers' As .x approaches 1 from the right'
. The portion
27
in
Fig.
UeJome Iarge positive numbers. The curve is sketched
As
x moves
RGURT 27
cJoser
to
sf cTloN
RETATIoNS. INfENCETTS. ASYMPTOTES
1.5
to the left of the
1,
negative values of
axis is a reflection of that to the right of the y axis since
x yield the same values of y as do positive values of x.
D
In Example 5, the vertical line through the point (1,0) shown inFig.27
plays a special role. The curve to the right ofthe line gets closer and closer to
this line as the curve becomes higher and higher. In fact, the distance between
the curve and the line tends to zero as the curve continues upward beyond all
bound. Such a line is called a vertical asymptote to the curve. Similarly, the x
axis is called a horizontal asymptote, since the distance between the curve and
the x axis tends to zero as x increases beyond all bound. A knowledge of the
location of the vertical and horizontal asymptotes is of great help in sketching
the curve. We now give a rule (which works in many cases) for finding the
asymptotes.
the vefiical asymptotes, solue the equationfor y in terms ol x. lf the
result is a quotient td two expressions inuoluing x,.find all those ualues of x for
which the denominator uanishes and the numerator does not. If a is such a ualue,
the uertical line through the point (a,O) will be a uertical asymptote . To locste
the horizontal asymptotes , xtlue for x in terms of y , and find those ualues of y for
which the denominator uanishes (and the numerqtor does not).If b is such a
oalue, the horizontal line throu{h the point (0, b) ,s a horizontal asymptote.
RUIE To locatc
Two examples illustrate the technique.
TXAMPLE
5
Solution
Find the intercepts, domain, range, and asymptotes, and sketch a graph of the
equation (x2 - 41yz :1.
a) Intercepts: The value -y: 0 yields no r
!2: *i, and there are no y intercepts.
b)
:0,
we have
Domain: Solving for y in'terms of x:
l':
The domain is all x with
c)
lxl>
*
2.
Range: Solving for x in terms of
except y:0.
dl
intercept; when x
y:x: +rff+4y,1y.
The range is all y
Asymptotes; To find the vertical asymptotes, we use the expression in part
(b) for the domain, Setting the denominator equal to zero, we have the
vertical lines x :2,
- 2 as asymptotes. To find the horizontal asymptotes,
we use the expression in part (c) for the range. Setting the denominator equal
x:
to zero, we have the horizontal line
sketched in Fig.28.
):0
as an asymptote. The graph ir
n
CHAPTEI
lrsQultfi ls. fu NcrrcN$
1
THE
urd
RGURE 20
E(AMPIE 7
graph of
Find the intercepts, domain, range, and asymptotes, and sketch the
the equation
x2Y:x-3'
Solution
a)
Intercepts:
If
y:0,
then
x:3'and
the x intercept is 3' Setting
x:0 yields
no y intercePt.
b)
Domain: Solving for y in terms of x, we find
'
v-
x-3
The domain is all x # 0.
c\ Range:To solve for x, we write the equation yx2 - x * is3 :0' The solution
of tnit equation in x obtained by the quadratic formula
, - ,r,
*:,
^- J:2y
\
The original equation shows that
FtGLnt 29
x:
1f v
3 when
*0.
):
0' The range is all y <
it;
that is, (-*,#).
is a
d\ Asymptotes: The line x:0 is a vertical asymptote' The line l:0 table:
following
the
graph
we
construct
the
horizontal asymptote. To draw
2l3l4l5
xl-+i-3l-21-l
y
l-*l -+l -i
I -a I -z I
The graph is sketched in Fig.29.
-* I o ltlil
n
31
sEcrloN
RELAnoNs. lNrEncETs.
1.6
6
PROBLEMS
Problems I through 33, find in each case the intercepts'
if any'
domain, range, and veitical and horizontal asymptotcs'
Also sketch the graPh.
In
y':2x
4 Y2 :2x'4
I Y=4*'
3 y': -4x
5 Y:1-*'
2
8Y':*'+5
13 xy2
: 18
14 v2:l-2x1
16 y2 - xY:)
l8 -v(,r2 - 1) : I
:6
15x2+xv*v2:12
11 y2(x + 1) :4
19 x2(y2 - 4) :4
2l (x2 + l\y' :4
23 y(x- 1)(x-l):a
-4\:2Y
27 y2 + 2x: x'
29 ya:4(x'- y'\
3l y2(x2 - ll: x + 2
3l Sy(x - 1)(x - 3) :2(5x
25 x{y2
.Y2
3x2
+
24 v2(x*
for intercepts, asyrnptotes, range, and domain' Draw the
graph.
37 Civen
ly'- tl:l.x+21. Discuss for domain'
range' and
intercePts. Draw the graPh'
26 x2\y2
28
|
and plot the graph of the equation
38
- Discuss for intercepts
if (x,, y,) and (xr, yr) are on the
ihat
x:y + ly'. Show
g."ptt uni lr < lz,then .x, <x2'(Hint: Consider the cases
you say that
6 <r' . yr, y, <0 < yr, !r 1!z< 0') Would
the solution set is a function?
39 Let n be anY integer. Define
f(x) = n
3)
*34 a) A curve may closs an asymptote, as demonstrated in
the x axis'
Fig. 29. where we see that the graph crosses
crosses its
graph
its
CJnsruct an equation such that
asymptote twice.
I
/-1x+31
l)(x-3):a
- 4\ -- 4y
Ya:Y'-x'
30 -x2-Y2 : x- 2
32 (x2 - 4)Y' -- x' -
I
lx-21
2y2
20 .r2(y -21 =2
22 y1(x - 2\ + 2: i)
+
is any positive
where the number of crossings is n' where n
integer.
36 Discuss the solution set of
lo9x2-4y2+36:0
12
x axls as an
the x axis 3
crosses
graph
the
asymptote and such that
equdtions
such
constructing
for
method
times. State a
b) Construct an equation which has the
of y
35 F ind the range and intercepts of the solution set I graPh.
: lx - 21. Draw the
6 xY:$
7 x2Y:g
9 2xx - 3y' :6
ll 2x2 + 3yz :6
AsY{rrofts
a)
b)
c)
for
n<x<n+ l,n:0' +l'
of /.
Plot the graPh
Plot the graPh of 8(x) : x -f
Plot the graph of l'{x + il.
+2'
""
(xl'
a
DISTANCE FORMULA. MIDPOINT FORMUTA
Pz@2, Y'z\
L-
I.t(rr,
-
o)
.-+
i
B(r2,0)
r'
We consider a rectangular coordinate system with the unit of measurement
two points in
the same along both coordinate axes. The distance between any
a
joining
shall.derive
We
them.
segment
the
line
it pt"n" is tf,e length of
points"
"
two
the
of
coordinates
of
the
in
terms
formula for this distance
Let Pr, with coordinates (xr, y1), and P2, with coordinates (x2, Iz), be
two points, and let d denote the length of the segment between P, and P,
(Fig.30). Draw a line parallel to the y axis through Pr, and do the same
itrriugtt Pr. These lines intersect the x axis at the points -A(xr,!) and
f(xr,-O).1fhe symbol ,{(x,,0) is to be read'the point 'n witlr.coordinates
(r,, d).';) i.tow diaw a line through P, parallel to the x axis; this line inters:cts
itre.uirticat through P2 and B at a point C(x2, y,)'
CHAPTER
lNEeurt.mrs. FuNcroNs.
1
Trr uxr
The length of the segment between P, and C (denoted lP, Cl) is equal to
the length lABl.lf x, is to the right of x, (as shown in the figure), then the
length of lABlis x2 - xr. If x, were to the left of x,, the distance would be
Xr-xz.In either case, the length of lABl is easily written by using abso-
lutevaluenotation: lABl: lx, - xrl.Similarly,thelengthof lPrCl :lyz- ytl.
It is a good exercise for the reader to check the correctness of these facts
when the points P, and P, are in various quadrants of the plane.
We note that triangle P tP 2C is a right triangle, and we recall from plane
geometry the Pythagorean theorem: "The sum of the squares of the legs of a
right triangle is equal to the square of the hypotenuse." Applying
Pythagorean theorem tc triangle PrP2C, we get
d2:lxz- x,l'+ lyr-- yrl'.
lPrPrl2 :lPtC12 +lP2C12 or
the
This may also be u'ritten
d2
: (xz- x,), + (y, -
yr),
and, extracting the square root, we obtain:
Formula) The distance
Pr(xz, y-r) is defined to be
DEFINITION (Distance
d
between two points Pr(.xr,
yr)
and
Notation.
The distance d between any two distinct points is always positiue
and, in keeping with this, we shall always use the square-root symbol without
any plus or minus sign in front of it to mean the positive square root. If we
wish to discuss the negative square root of some number, say 3, we write
-v3.
EXAMPIE
I
Find the distance between the points P,(1.
Solution We substitute in the distance formuia and
d
EXAtvtPlf
2
l) and P:(5.r-2).
get
: 16 * 31t;1:1:1V :
J4
The point P,(5, -2) is 4 units arvay from
coordinate is 1. Locate the point Pr.
+e
:
rn
a second pcint Pr, whose y
Solution The point P, will have coordinates (x2, 1). From the distance formula
have the equation
4:
To solve for
x,,
Ja,;* -, 1t:-[]l/
we square both sidcs and obtain
t6:(-rz-5)r+9
or
x,-5:+tn.
-
There are two possibilities for xr:
xz:5+{,
u
5-!n.
_v
we
sEcrloN
Dsrrxce ForMur. AffDroNr r(ltrtrur
1.7
In other words, there are two points Pr, one at (5 + ,fi,1) and the other at
6 - ,f;,l), which have y coordinate 1 and which are 4 units from P,' D
.t(rr,0)
Let A(xr,0) and B(x2,0) be two points on the x axis' We define the
directed distance from ,4 to B to be x, - xr. lf B is to the right of z{, as shown
in Fig. 31(a), the directed distance is positive. tf B is to the left of .4, as in
ng. ir(Ul, iire directed distance !s negative. If two points C(0, yt) and
D6, yJ ui. on the y axis, the rlirected distance from C to D is similarly defined
to'# yr-yr. Directed distances are defined only for pairs of points on a
coordin-ate aiis. while ordinary distance between any two points is always
positive, note that directed distance may be positive or negative.
l*t Pr(x',yr) and Pr(xr,)z) be any two points in the plane' We show
B(.r2, 0)
(a)
A(r1,0)
B(c2,0)
ih)
R6IRE
31
Pz@z' az)
Pr(t1,
ciordinates of the midpoint of the line segment ioining Pt atd
P2. Le; P, with coordinates (i, l), be the midpoint" Through Pr, P, and P2
dr-aw parallels to the y axis, and through P, a parallel to the x axis, forming
the triangles shown in Fig. 32. We recall from plane geometrY lhe statement,
..A
line parallel to the base of a triangle which bisects one side also bisects the
other side." Since PC is parallel to-PzD and P is the midpoint of P1P2, it
follows that C is the midpoint of PrD. We see at once that Ais the midpoint
of ArBr.The coordinates of ,4, are (x1,0), the coordinates of Aate (i,0), and
thosi oi B, are (xr, 0)" Therefore the directed distance z{1,4 must be equal to
the directed distance .48,. From the definition of directed distance we have
how to
y v)
c
HGLNE 32
fini tie
i-xr:xz-i
and, solving for x, we obtain
-H##HftH#$#trtr
If we perform the same argument for the y coordinates, the result by analogy
is
Thus we have established the Midpoint Formule. The midpoint
line segment joining Pr(rr, yr) and P2(x2, y2) is
(i,
E;,qMpt"E
3
y)
: (l(x, * xz),i(y,
(i, y) of the
+ yz)).
Locate the midpoint of the line segment joining the points P(3,
-2)
and
Qe4,5).
Solution From the above formula,
- 3-4
f:-A-:-i
g
Mpf"E
I
I
3
and l:- -2+5
2 =1.
Find the length of the line segment joining the poinr A(7,
D
-2\ to the
midpoint of the line segnent betwecn the points B(41) and C(3'
-5).
CHAP.TIT
lNEeuAuflcs. FuNcnols' THE ut{t
1
Solution The midpoint of the line segment between B and C is ar
_ 4+3
_ 1-5 : ^
y:--2
-2.
From the distance formula applied to / and this midpoint we have
7
^- 2 -2'
a:r/W:1r.
tl
A line segment AB has its midpoint at C(5,
(2,3); find the coordinates of B.
EXAMPTE 5
In the midpoint formula, we know that
Solulion
i:
5,
-
1). Point
,4
has coordinates
| = - l, 11 = 2, lr:3,
and we
have to find x2, y,r. Substituting these values in the midpoint formula, we get
2+ x.
2'
-1-
3+v,
T,
and
x2:8,
7
PROBLEMS
ln Problems I through 4. find the lengths of the sides of the
triangles with the given points as vertices.
t
A(4.1). B(2,
2
3
4
A(
-
-
I),
1,0i. B(5. 2), C(3. -21
-2)
A(4,0),8(0, -2),C(5,7)
P,(2.5).P2(7.
-4)
ln Problems 7 and
6
C(4, 8) is a right triangle.
following vertices bisect each other:
In Problems 5 and 6, locate the midpoints of the line segments
joining the given points.
s
15 Show that the triangle with vertices at A(O,2), 8(3,0),
16 Same as 15, with A(3,2),8(1,l), C(-1,5).
17 Show that the quadrilateral with vertices at (3,2), (0,5),
( -3,2), (0, - l) is a square.
'18 Show that the diagonals of the quadrilateral with the
C(- I,s)
A(3, -4t. BQ,r),C(6,
A(3,71, B(8,
-12t
(8,
t'i(1,1),
9
The.midpoint of a line segment AB is at the point P(-4,
-3). The point .,1 has coordinates (8, -5). Find the
coordinates of B.
Thc midpoint of a line segment ,48 is at the point P(-i,2).
l0
Pr(4, O), P2( -1,
-2,
The .r coordin ale ol A is at 5, and the y coordinate of B is at
t
t
l2
l3
l4
-9.
Find the points A and B.
iind
the lengths of the medians of the triangle with vertices
at At4. I ). B( - 5, 2). C(3, - 7).
Same as Problem ll,for A(-3,2), B(4,3),
Show that the triangle with vertices at A(1,
C( l, 6) is isosceles.
Same as 13,
with A(2,3), 8(6,2), C(3,
-l).
s), (2, 3).
-4, *2),
(2, -- 10),
the
vertices of a quadrilateral. Show that the midpoints of the
sides are the vertices of a parallelogram.
line segment joining P, and P, into four equal parts.
7
-
(
19 Same as 18, for (5, 3), (2, -3), ( -2, l), ( 1, 7).
20 The four points A(l,l), B(3,2), C(7,3), D(0,9) form
8. locate the three points which divide the
P2{-2,5) 8
it
lz: -5.
2l
a)
Show how directed distances may he defined in a natural
way for pairs of points on a line parallel to a coordinate
axis.
b) Using the defrnition in (a), carry through the proof of
the midpoint formula, using directed distances along lhe
Iine P,D in Fig.32.
22 The formula for the coordinates of thc point QGo, yn)
which divides ttre line segment from P,(x,, yr) to P r(xr, yr)
in the ratio p to q is
C(- l, -41.
-2), B(-4,2),
Derive this formula.
sEcTloN
DISTANCE FONMULI MIDPOINT FONMULA
1.7
be the vertices ol 1nV
joining
triangle. Show that the length of the line segment
j
length
is
the
triangle
of
this
the midpoints of any two sides
23 l-et A(xr, yr\, B(xr, yz\, C(xr, y.)
of the third
26 Find the set of all points P(x, y)
27
a distance
of
5
Find the set of all points P( x, y) such that the distance of P
from (4, l) equals the distance of P from the y axis' Draw
-
side.
the set of ail points P(x, y) which are equidistant from
(-1,2) and (3, 41.(Hint: Let d, be the distance from P to
and d2 the distance from P to (3,4)' Then the
the graph.
24 Find
i-r,zi
J6 +F
+
o:*
2E
Find the set of all points P(x, y) such that thc distance of P
from (1,2) equals the distance of P from the - axis Drarv
the graph.
condition il r = dz becomes
25
which are at
from (3,4). Draw the graPh.
=
,q;-j1'
a 1v -
+li
Squaring this equation and simplifying yields the equation
of the solution set.) Draw the graph.
Find the set of all points P(x, y) which are at a distance of 2
from (1, l). Draw the graPh.
29 Find
ofall points P(x, y) such that the distance of P
l) is twice the distance of P from (4, - 1)' Draw
the set
frorh (1, the graPh.
30 F'ind the set of all points P(x, y) such that thc sum of
thc
squares of the distances of P from the coordinate axes is I 6'
Draw the graPh.
SIOPT OF A IINE. PARATTEL AND PERPENDICULAR TINES
A line /l, not parallel to the x axis, intersects it. Such a line and the .t axis lorru
two angies *i,i"h ur" supplementary. To be definite, we denote by a the angle
formeJby starting on the side of the x axis to the right of L and going
countercllckwise until we reach the line L. The angle fl will have a value
betfeen 0 and 180'. Two examples are shown in Fig. 33. The angle a is called
the inclination of the lin€ L. All lines parallel to the x axis are said to make a zero
angle with the x axis and therefore have inclination zerp. From plane geometry
we- recall the statenrent: "I[, when two lines are cut by a tlansve ist l.
corresponding angles are equal, the lines are parallel. and conversely." In
Fig. 3i,lines i, uid freach have inclination a. Applying th; theorem of planc
geomedry, wittr the x axis as the transversal, we conclude that Lr an.1 / , are
(g)
!
(b)
HGtIf I}
RGUTE
3'
CHAPTER
lNEQUAtlrlEs.
1
FuNcnoNs' THt
Darallel. More generall
y'
UNE
are parallel arnd'
all lines with the some inclination
haoe the same inclination'
conversely, all parallel'liis
understand'
simple one which is easy to
The notion of in"fioution is a
cumbersome
geometry.and calculus it is
of a
However, for purpotes of analytic
inooaoo the notion of the slope
the
and difficult to use' F*;;;;-*t
of
terms
in
aenoted by * ila is defined
line L. The slope i' f"'"UV
inclination as follows:
D$lNrnoN
1
(Trigonometric)
lf
L' ilenoted bY m' is
Lis aline of inclinati
gioen bY
us recall some
Before discussing slope, let
:
runction.* rhe tangent'il;il
3f
the properties of the tangent
;;;"'l I' i:^t:::::Y':t$
:'j ;,oi i:It
lSlt;l',;rui,u.., .t""a,v, i"u'.tine 1 yl.n d:45'.
:l#T.,:ff increase t"'i'-"t
continues
to
the function
q"]:.approachis 90"
speaking'
;**,'t'" of 9-0" is;;t defined. Loosely
90' and 180' th€
i"rg*i "igO' is infinite. Between
with the help of the
increases withour bou"d:i;
we sometimes say that'the
and its values are obtained
tangent function is negative'
relation
tan(180'-a): -tand'
of the
angle is the negative of that
That is, th€ tangent of an obtuse
anqle, (See Fig' 35')
corresponding acute 'upplementurV
parallel
iunction, wI ... ihut any line
From these fu.t, ul'o'uilt e tanlent
0" and
between
inclination
tin' *itr' un
to the x axis has "'" ;G:;hil'u
strictly
slope'
no
has
to
v axis
90' has positive 'ropu"Xjint farallel such
'the
has infinite slope' If the
iine
a
say
soeaking, although *; ;;iiles
in Fig' 33(a) has
;ht
inclination is an obtu; "'gl;'
'l-o-p::ln"cutive"The'line
slope'
Fig. lj(U) tras negative
oositive slope, and 16";;;in
[now two points on it' To see
if
Jr"p"
We can always frnd;'he
.we
"i"il'9 the points P(2,1) and 0(5,3), as
ti;;';;Igitrougt
,h.
consider
we
this
x axis
of t[is tine, Araw a parallel to.the
shown in Fig.36. To ffiilGe
formlng the right triangle
to
through P and a p"'"ii"f
y axis through Q'
of
ir"tin"t"ion]ty correiponding angles
poR. The angte a
", aennition of tangent run"iion in. a right triangle is
The
lines.
pa-rallel
given by
'f"
;1;;";i1o.ttr.
FIGUTE 35
*opposite
or., uo;^"tii';
The slope is therefore
m:tan
o: lR0l
lF-Rl
Since|R0l:3_|=2al'dlPRl:5-2-3,theslopemisJ.
the plane'
p(r,]i; ;ro QGr,,vz) lleranv two points intwo
Suppose tr,ut
points'
th;'i;;',1 olint rint pu"initn'ough'these
Fig' 37(a) we
ti'" *tt" as the one jusidescribed' In
and we wish to tud
The procedure is exactli
rrcwt
35
in Appendix
.We providc a rcviev of the elcmens of trigonometry
l'
t7
SECTION
Sropt oF A uNE PArA[s. AND PnPslucuun uNEs
1.E
see that
m:
tan
lt
d: !2:
xz- xt
,
and in Fig.37(b)
m
rz, j't
:
tana
= -tan(lgo
- a): -
)
,
#:
#.
A difficulty arises if the points P and Q are on a vertical line, since then
xt: xz aid the denominator is zero. However' we know that a vertical line
when
has no siop", and we state that the formula holds for all cases except
In
line.
horizontal
a
on
lie
P
and
if
difficulty
Q
xr: Xz.Ntte that there is no
t
case, lr:lzand the slope is zero, as it should be'
that
""-'W"
."" tt ri itr"rr is an algebraic formula for the slope of a line which is
equivalent to the definition in terms of the inclination'
(^)
The slope m of a line L through the two points
Q{xr, yr) with x, * x2 is giuen by tke formula
, (AlC"b*t")
HCLnt
P(xr,yr)
and
If we think of a particle traveling along the line L from P(xr,Ir) to
Q(*r,yr), then the particle will rise an amount !z-!r while running
horiionially an amount Xz-xr- We say,. intuitively, that the slope is the
37
"rise" divided by the "run" or, more succinctly,
sioPe
:
rise
-
'
yr-y, is negative then the "rise"
is to be taken loosely'
rise/run
expression
Of course, if
is actually a fall so
the
a
4+
EXAMPTE
,f
1
Solution
2+
g
I
1+
P(4,1)
E(5,1)
Find the slope of the line through the points (4, -Z) and (7,3).
First wg note that in the formula for slope it doesn't matter which point we
label (x,, y1) [the other being labeled (xr, yr)). We let (4, -Z) h (xr, y,) and
(7,3) be (*r,yr). This gives
m:
n/
rct f 3a
EXAMPTE 2
Solution
3-(-2)
-1-T-:3
s
Through the point P(4, 1) construct a line with slope equal to
n
!.
Starting at P, draw a parallel to the x axis extending to the poin-t R one unit
to the right (Fig. 38). Now draw a parallel to the y axis, stopping I units above
R. The coordinates of this point Q are (5, i). The line through P and Q has
slope
!.
tr
We have seen that parallel lines always have thc same inclination
Therefore, we have the following result.
CHAPTER
lNEeuAunEs. FuNcnoNs. THE ttNE
1
THEOREM
7
Proof
two lines with the same
Two porollrl lin,' haue the same slope' Conuersely'
slope are parallel.
Sinceparallellineshavethesameinclination,thedefinitionofslopeinterms
two lines have
oiin.rinution shows they have the same slope. Now suppose
angle between 0"- and 180"'
the same slope, say ,. th.n there' is exactly: one
of the tangent
;;il;;r, t"th that ian fi : m'To see this observe that thern graph
range of the
the
in
value
iur.tlo, in Fig. 34 shows clearly that for any
one value of
exactly
is
there
ir^g.rifrr.tiJn, i'e., for any value on the- y axis'
slope' their
same
the
have
lines
[r,i.fl ,fr", tan I :rm. We conclude that ifthetwolines
(This
parallel'
are
therefore
inclinations must also be the same, and
to
perpendicular
are
lines
the
works even if the slopes are infinite, since then
T
the x axis and so Parallel')
Whenaretwolinesperpendicular.llfalineisparaileltothcxaxis,ithas
rnfinite slope. Suppose,
zero slope and is perpeniicular to'any line which has
(Fig'39)' Its inclination
zero
is
not
which
n,
slope
hu,
t o*.uri, that a line
Lt and have
perpendicularto
be
L,
L,et
from-zero.
*lff U. o1,, also different
> 0. We
inclination n2, &S Shown; ii is assumed here that m,
i,
;il;;;';;d
recallfromplanegeometry:,..Anexteriorangleofatriangleis.equaltothe
(in Fig. 39)
sum of the iemote interio; angles.,' This means that
and lan d'2: tan (90' * a1)'
a: : 90' * ar
We recall from trigonometry* the basic formulas
* B) :
cos (A + B):
sin '4 cos B + cos '4 sin B
sin ('4
cos '4 cos B
-
sin '4 sin B'
We find
mz:
lan
sin (90" + a')
dz:lan(90. + dr) : cos(90. + sr).
Nowapplyingthebasictrigonometryformulastotheaboveequation'weget
sin 90'cos a, * cos 90" sin o,
m2
c* gO'=ot r, --sin 9tr si, o, ,
We use the fact that sin 90'
:
1; cos
cos a1
m.:
" ---:--:
-stn 1,
90'
:0
to obtain
i
a,
-cot '
The last formuia, in terms of slopes, states that
We have established the next result'
l.
'See Appendix
tan al
39
sECTloN
Sro?t
1.8
of
A uNE. PAnALLE AND
mPENDlcuun
LINES
r if an'| onty if
their slopes are the
negatiue
reciprocals of each other.
EXAMPIE
I
Show that the line through
P,(]'--4)
*2'6)
and Q'(
is parallel to the line
through Pr.(-3,6) and 02(9' -18)'
to the formula'
Solution The slope of the line through Pt and Q" according
-ts
2-3
^,:#^l:
-5
is
- -2'
Similariy, the line through P, and Q2 has slope
m': -!-3:
9+l
-
2'
parallel'
Since the slopes are the same' the lines must be
ExAMptE
4
Determine whether or not the three points
iie on the same straight line'
P(-l' -5)'8$'3)'
and R(7'
12)
Solution The line through P and Q has siope
3-( -5) :*'
'':l-(_l.)
a
very same
If R were on this line, the line joining R and Q would have to be the
R
through
line
the
of
slope
The
slope.
same
the
have
would
line; therefore it
and Q
is
*r:H:1,
z
l-l
same
and this is different from m,. Therefore, P, Q, and R do not lie on the
fl
line.
ExAMPIE
s
Is the line through the pr;ints Pr(5,
line through the points
Pr(-3,1)
-
1) and
Q,( -3'2) perpendicular to the
and 02(0' 9)?
Solution The line through Pt and Qt has slope
2-(-1) : _ 3
mr:_;3_5
g.
The line through P, and Q, has sloPe
,r:ffi:l
The slopes are the negative reciprocals of each other, and the lines are
D
PerPendicular.
EX
}{ru 6
Given the isosceles triangle with vertices at the points P(- l, 4), QQ: t), a1d
R(2, 5), show that the median drawn from P is perpendicular to the base QR
(Fie. a0).
CHA?T'I
lMeuruTtB. FuNcnoNs. Tm
T
Solution
l,et
M bethe point
uNE
where the median from P intersects the base QR. From the
definition of median, M must be the midpoint of the segment QR. The
coordinates of M, from the midpoint formula, are
.t:012:r,
,=511:1.
z2
Now we check the slopes of PM and QR. The slope of PM is
3-4
n,:t_(_D
The slope of QR is
I
*':sfi:z'
Since
HGUNE 10
I
frl: _
^r,
D
the median is perpendicular to the base.
8
PROBLEMS
In Problems I through 6, check to see whether the line through
the pair of points P1, Q, is parallel or perpendicular to the line
through the pair of points Pz, Qz.
I Pr(-5,2),QrQ, -l) and P2@,2\,Qr02, -l)
2 P13,l), Qr{-2,7) and P2$, *3)., Qr(- 1, -8)
3 Pr(5,3), 0r(8,3) and Pr(7, 4), Q2Q, -4)
4 P t(-4,5), Qll4,2) and Pr(6, 0), Q2(9, -4)
5 PJl2,8), 0,(4, 8) and Pr(3, 1), 0r(-6, 1)
5 Pt(7,-1), Cr(10,2) and Pr(O, -4),0r(1, -5)
10, determine whether or not the three
points all lie on the same straight line.
In Problems 7 through
7
P!3,4), 0(8,5), R(13,6)
t
P(2,
-
t4 P(r,3), Q(3,7), R(3, 11), S(1, 7I
rs P(2,3), Qg, l), R(0, - 8), S( - 5, - 3)
16 p(5, 6), 0( -8, -7), R(-8, - 10), S(5, 3)
t7 P(-1,2),QQ, -2), R(?8, -t), s(-+3,*)
rE P(3, +), Q$2), R(r, -4), S(5, 12)
P(3, -\, Q(2, -6), R(1,7), S(0, -8)
20 P(5, 3), 0(8, 3), R(4,4), S(7,4)
lg
2t
P(3,0), 0(3, 5), R(0, e), s(0,4)
22 P(2,1), Qfi,13), R( - 5, l8), S( - 10, 6)
23 P(6,2), QQ, - 1), R(7, 5), s(8, -4)
24
The points .4(3, -2), B(4,1), and C( * 3, 5) are the vertices
of a triangle. Show that the line through the midpoints of
the sides AB and /C is parallel to the base 8C of the
25
The points.4(0, O), A(o,0), and C(|a, b) are the vertices of a
triangle. Show that the triangle is isosceles. Prove that the
median from C is perpendicular to the. base ,{8. How
general is this proof?
l), 0(5,3), R(-7,4)
e P(7,1),Q(-7,2), R(4, -5)
t0 P( - 3, 0), 0(4, 1), R( 11, 2)
triangle.
ll
Construct a line passing through the point (5, -2) and
having slope fl..
12 Construct a line passing through the point (-3, l) and
having slope
- ].
13 through 23, the points P, Q, R, S are the vertices
of a quadrilateral. In each case, determine whether the figure is
a trapezoid, parallelogram, rhombus, rectangle, square, or irone
25 The points /(0,0), B(a,0), C(c +
b, c), D(b, c) form the
vertices of a parallelogram. Prove that the diagonals bisect
each cther.
In Problems
of
these.
13
P(1, 3\, QQ,5), R(6, l7), S(5, l5)
27
The points
/(Q
0), 8(a,0), C(b, c),
D(e,/)
are the vertices
of a quadrilater[l. Show that the line segments joining the
midpoints of cpposite sides bisect each other. How generat
is this proof?
'
SECTION
Srorr or
r uxe prnru.a
2t
Show that in apy triarigle the length of one sidc is
no largcr
than the sum of the lengths of the other two sides.
29
Let
P(-ta,0)
and e(!a,O) be two a_djacent.vertices of a
regular hexagon above the side pp. Find the coordinates
of
AND
pmptNDtcuun
**
joining two opposite vertices.
30 Let P(-io,O)
the remaining vertices and the length of the diagonat
and e(!a,0) be two adjacent vertices of a
regular octagon situated above the .ia. fg.
Find the
coordinates of the remaining six vertices and the
length of
the diagonal joining two opposite vertices.
THE STMIGHT LINE
It is.easy to verify that.the equation y:4 represents all points
on a line
parallel to the x axis and four units above
it. similarry, d"i;;;,
x= -2
represents all points on a line parallel
to the y axis and two units to the left. In
gcneral, any line parallel to the y axis
has an equation of the form
X:
Q,
where c is the number denoting how far the
line is to the right or left of the y
axis: The equation
!=b
describes a line parailel to the x axis and
b units from it. In this way we obtain
the
equations of a[ rines with zero or innnite
srope. A rine which is not
parallel to either axis has a slope m which
i, aimer"ni'rro;r;;; i;;;;#;
the line passes through a poinf denoted p(-x;
yr ). To be sp""inc, *'e'consider
dqry m t9 tte -] andthe point p to t ur. coordinates
(4, _3).
1:
Q6,
i
is on this line, then the srope as carcurated
from
shown in Fig.41. That is,
rcUE{1
y+3 2
i--7: -,
or
If a point
p to must be
e
- !, as
y+3: -lO-oy.
frus rs the equation of tle tiye paslilr through
the point(4, _ 3) with stope _
In the general case of,a line withltop" i pu.rirg
t;;od; iir,,y,t, tt.!.
statement that Q$,y) is on the line is the
same as thi statem-ent that^ihe slope
m as ccimputed from p to
e (Fig.42) is
!-lr
r-rr
Of
-:ttl
FGIIf €
We summarize this general statement
as follows:
THEOREI|
r
@oint-Slope Formuh for the Equetion of a
passing through the point P(xr,
yr) with
Une) fru
slope m is
,iiffi
CHAPTER
lNreuAuTrrs. Fuxcroxs. Tnr uxr
1
If we are given the coordinates of a point and the numerical value of tha
slope, substitution in the above formula yields the equation of the line going
through the point and having the given slope.
EXAMPTE
1
Find the equation of the line passing through the point
slope f
(-2,5)
and haring
.
Solution Substitution in the above formula
gives
.,-5:*t-x-(-2)l
3y:4x+23.
or
We know that two points determine a line. The problem of finding thr,a
equation of the line passing through the points (3, -5) and (-7,2) car,re
solved in two steps. First we employ the formula for the slope ol'a line.
".
given in Section 8, to obtain the slope of the line through the given points. \\'e
get
n
7-.
:?-!:5)
_7_ -t - -- 10'
Tlren, knorving the slope, we use either point, togethcr with the slope, in th;
point-siope formula. t his gives [using the point (3, -5)]
or
I-(-5): -f6(.x-3)
i0],- -7x-29.
We verify quite easily that the same equation is obtained if we use the
(-7,2)
porll
instead of(3, -5).
The above process nray be transformed into a general formula brl
applying it to any two points, Pr(xt, y,)and Pr(*r,yr). The slope of the lice
through these points is
lt
" n': lz- Xt
Xz-
Substituting this value for the slope into the point-slope form, we get the twe
point form for the equation of a line:
Note that this is reaily not a new formula, but merely the point-slope fonr:
with an expression for the slope substituted into it.
Thus we have the following corollary.
COROttARy TO THEOREM
9
('I'wo-Point Form for the Equation of a Line) The equation of the line passine
throu,gh the two points Prlx1,),r) and Pr(r2,y) with xr*x2is giuen by
v...-!:V irir*t
:II:ili1M,,t*::ia
:it ;
Iil
Another variation of the point-slope form is obtaincd by introducing a
number called the y intercept. Every line not parallel to the y axis musi
intersect it: if we denote by (0, D) the point of intersection, the number b b
SECTION I.'
THt STIAIGHT
€
TINE
ca'ed the y intercept.* Suppose a rine has slope
m and y intercept D. we
substitute in the point-slofe form to get
or
t- b :m(x -o)
fldf#lffir
This is cailed the srope,intercept form for th6
equation of a straight rine.
EXAM'.E
2
A rine has srope 3 and y intercept
-4.
Find its equation.
Solution Substitution ln the slope-intercept formula
gives
!:3x-4.
The important thing to notice is that the point-slope
form is the basic one
for the equation of a straight rine. The other iormur"r;";
;.;;;; ;;,t.pi;
variations or particular cases.
Examples I and 2led to equations of lines which
could be put in the form
Ax+By+C=0,
where l, B, and c are any numbers. This
equation is the most general
equation of the first degree in x and y. \Ve
shali establish the theorem;
THTOREM
10
Euery equation of the form
r-ifi[fr[11$fl*i#
so long as A and B are not
Proof
We consider two cases, according as
have A * 0, and the above
"quutlon
B=0 or B#0. If B=0,
becomes
,: -
then we must
C
A,
which we know is the equation of a straight
line parallel to the y axis and
- C/,4 units from it.
If B + 0, we divide by B and
solve for ;,, getting
AC
t: -Ex-8.
From the slope-interceqt.foy for the equation
of a line, we recognize this as
the equation of a line with slope _ AIB Lnd y
_
intercept
ClB.
Ij
In the statement of the theorem it is necessary
to make the requirelncnt
that ,{ and B are not both zero. Ifboth ofthem
vanish and c is zero, theringar
equation reduces ro the triviarity 0=0, which
i";tir# by'J"rry point
in the plane. lt C +0, then no point p(x, y).urirn"r"ri.
[!,i,
.The point (O 6) is
atso called the y intercept.
"ifi;;
CHAPTET
lN[eu LrnE tuNcltoNs' Trr uxe
1
D(AIVIPI..E 3
Given the linear equation
3x+2Y+670,
find the sloPe and Y intercePt.
Solving for Y, we have
y: -jx - 3.
:
From this we simply read off that m - ] and b: -3'
tl
a linear equation' When
An equation of the first degree in x and-y is called
form' y becomes a function of
we solve for y in terms of x, u, i-n th' point-sl'ope
x.Suchafunctioniscalledalinearfunction.Anylinenotparalleltotheyaxis
may be thought of as a function'
E(Al\{ru d
Solution
bisector of the line
Find the equation of the line which is the perpendicular
Fig'43')
r.g*"r, joining the points P('-3,2) and 0(5,6)'(See
We give two methods.
Method
1.
Wp first find the slope m of the line through P and Q'
6-2
^:5al:1'
It
is
I
Theslopeoftheperpendicularbisectormustbe-2'thenegativereciprocal.
segment PQ.They:are
N.*a *. get the *oidinut., of the midpoint of the line
5-3
'=;:
).:
6t2:4.
2
The equation of the hne through (1,4) with slope
y- 4: -2tx- l)
+
-2
is
2x*Y-6:0'
which is the desired equation'
Method2.WeStartbynotingthatanypointontheperpendicularbisector
let d'
2 Let R(x, v) be any such-poing and
equidistant from P
"ia
the
From
to
R
from
distance
Q'
J."",. the distance from R to P, and d, the
dr:dz,we
condition
the
and
roi*uru for the distance between two points
;;;
have
$ + 3tr+lil : Jir - tl';
Iv
out, we obtain
fuuanng both sides and multiplying
x2 +6x +g + y2 - 4y +4: x2* 10x* 25 +
- r'
v2
-
'
12y
+
36.
terms combine to
The terms o[ the second degree tancel and the remaining
give
6x-4y* 13: -10x- 12v+61 +
l6x+8y-48:0'
Dividing by E, we obtain the same answer as in the first method:
2x+Y-6:0'
tl
sEcTtoN
{5
1.9
THE sTnArcHT UNE
9
PROBLEMS
In Problems I through 17, find the equation of the iine with the
given requirements.
I
Slope 2 and passing through (
-
1,4)
14 Slope
-j
segment
segment
3I
Slope *3
-''l)
-
Find the slope and y intercept of the line 2x 3y 7 :0.
- 20 Find the equation ofthe line through the point (1, _4) and
parallel to the line x + 5y 3 :0.
2l Find the equation of the line through the point ( _2, _3)
and parallel to the line 3x 7y + 4:0.
-
23
24 Find the equation of the line through the point ( l, _3)
and parallel to the line ihrough the points (3,2)
and
-
5.7)
25 Find
1, 3).
- l)
bisector of the linc
and (5,2).
the points p(0, O), e@,0), R(a, b), S(0, b) are the vertices
of a rectangle. Show that if the diagonals mect at right
a: *b and the rectangle must be a square.
the figure is a rhombus.
34 The points
.4(1, o), B(7,0), c(3,4) are the vertices of a
triangle. Find the equations of the three medians. Show that
the three medians intersect in a point.
35
the equarion of the line passing through (4, _2) and
parallel ro the line through the points (2, _l) and (5,7).
2.6 Find the equation of the line through the
midpoint of the
Iine segmenr joining (2, l) and (6, _4) and ihrough the
point which is j of the way from (3,2) to (7, _6).
Tlre points , (0,0), B(a, 0), C(b, c) arevertices of a trianglc.
Show that the three medians meet in a point.
36 The points .4(1,0), B(5,1), C(3,g) are the vertices of
a
37 The points ,<10,01, B(a,0), C(b,c) are the vertices of
a
triangle. Find the equation of the perpendicular from each
vertex to the opposite side. Show that these three lines
intersect in a point.
triangle. Find the equation of the perpendicular from cach
vertex to the opposite side. Show that these thrce_lines
intersect in a point.
and
F-ind the equation of the line through the poinl (1,5) and
perpendicular to the line 5x 4y + I :0.
-
(
-
33 P, Q, R, S are the vertices of a parallelogram, Show, by
analytic geometry, that if the diagonals arc perpendiculai
and y intercept 0
-2)
(
parallelogram must be a rectangle.
1)
rhe equation cfthe line through the point (3.
perpendicular to the line 2x + 3y + 4:0
joining (3,
angles, then
Parallel to the y axis and passing through the point (4, _- 3)
Parallel to the x axis and passing rhrough the point(6, _5)
Find the slope and -y inrercept of the line 2x +1y + 4 :0.
22 Find
and
32 Let P, Q, R, S be the vertices of a parallelogr:rm. Show, by
analytic geometry, that if the diagonals are equal the
(4,7)
and (5,
joining (6, 2) and
30 Find the equation of the perpendicular
and passing through the midpoint of the line
segment connecting (2,
15
16
17
18
19
-2)
to the line through the points {7,0)
28 Find the equarion of the line passing through (2, 1) and
perpendicular to the line through the points (3, l) and
(-2,5).
29 Find the equation of the perpendicular bisector ofthc linc
13 Slope 2 and passing rhrough rhe midpoint of the line
aruJ
the equation of the line passing through ( * 5, 3) and
perpendicular
(-8. r).
2 Slope 4 and passing through (3, - l)
3 Slope -J and passing through (-2,5)
4 Passing through the poinrs (5, 2) and ( - 1, -6)
5 Passing through the poinrs {2, -3) and (0, -4)
6 Passing through the points (1,4) and (-2, -7)7 Slope 0 and passing through (-2, -7)
8 Passing through the points (3, 8) and (3. - j)
9 Passing through the points (4, _Z) and (-7, - 2)
l0 Slope - f and y intercept 3
ll Slope 0 and y intercept -5
12 Slope ! and y intercept 8
segmenr connecting (3,
27 pind
3t !:l a > 0, c > 0 be given numbers. Show that the poinrs
,{(0,0), B(a,0), C(a+b,c),.D(b,c)
are rhe verticei of a
39
parallelogram, and show that its area is cc.
Let c > 0 and d > b be given numbers. Show that the point$
,4(0, 0), B(a, 0), C(d, c), D(b, cl are the vcrticcs of a irapc.
zoid and that its area is jc(a + d _ b).
40 Let a>0, c>0 be
given. numbers- The points ,{(0,0),
B(d,9), C(6, c) are rhe verrices of a triangle. Supposc'that
b
arrd E are points on sides AC andaC, respectivciy,
*ittr tinc
segment DE parallel to AB. Show that
lcAVlABl.
1OO1lOep
CHAPTTN
lNreuAunEs. FuNcnoNs. THr UNE
1
CHAPTER
1
REVIEW PROBTEMS
In Problems
1 through 6, solve
| 3+2x<4-x
3 2*'.4
x-l
In Problems
7
f
2 2(6-4)<3x+5
x*l 3
4. _<_
2-x 2
1r-
5 ''^<3
x
.')
6 =-:-<3
I*x
for x.
and
and
3+x
2
Plot lhe graph of
<-4
9 l2x(l - x)l < 2
t0 lx + 2l < lx(x - l)l
I I Find all solutions of the simultaneous equations
lx - yl : l,
l2x-Yl:1.
12 Write a complete proof of the srarement: la. bl: lallbl for
a, b.
the graph
11.
and (5,6).
b) Find the equation
of the line through this midpoint and
(
-
l,
point
3) to (4. l).
(2, l) is always equal to the distance from
this set.
(-1,3).
Sketch
Suppose a line has x intercept (a,0) and y intercept {0, b).
Show that the equation of the line is
xv+-:1.
ab
in [o' zi
This equation is known as the two-intercept forrn of the
straight line.
12+l
show that la,
* a2*
..
19 In the inequality la + bl < lal + ibl, state all conditions in
which the equality sign holds.
20 Civen i{x\: x2 -2x * a, find /(
f(a + 3).
2l Civen f{r) : x2 +2, show that
-
3l Describe the set of ai! poinrs such that the distance from
I
t7 ;r;i in [3,7l
l8 If a,, a2, ..., an are any numbers,
* onl < larl + la2l * -.. * lo,l.
lx + 1l + lx
Find the domain, range, intercepts, and asymptotes.
a) Find the midpoint of the line segment L joining (2, l)
14 lx(x + l)l in t- 2. 3l
15 lx2 + 2x - 3l in [-a,2]
ix+21
the
r@:l-.
lx- ll
one-third of the way from
In Problems 14 through 17, find an estimate lor how large the
given quantity can become in the given interval.
lr-*-r, * i,
:
Use the formula in Problem 22 of g6p1i61 7 to find the
: lll
lb
PIot
of
perpendicular to L.
Write a complete proof of the statement:
for all real numbers a,b with b +0.
tu
/(x)
E 12+xl<14-xl
tbt
(ab)
24 y2(x - 3):a
25 y= f ar2
26 y'-3xy+2xz:O 27 la:x3
28 Draw
t,r
-f
In Problems 24 through 27, find the intercepts, domain, range,
and asymptotes, if any. Sketch the graph.
through 10, find the values of x, if any, for which
all real numbers
(b)
(a)
graph.
the inequalities hold.
l3
f
+I0l - 3 -1
What is the domain of /(x) =U@aya77
2x-6<5
,ffi-,
(a)
f
Find the equation of the line through the point A(2,
which is parallel to the line 2x - 3lt - i : 0.
-l)
Find the equation of the line through the point B(3, -Z)
which is perpendicular to the line x + 2y + 3:A.
Given the line 2x * ky -3:0. find the value offt such that
the point (1,4) is on the line.
-
l),,f(0), f
(2},
f {a),and
Given the vertices of the triangle A(2, 1), B(4,0), C(5,4),
find the equation of the line passing through / and the
point which is one-iourth of the way from .B to C.