_tl9P53EINhZ

Lo**ryotl

i.J '-'

t l

'"' 20.

f du f ---::r J u\/a+nf itu

[-1-r" 16lT + {Ll *, - ir a > o \/i "' l\/a +Tn lal |-

lZr^n_rl!-a \/;T6

r| @-_ a

q6 ifa 8,2> -10 < -7.In particular, a 1b and a > b are called strict inequal5 = 3, ities, whereas a -b anda= b are called nonstrict inequalities.. The following theorems can be proved by using L.L.6 through 1.1'.19. "1..1.20 Theorem (i) (ii) (1ii) (iv) a a a a > < > < 0 0 0 0 if if if if and and and and only only only only if if if if a is positivq. a is negative. -a < 0. -a > 0. if andonlyif b isless

T 1.7.21 heorem

If.a-5; so7*3>2+(-5). o

L.1.30Theorem

If a > b and if c is any positive number, then ac ) bc. o rLLUsrRArroN 12: -3 > -7; so (-3) > eDA.

1.1 SETS,REAL NUMBERS, AND INEQUALITIES

1.1.31Theorem lf. a > b and if c is any negative number, then ac I bc.o rLLUsrRArroN 13: -3 > -7; so (-3)(-4) < (-7)(-4). o

1.1.32Theorem If a > b > }and c > d > 0, then ac > bd.o r L L U S r R A r r o1 4 : 4 > 3 > 0 a n d 7 N >5 ) 0; so 4(7)>3(6). o

n

79

tz24

lllttrttllrll -4t -2 1 Figure .1.1

0 |

zt

4

So far we have required the set Rl of real numbers to satisfy the field axioms and the order axiom, and we have stated that because of this requirement Rr is an ordered field. There is one more condition that is imposed upon the set R1. This condition is called the axiom of completeness (Axiom 16.2.5). We defer the statement of this axiom until Section 16.2 because it requires some terminology that is best introduced and discussed later. However, we now give a geometric interpretation to the set of real numbers by associating them with the points on a horizontal line, called an axis. The axiom of completeness guarantees that there is a oneto-one correspondence between the set Rl and the set of points on an axis. Refer to Figure 1.1.1. A point on the axis is chosen to represent the number 0. This point is called the origin A unit of distance is selected. Then each positive number x is represented by the point at a distance of r units to the right of the origin, and each negative number x is represented by the point at a distance of -x units to the left of the origin (it should be noted that if r is negative, then -r is positive). To each real number there corresponds a unique point on the axis, and with each point on the axis there is associated only one real number; hence, we have a one-to-one correspondence between Rl and the points on the axis. So the points on the axis are identified with the numbers they represent, and we shall use the same symbol for both the number and the point representing that number on the axis. We identify Rl with the axis, and we call Rl the real number line. We see that a < b if and only if the point representing the number a is to the left of the point representing the number b. Similarly, a > b if and only if the point representing a is to the right of the point representing b. For instance, the number 2 is less than the number 5 and the point 2 is to the left of the point 5. We could also write 5 ) 2 and say that the point 5 is to the right of the point 2. A numberx is between a andbif a ( rand x "1..2.3 Corollary Irl = a if. and only if -a < )c < a, where a > \.2.4 Theorem ltl > a if andonly if x > a or x I -a,where 7.2.5 Corollary ltl = a if andonly if x > a or x s -a,where

The proof of a theorem that has ant"if and only if " qualification requires two parts, as illustrated in the following proof of Theorem 1.2.2. plnr l.: Prove that lrl < a if.-a < x I a,where a ) O.Here,we haveto .x consider two cases: > 0 and r < 0. Casel: x=0. r Then ltl : x. Because < a,we concludethat lrl < a. C a s e 2 :x < 0 . -a and Then lrl : -r. Because < x, we aPPlyTheorem L.1,.25 obtain -x: -x < a.But because we have l*l < o. a)-x or, equivalently, lrl, then, In both cases, wherea>0 0andr(0. CaseT: x>0. Then ltl : r. Because lxl < a,we conclude that r < a. Also, because -a 10 a ) 0, it follows from Theorem 1.1.31 that-a < 0. Thus/ we have 5 or 3x*2 l3r+21 5

That is, the givgn inequality will be satisfied if either of the inequalities in (1) is satisfied. Considering the first inequality, we have 3x*2>5 or, guivalently, x)l Therefore, every number in the interval (1, +oo; is a solution. From the second inequality, we have 3xt2 7

= !. lr+41 lzx-61 ) a 3 + - l 2l - - ' l.1 o x s-r= 127. Prcve Theorem 1.2.4.

16. l2x-sl < 3 1 9 .l 2 x - s l > 3 22. l3rl > 16 }xll 2s.v * ? l .n lffil

1 7 .l 3 x - 4 1- z 2 0 .1 3 + 2 x l < l a - r l 23.le- zxl =- l4xl rc I l1_l

=1,-rr 26. !ffiI

28. Prove Theorem 1'.2.7.

In Exercises29 through 32, solve for r and use absolute value bars to write the answer.

29.31..

12'8') then 33. prcvethatif , andb areany nurnberg, lc-bl = l4l +lbl.(HINI:Writea-basa'+(-b) anduseTheorem 1.2.8.) Let 34. prcve that if a and b are any numben, then lal lbl < la bl. (Hrr.rr: lal: l(a b) + bl, and useTheorem s>b+ ca d a> b- c? to 35. What singleinequalityis equivatent the followingtwo inequalities:

1.3 THE NUMBER PLANE AND GRAPHS OF EQUATIONSwriting them in parentheses with a comma separating them as (I/ y). Note that the ordered pair (3, 7) is different from the ordered pair (7,3).

Definition 1..3.1.

The set of all ordered pairs of real numbers is called the number Plane' and each ordered Pafi (x, D is called a point in the number plane. The number plane is denoted by R2. fust as we can identify Rl with points on an axis (a one-dimensional space), we can identify R2 with points in a geometric plane (a two-dimendional space). The method we use with R2 is the one attributed to the French mathematician Rene Descartes (1595-1650), who is credited with the invention of analytic geometry in L637. A horizontal line is chosen in the geometric plane and is called the r axis. A vertical line is chosen and is called lhe y axis. The point of intersection of the r axis and tlne y axis is called the origin and is denoted by the letter O. A unit of length is chosen (usually the unit length on each axis is the same). We establish the positive direction on the r axis to the right of the origin, and the positive direction on the y axis above the origin.

REAL NUMBERS, INTRODUCTION ANALYTIC TO GEOMETRY, AND FUNCTIONS

Figure .3.1 1

We now associate an ordered pair of real numbers (x, y) with a point P in the geometric plane. The distance of P from the y axis (considered as positive if P is to the right of the y axis and negative if P is to the left of the y axis) is called the abscissa(or x coordinate) of P and is denoted by *.The distance of P from the r axis (considered as positive if P is above r the x axis and negative if P is below the x axis) is called the ordinate (or y coordinate)of P and is denoted by y. The abscissa and the ordinate of a point are called the rectqngular cartesian coordinatesof the point. There is a one-to-one correspondence between the points in a geometric plane 7) and R2; that is, with each point there corresponds a unique ordered pair (x,y), and with each ordered pafu (x, y) there is associated only one point. This one-to-one correspondence is called a rectangular cartesian coordinate system, Figure 1.3.1 illustrates a rectangular cartesian coordinate system with some points plotted. The x and y axes are called the coordinateaxes,They divide the plane into four parts, called quadrants, The first quadrant is the one in which the abscissa and the ordinate are both positive, that is, the upper right quadrant. The other quadrants are numbered in the counterclockwise direction, with the fourth, for example, being the lower right quadrant. Because of the one-to-one correspondence, we identify R2 with the geometric plane. For this reason we call an ordered pair (x, y) a point. Similarly, we refer to a "line" in R2 as the set of all points corresponding to a line in the geometric plane, and we use other geometric terms for sets of points in R2. Consider the equation A:x2-2 (1)

where (x,y) is a point inR2. we call this an equation inR2. By u solution of this equation, we mean an ordered pair of numbers, one for r and one for y, which satisfies the equation. Foi example, if r is re p l a c e d by 3 i n E q. (1), w e see that A :7; thus, x:3 and,y:7 const itutes a solution of this equation. If any number is substituted for x in the right side of Eq. (1), we obtain a corresponding value for y.It is seen, then, that Eq. (1) has an unlimited number of solutions. Table 1.3.1 gives a few such solutions. T a b l e .3 .7 1

x

l0

L2

3

4-1.-Z

-3-4

a:x2-zl-z

-1

2 7 74 -1

2

7

14

F i g u r e1 . 3 . 2

If we plot the points having as coordinates the number pairs (x,y) satisfying Eq. (1), we have a sketch of the graph of the equation. In Fig. 1'.3.2we have plotted points whose coordinates are the number pairs ob-

1.3 THE NUMBERPLANEAND GRAPHSOF EQUATIONS

tained frorn Table 1.3.1. These points are connected by u smooth curve. Ary point (x, y) on this curye has coordinates satisfying Eq. (L). Also, the coordinates of any point not on this curve do not satisfy the equation. We have the following general definition. 1.g.2 Definition The graph of an equation in I is the set of all points (x, y) in R2 whose coordinates are numbers satisfying the equation. We sometimes call the graph of an equation the locus of the equation. The graph of an equation in R2 is also called a curae. Unless otherrryise stated, an equation with two unknowns, x and y, is considered an equation in R2.

sxelrpr-n L: Draw a sketch of the graph of the equation' Y'-x-2:0 Q)

soLUrIoN:

Solving Eq. (2) for !, we have

y:!\tX+2 Equations (3) are equivalent to the two equations

(3)

Y:lx+2 Y-_\'x-2

(4 )

(s)

The coordinates of all points that satisfy Eq. (3) will satisfy either Eq. (4) or (5), and the coordinates of any point that satisfieseither Eq. (4) or (5) will satisfy Eq. (3). Table 1.3.2 gives some of these values of x and y. 1-.3.2 Table

x

00

1

"1. 2

2 -2

3

3-L-1-2

v \/2 -\/2

\/i

-\/5

2

\/5 -\tr

1, -L

0

Note that for any value of r -2 there are two values for y. A sketch of the graph of Eq. (2) is shown in Fig. 1.3.3.The graph is a parabola.


2: ExAMPLE Draw sketchesof the graphs of the equations y:\/x+2 andY:-1/Yt'2

soLUrIoN: Equation (5) is the same as Eq. (a). The value of y is nonnegative; hence, the graph of Eq. (6) is the upper half of the graph of Eq. (3). A sketch of this graph is shown in Fig. 1..3.4. (5) Similarly, the graph of the equation

Y:-\E+2a sketch of which is shown in Fig. 1.3.5, is the lower half of the parabola o f F i g . 1 .3.3.

F i g u r e1 . 3 . 4

F i g u r e1 . 3 . 5

rxltrpr.n 3: Draw a sketch of the graph of the equation

soLUTroN: From the definition have y:x*3 and if r*3>0 if r*3-3 if x1-3

Table 1.3.3 gives some values of x and y satisfying Eq. (Z). T a b l e .3.3 1Figure1.3.6

x

0 3

L 4

2 5

3 6

-L 2

-2 1

-3 0

-4 1,

-5 2

-6 3

-7 4

-8 5

-9 6

v

A sketch of the graph of Eq. (7) is shown in Fig. 1,.9.6. rxetvrplr 4: Draw a sketch of the graph of the equation (x-2y+3)(V-x'):0 (8) soLUTroN: By the property of real numbers that ab:O a : 0 o r b :0, w e have from E q. (8) x-2y*3:0 if and

1.3 THE NUMBERPLANE AND GRAPHSOF EQUATIONS

y-xz:o

(10)

The coordinates of all points that satisfy Eq. (8) will satisfy either Eq. (9) or Eq. (L0), and the coordinates of any point that satisfies either Eq. (9) or (10) will satisfy Eq. (8). Therefore, the graph of Eq. (8) will congives some values of r sist of the graphs of Eqs. (9) and (10).Table 1..3.4 and y satisfying Eq. (9), and Table 1.3.5 gives some values of x and y satisfying Eq. (10).A sketchof the graph of Eq. (8) is shown in Fig. 7.3.7.1 Figure .3.7

L.3.4 Table x 0L23-1-2-3-4-5

v

821r31,+o-+-1

1 T a b l e .3 .5

x

01,23-7-2-3 0L49149

v1.3.3 Definition

An equation of a graph is an equation which is satisfied by the coordinates of those, and only those, points on the graph. is an equation whose graph consists of 1: o ILLUSTRATIoN In R2, y:8 those points having an ordinate of 8. This is a line which is parallel to the o x axis, and 8 units above the r axis. In drawing a sketch of the graph of an equation, it is often helpful to consider properties of symmetry of a graph.

1.3.4 Definition

Two points P and Q are said to be symmetric with respect to a line if and only if the line is the perpendicular bisector of the line segment PQ. Two points P and Q are said to be symmetric with respectto a thirdpoint if and only if the third point is the midpoint of the line segment PQ. o rLLUsrRArroN 2: The points (3,2) and (3,-2) are symmetric with respect to the x axis, the points (3, 2) and (-3, 2) are symmetric with respect to the y axis, and the points (3,2) and (-3, -2) are symmetricwith o respect to the origin (see Fig. 1.3.8). In general, the points (x, y) and (x, -y) are symmetric with respect to the x axis, the points (r, y) and (-x, y) are symmetric with respect and the points (x,y) and (-r, -y) arc symmetric with respect to the y axLS, to the origin.

(3'2\o

F i g u r e1 . 3 . 8


1.3.5 Definition

are symmetric with respect to R. From Definition 1.3.5 it follows that if a point (x, y) is on a graph which is symmetric with respect to the x axis, then the point (x,-y) also must be on the graph. And, if both the points (x,y) and (x,-y) are on the graph, then the graph is symmetric with respect to the x axis. Therefore, the coordinates of the point (x,-y) as well as (x, y) must satisfy an equation of the graph. Hence, we may conclude that the graph of an equation in r and y is symmetric with respect to the r axis if and only if an equivalent equation is obtained when y is replaced by -y in the equation. We have thus proved part (i) in the following theorem. The proofs of parts (ii) and (iii) are similar. L.3.6 Theorem(Testst'or Symmetry)

The graph of an equation in r and y is (i) symmetric with respect to the x axis if and only if an equivalent equation is obtained when y is replaced by -y in the equation; (ii) symmetric with respect to the y axis if and only if an equivalent equation is obtained when x is replaced by -x in the equation; (iii) symmetric with respect to the origin if and only if an equivalent equation is obtained when r is replaced by -x and y is replaced by -y in the equation. The graph in Fig. 1.3.2 is symmetric with respect to the y axis, and for Eq. (1) an equivalent equation is obtained when x is replaced by -r. In Example 1 we have Eq. (2) for which an equivalent equation is obtained when y is replaced by-y, and its graph sketched inFig. L.3.3 is symmetric with respect to the r axis. The following example gives a graph which is symmetric with respect to the origin.

rxlrvrpr.r 5: Draw a sketch of the graph of the equation xy:1

solurroN: We see that if in Eq. (1,1) is replaced by -x and y is replaced x by -y,an equivalent equation is obtained; hence,by Theorem 1.3.6(iii) the graph is symmetric with respect to the origin. Table L.3.6 gives some (11) values of x and y satisfying Eq. (11). T a b l e .3.6 L

1.3 THE NUMBERPLANEAND GRAPHSOF EQUATIONS

'l.lx. We seethat as r increases through From Eq. (11) we obtain y : positive values and gets closer and through positive values, y decreases through through positive values,y increases closerto zero.As x decreases positive values and gets larger and larger. As I increasesthrough negative values (i.e.,r takeson the values-4, -3, -2, -t, -+, etc.),y takes on negative values having larger and larger absolute values. A sketch of the graph is shown in Fig. 1..3.9.

F i g u r e1 . 3 . 9

1.3 ExercisesIn Exercises 1 through 6, plot the dven Point P and such of the following PoinB as may aPPly: (a) The point O suitr ttrit ttre tnJ through Q and P is perpendicular to the r axis and is bisected by it. Give the coordinates of Q. (b) The poiniR such that the line through P and R is perpmdicular to and i6 bisected by the y axis. Give the coordinates of R. (c) The point s such that the line through P and 5 iE bisected by the origrn. Give the_coordin"F-"-9f s.by the als' line though the origin 1ai fn" poi"t T such that the line *Eough P and 7 is perpendicular to and is bisected coordinates of T. bisecting the first and thtd quadrants. Give the 1 . P ( 1, - 2 ) 4. P(-2, -2)

2. P(-2,2) 5 . P ( - 1, - 3 )8'Y:4x-3 11.y':x-3 'I.. y2 * 1 7 .y : l r l - 5 20. 4x29Y':36 L4. x: 23. 4x2- Az:0 2 6 .( 2 x + V - l ) ( 4 V + r 2 ) : 0

3. P(2,2) 5 . P ( 0 ,- 3 )

In Exercises7 through2S, draw a sketch of the graph of the equation.

7'Y:2x*5 l0.y:-nffi 1 3 .x : - J 15.y:-lx+21 1 9 .4 f * 9 Y 2 : 3 622. Y' : 41cs 25. xz*y',:O 2 8 . ( y 2- x * z ) ( V + t E -

Y: \81 y:5 y:lx-51 y--lxl+zA:4x3 3x2- L3xy- L0y2: g xa- 5x2yI 4y2: g

+):O 29. Draw a sketch of the graph of each of the following equations:

(a) V: lzx(a) Y: !-2x

(b\ v : -{2x(b\ y : -!-2x

(c) y' :2x (c\ y' : -2N

30. Draw a sketch of the graph of each of the following equations:


3L. Draw a sketch of the graph of each of the following equations:

( a )r * 3 y : 0

(b)r-3y:0

(c) r' - 9y' :0

32. (a) Write an equation whose gtaPh is the r axis. (b) Write an equation whose graph is the y axis. (c) Write an equafion whose graph is the set of all points on either the .x axis or the y axis. 33. (a) Write a,n__equ{iol whose graph consists of all points having an abscissa of 4. O) Write an equation whose graph consistE of all points having an ordinate of -3. 34. Prove that a graPh that i5 symmetric with respect to both coordinate axes is also symmetric with rcspect to the origin. 35' Pro-eethat a graPh that is symmeFic with rcspect to any two perpendicular lines is also sym.rretric with respect to their point of intersection-

1.4 DISTANCE FORMULA AND MIDPOINT FORMULA

If A is the point (x,.,y) and B is the point (xr,y) (i.e., A and B havethe same ordinate but different abscissas), then the directed distance from A to B, denoted by AB, is defined ds x2 - x1. o ILLUSTRATToN Refer to Fig. 1.4.1(a),(b), and (c). 1: u B(r,z) A(4,2)

AB:'/.,4 (b)

Figur1.4.1 e If .4 is the point (3, 4) and B is the point (9, 4), then AB :9 - 3: 6. rf A is the point (-8, 0) and B is the point (6,0), then B :6 - (-s) : 14. If A is the point (4, 2) and B is the point (L,2), then AB : 1.- 4: -3. we see that AB is positive if B is to the right of A, and AB is negative if B is to the left of A. oD(-2,4)

v

If C is the point (x,yr) and D is the point (x, yr), then the directed distance from C to D, denoted by CD, is defined as!z!r. o rLLUsrRArroN 2: Refer to Fig. La.2@) and (b). If C is the point (L, -2) and D is the point (1, -8), then CD : -g -J-2)--6. If C is the point (-2, -3) and D is the point (-2,4), then (-3) :7. Tl r.enumber C p i s posi ti vei f D i s above C , and CD C D :4 is negative if D is below C. o We consider a directed distance AB as the signed distance traveled by u particle that starts at A(rr, y) and travels to B(x2, !). ln such a case, the abscissa of the particle changes from 11 to x2, dfld we use the notation Ar ("delta x") to denote this change; that is, A,x: xz- xt Therefore, AB: Lx.

C(-2, -g) CD: -6(a)

CD:7 (b)

Figure 1.4.2

FORMULA AND MIDPOINT FORMULA 1.4 DISTANCE

Pr(x", vz)

It is important to note that the symbol Ar denotes the difference between the abscissa of B and the abscissa of A, and it doesnot mean "delta multiplied by x." Similarly, if we consider a particle moving along a line parallel to the y axis from a point C(x, yr) to a point D(x, yr), then the ordinate of the particle changes from Ar to Az.We denote this change by Ly ot

Ly: az- ar Thus, CD: Ly. Now let Pr(r1, yr) and Pr(xr, a) be any two points in the plane. we wish to obtain a formula for finding the nonrtegative distance between these two points. We shatl denote this distanceby lPtPrl. W" use absolute-value bars becausewe are concernedonly with the length, which is a nonnegative number, of the line segmentbetween the two points P1and Pr. To derive the formula, we note that lffil is the length of the hypotThis is illustrated in Fig. 1.4.3for P1and of a right triangle PLMP2. "rrrrr" of which are in the first quadrant. Pr, both Using the Pythagoreantheorem, we have

F i g u r e1 . 4 . 3

+llvl' EF"f: lArl2So

lF-nl:\MV+WPThat is,

lF:P,l:

(1)

Formula (1) holds for all possible positions of Pt and P, in all four qqadrants. The length of the hypotenuse will always be lFfrl, and the lengths of the two legs will always be [Ar[ and lAyl (seeExercisesL and 2). We state this result as a theorem. '1,.4.1 Theorem The undirected distance between the two points Pr(xr, Ar) and Pr(xr, yr) is given by

EXAMPLEl.: If a point P(x, U) is such that its distance from A(3,2) is always twice its distance from B(-4, L), find an equation which the coordinates of P must satisfy.

solurroN:

From the statement of the problem

lPTl:2lPBlUsing formula (1), we have

@:z@Squaring on both sides, we have x 2- 6 x + 9 + y ' - 4 y I 4 : 4(xz*8r * t6 + y2- 2y * 1)

or, equivalently, * 3 x 2 3 y ' * 3 8 r- 4 y + 5 5 : 0


ExAMPLE Show that the tri2: angle with verticesat A(-2, 4), B(-5, L), and C(-6,5) is isosceles.

SOLUTION:

The triangle is shown in Fig. L.4.4.

: lBel @ : \ 4 + 1 6 : f r : la-e; @ : \ t r 6 + 1 : { l T IEAI:W : \ / N : j f iTherefore,

A(-2,4)

Hence, triangle ABC is isosceles.

B(-5, i.)

Figure1.4.4

nxervrrrn3: Prove analytically that the lengths of the diagonals of a rectangleare equal.

C(0, b)

B(a,b)

SOLUTION: Draw a generalrectangle.Because can choosethe coordiwe nate axes anywhere in the plane, and becausethe choice of the position of the axes does not affect the truth of the theorem, we take the origin at one vertex, the x axis along one side, and the y axis along another side. This procedure simplifies the coordinates of the vertices on the two axes. Referto Fig. 1..4.5. Now the hypothesis and the conclusion of the theorem can be stated. Hypothesis: OABC is a rectanglewith diagonalsOB and AC. Conclusion: IO-B : lrel . |

(0,0) F i g u r e1 . 4 . 5

l o B:lf f i : r t + 6 2 : l A C lW : \ f r ] A tTherefore,

lml : ld-clLet P, (xr, yr) and P, (xr, yr) be the endpoints of a line segment. We shall denote this line segment by PtPr. This is not to be confused with the notation PrPr, which denotes the directed distance from P, to Pr. That is, PrP, denotes a number, whereas PrP2is a line segment. Let P(x, y) be the midpoint of the line segment P,Pr. Refer to Fig. L.4.6. In Fig. I.4.6 we see that triangles P'RP andPTPrare congruent. Therefore, l-PtRl: lPTl, and so r - xr: xz- x, giving us

FORMULA AND MIDPOINT 1.4 DISTANCE FORMULA

Similarly, lRPl :

Then A - Ar: Az- A, andtherefore lTPzl. (3)

Hence, the coordinates of the midpoint of a line segment are, respectively, the average of the abscissasand the average of the ordinates of the endpoints of the line segment.P z( x z ,v z )

P(x,y)

T(xr, Y)

u__ R(t, yt)

__JS(rr,yr)

1 F i g u re .4 .6 In the derivation of formulas (2) and (3) it was assumed that x, ) xt and y, ) yr.The same formulas are obtained by using any orderings of these numbers (see Exercises 3 and 4).

nxaurr,E 4: Prove analytically that the line segments ioining the midpoints of the opposite sides of any quadrilateral bisect each other.

Draw a general quadrilateral. Take the origin at one vertex solurroN: and the x axis along one side. This method simplifies the coordinates of the two vertices on the r axis. See Fig. L.4.7. Hypothesis: OABC is a quadrilateral. M is the midpoint of OA, N is the midpoint of CB, R is the midpoint of OC, and S is the midpoint of AB. Conclusion:MN and RS bisect each other. pRooF: To prove that two line segments bisect each other, we show that they have the same midpoint. Using formulas (2) and (3), we obtain the coordinates of M, N, R, and S. M is the point Ga, 0), N is the point ( + ( b + d ) , t r ( c * e ) ) ,R i s t h e p o i n t 1 ! d , t e ) , a n d S i s t h e p o i n t G @ + b ) , i c ) . + d). o The abscissa f the midpoint of MNis ilLa ++(b + d)l:*(a+b : ik + e). The ordinate of the midpoint of MN is +[0 + Lk + e)f Therefore, the midpoint of MN is the point (ifu + b + d), ik + e)). + d). The abscissa f the midpoint of RS is ilia+*@ + b)l :*(a+b o The ordinate of the midpoint of RS is tlie + *cf : Ik + e). Therefore, the midpoint of RS is the point (ifu + b + d), i(c + e)). Thus, the midpoint of MN is the same point as the midpoint of RS. I Therefore, MN and RS bisect each other.

B(b,c)

F i g u r e1 . 4 . 7

92


Exercises 7.41. Derive distance formula (1) if Pr is in the third quadrant and P, is in the second quadrant. Draw a figure. 2. Derive distance formula (1) if Pr is in the second quadrant and P, is in the fourth quadrant. Draw a figure. 3. Derive midpoint formulas (2) and (3) if Pr is in the first quadrant and p, is in the third quadrant. 4. Derive midpoint formulas (2) and (3) iI4 (rr, yr) and &(rr, yr) are both in the second quadrant and x, > xrand,y, > yr. 5. Find the length of the medians of the rriangle having vertices A(2,3),8(3,-g), and C(-1,-1). 5. Find the midpoints of the diagonals of the quadrilateral whose vertices are (0,0), (0,4), (3,5), and (3, 1). (.\Prove that the trianSle with vertices A(3,-6), B(8,-2), and C(-1, -1) is a right triangle. Find the areaof the triangle. \'-l(HrNr: Use the converseof the Pythagoreantheorem.) 8. Prove that the Points A(6, -lS) , B(-2, 2), C(13, 10), and D(21, -5) are the vertices of a square. Find the tength of a diagonal. 9. By using distance formula (1), prove that the points (-3, 2) , (1, -2), and (9, -10) lie on a line. 10. If one end of a line segment is the point (-4, 2) and the midpoint is (3, -L), find the coordinates of the other end ofthe line segment.

11. The abscissa of a Point iB -6, and its distance from the point (1, 3) is V-74. Find the ordinate of the point. 12. Determine whether or not the poifis (14,7) , (2, 2) , and,(-4, -1) lie on a line by using distance formula (1). /;:\ '\3.llf two vertices of an equilateral hiangle are (-4,3) and (0,0), find the third vertex. d Findan equation that must be satisfied by the coordinatesof any point that is equidistant from the two points (-3, 2) and (4,6). 15. Find an equation that must be satisfied by the coordinates of any point whose distance from the point (5, 3) is atways two units Feater than it8 distance frcm the point (-4, -2). 16. Giver the two Points d(-3,4) andB(2,5), find the coordinates of a point P on the line through A and B such thatp is (a) twice as far from 4 as frcm B, and (b) twice as far frcm B as from _A. 17. Find the coordinates of the three points that divide the tine segment ftom l(-5, 3) to 8(6, g) into four equal parts. 18. If rr. and r, are positiv_ intgg:ers,-prove that the coordinates of the point p(r, y), which divides the line segment plp, : in the ratio rr/r2-that i3, IP-,,PUlPrEI r,/rr-are given by ,: (t" - rt)x, + rrxufz"t2

ur4 , :

(r, - ,r)y, + ,rv"

In Exercises 19 through 23, use the formulas of Exercise 18 to find the coordinates oI point p. 19' The Point P is on the line segment between points Pr (1 ,3) ?flrd, P26,2) and is three times as far from P, as it b ftom pr. 20. The Point P is on the line segmentbetween points Pl(1 ,3) and Pr(6,2) and is three times as far from Pr as it is frcmpr. 21. The Point P i8 on the line through P' and P, and is three times as far ftom Pr(6, 2) as it is from P, (1,3) but is not between Pr and Pr. 22. The Point P is on the Une though P, and P, and is three times as far from P1(1, 3) as it is ftom Pr(6,2) but is not between P1and P2.

23. The point P is on the line through Pr(-3,5) and Pr(-L,2) so that ppr : + . p.,pr. 24. Find an equation whose graph is the circle that is the set of all points that are at a distance of 4 units from the point(1, 3).

OF 1.5 EQUATIONS A LINE

25. (a) Find an equation whose graph consist8 of all points equidistant frcm the Points (-1,2) and (3,4). (b) Draw a skekh of the gnph of the equation found in (a)' 26. prove analytically that the sum of the squares of the distanc$ of any Point from two opposite vertics of any rcctangle is equal to the sum of the squares of its distances from the other two verticeE' 22. prcve analytically that the line segment ioining the midpoints of two opposite sides of any quadrilateral and the line segment joining the midpoints of the diagonala of the quadrilateral bieect each other' 28. prove analytically that the midpoint of the hypotenuse of any right triangle is equidistant frcm each of the three vettices. 29. Prove analytically that if the lengths of two of the medians of a triangle arc equal, the triangle is isoeceles.

1..sEQUATIONS OF A LINE

Letl be a nonverticalline and Pr(x' U) andP"(xr, A)be any two distinct points on l. Figure L.5.1shows such a line. In the figure, R is the point Qc,Ur),and the points Pr, Pu and R are vertices of a right triangle; furthermore, P,.R: xz-xr and R-Pr: Az-Ar The number Az-Ar gives the measure of the change in the ordinate from Pr to Pr, and it may be positive, negative, or zero.The number x2- xl gives the measureof the change in the abscissafrom P, to Pr, and it may be positive or negative.The number x, - rr may not be zero becaus?x2 * r, since the line I is not vertical. For all choices of the points P, and P, on l, the quotient Az- Ur Xz- xr is constanq this quotient is called the "slope" of the line. Following is the formal definition.

R(rz,yJ

,F i g u r e1 . 5 . 1

1.S.1 Definition

If Pr(xr, Ar) and Pr(xr, yr) arc any two distinct points on line l, which is not parallel to the y axis, then the slopeof l, denoted by m, is given by

;:yE * Xz

(1)?Cr

-atP, (xr, y, R(xr,Yr)

In Eq. (L), x, * x1 since / is not parallel to the y axis. The value of m computed from Eq.(t) is independent of the choice of the two pointr_P, and P2 on L To show this, suppose we choose two different points, tr(n, yr) and Pr(ir,yr), and compute a number nr from Ee. (1). * - :_ A z - a r m -xz- xt We shall show tlnilt n - m. Refer to Fig. 1.5.2. Triangles PrRPz and P1RP2 are similar, so the lengths of corresponding sides are proportional. Therefore, V:Y_t Az Xr

Pr(7r,V,)

7z,Aix

-

ArXt

F i g u r e1 . 5 . 2

Xz-

Xz-

or m:m

34

REAL NUMBERS, INTRODUCTIONO ANALYTIC T GEOMETRY, AND FUNCTIONS

Hence, we conclude that the value of m computedfrom Eq. (1) is the same number no matter what two points on I are selected. In Fig. 1.5.2, x2 ) x1, Uz) Ar, x, > n, and y2 > fr. The discussion above is valid for any ordering of these pairs of numbers since Definition 1.5.1holds for any ordering. In sec. 1.4 we defined Ly:uz-ur and Ar:xz rr. substituting thesevalues into Eq. (L), we haveF i g u r e1 . 5 . 3

Multiplying on both sidesof this equationby Lx,we obtain Ly: mLx

e)

F i g u r e1 . 5 . 4

It is seen from Eq. (2) that if we consider a particle moving along line l, the change in the ordinate of the particle is proportional to the change in the abscissa, and the constant of proportionality is the slope of the line. If the slope of a line is positive, then as the abscissa of a point on the line increases, the ordinate increases. Such a line is shown in Fig. 1.5.3. In Fig. "1..5.4, have a line whose slope is negative. For this line, as the we abscissa of a point on the line increases, the ordinate decreases. Note that if the line is parallel to the x axis, then Uz: Ur and so m:0. If the line is parallel to the y axis, xz: xr, thus, Eq. (1) is meaningless because we cannot divide by zerc. This is the reason that lines paiallel to the y axis, or vertical lines, are excluded in Definition 1.5.L. We Jay that a vertical line does not have a slope. . rLLUSrRArroNl.: Let I be the line through the points P,,(2,3) and p2(4,7). The slope of I, by Definition "!..5.'!., given by is

*:7n!:zL

Referto Fig. 1.5.5. P(x,y) and Q@+ L*,y + Lil are any two points rf on I, then L A_ . , Ax-'Ay :2 A,x

Pr $ , 9 )

P:-(2,3)

P+(-1

F i g u r e1 . 5 . 5

Thus, if a particle is moving along the line l, the change in the ordinate is twice the change in the abscissa. That is, if the particle is at pr(4,7) and the abscissa is increased by one unit, then the ordinate is increased by two units, and the particle is at the point Pr(5,9). Similarly, if the particle is at Pt(2,3) and the abscissa is decreased by three units, then the ordinate is decreased by six units, and the particle is at Pr(-L, -3). o Since two points Pr(xr, yr) and Pr(xr, y) determine a unique line, we

1.5 EQUATIONS F A LINE O

should be able to obtain an equation of the line through these two points. Consider P(x,y) any point on the line. We want an equation that is satisfied by r and y if and only rf P(x, D is on the line through P1(rr, yr) and Pr(xr, Ar).We distinguish two cases. C a s e7 : x z : x r . In this case the line through P, and P, is parallel to the y axis, and all points on this line have the same abscissa. So P(x,,U) ts any point on the line if and only if X: Xr (3)

Equation (3) is an equation of a line parallel to the y axis. Note that this equation is independent of y; that is, the ordinate may have any value whatsoever, and the point P(x,y) is on the line whenever the abscissais rr. Case2: x2 * x1. The slope of the line through Pt and P, is given by Ar ry:Uzxz- xr lf P(x, g) is any point on the line except (\, given by At 1--A x- xr (4) !r), the slope is also

(s)

The point P will be on the line through P, and Pz if and only if the value of m fuomEq. (4) is the same as the value of m ftom Eq. (5), that is, if and only if U A t- A z - A rXXr Xz- Xr

Multiplying on both sides of this equation by (x - xr),we obtain(6) Equation (5) is satisfied by the coordinates of P, as well as by the coordinates of any other point on the line through P1 and Pr. Equation (6) is called tkte two-point fonn of an equation of the line. It gives an equation of the line if two points on the line are known. -3) o rLLUsrRArroN 2: An equation of the line through the two points (5, and (-2,3) is

y-(-3):*(r-5)

y+3--2@-6)3x*4Y:6

INTRODUCTION ANALYTIC REALNUMBERS, TO GEOMETRY, AND FUNCTIONS

If in Eq. (6) we replace(y, - !r)l(x, - x) by *, we getI i.:r )i' I.'

} i Yfl:gqt{,&Es'*"atu

t I

+ l- o :

';

...

e)

Equation (7) is called the point-slope form of an equation of the line. It gives an equation of the line if a point Pr(xr, a) on the line and the slope m of the line are known. . rLLUSrRArroN 3: An equation of the line through the point (-4,-5) and having a slope of 2 is

y - (-s) :Zlx - (-4)l2x-y*3:0 o If we choosethe particular point (0, b) (i.e., the point where the line intersectsthe y axis) for the point (h, A) in Eq. (7), we have y-b:m(x-0) or, equivalently,

+b

(8)

The number b, which is the ordinate of the point where the line intersects the y axis, is called the y intercept of the line. Consequently, Eq. (8) is called the slope-intercept form of an equation of the line. This is especially important becauseit enables us to find the slope of a _form line from its equation. It is also important becauseit expresses ttre y coordinate explicitly in terms of the r coordinate.rxevrpln L: Given the line having the equation 3r I 4y :7, find the slope of the line. sol.urroN: Solving the equation f.or !, we have

,:-fix*IComparing this equation with Therefore, the slope is -*. Eq. (8), we see that and b:*.

Another form of an equation of a line is the one involving the intercepts of a line. We define the r intercept of a line as the abscissa of the point at which the line intersects the r axis. The r intercept is denoted by o. If the r intercept a and the y intercept b arc given, we have two points (a, 0) and (0, b) on the line. Applying Eq. (6), the two-point form, we have

Y-o:6ikb x* a A : a b

h- 0 .

a)

-aY:bx-ab

OF 1.5 EOUATIONS A LINE

Dividing

by ab, if. a * 0 and b + 0, we obtain

(e)Equation (9) is called the intercept fonn of an equation of the line. Obviously it does not apply to a line through the origin, becausefor such a line both a and b arc zero. nxlupln 2: The point (2,3) bisects that portion of a line which is cut off by the coordinate axes.Find an equation of the line.Refer to Fig. 1.5.6. lf. a is the r intercept of the line and b is the y intercept of the line, then the point (2, 3) is the midPoint of the line segment joining (a, 0) and (0, b). By the midpoint formulas/ we have soLUrIoN:A atl

t:ry andand b:6

A:4

The intercept form, Eq. (9), gives us x +!-t 4'6 or, equivalently,F i g u r e1 . 5 . 6

3x*2y-12 t.5.2 Theorem The graph of the equation Ax*By*C:0 (10)where A, B, and C are constants and where not both A and B are zero, is a straight line. PRooF: Consider the two cases B + 0 and B : 0. CaseL: B+0. Because B + 0, we divide on both sides of Eq. (10) by B and obtain

y--i*-,

AC

(11)

Equation (11) is an equation of a straight line becauseit is in the slope-- -CIB. intercept form, where m: -AlB and b Case : B :0. 2 B:0, Because Ax*C:0 we may concludethat A * 0 and thus have


or/ equivalently, C *:- a (12)

Equation (12) is in the form of Eq. (3), and so the graph is a straight line parallel to the y axis. This completes the proof. r Because the graph of Eq. (10) is a straight line, it is called a linear equation.Equation (10) is the general equation of the first degree in x andy. Because two points determine a line, to draw a sketch of the graph of a straight line from its equation we need only determine the cooidit utet of two points on the line, plot the two points, and then draw the line. Aty two points will suffice, but it is usually convenient to plot the two points where the line intersects the two axes (which are given by the intercepts). rxeupm 3: Given line lr, having the equation 2x - 3y : 12, and line Ir, having the equation 4x * 3y:6, draw a sketch of each of the lines. Then find the coordinates of the point of intersection of l, and Ir. solurroN: To draw a sketch of the graph of lr, we find the intercepts a and b. In the equation of Ir, we substitute 0 for r and get b - -4.In the equation of Ir, we substitute 0 for y and get a: 6. Similarly, we obtain the intercepts a and b for lr, and for l, we have a:8 and,b: 2. The two lines are plotted in Fig. 1..5.7. To find the coordinates of the point of intersection of /, and Ir, we solve the two equations simultaneously. Because the point must lie on both lines, its coordinates must satisfy both equations. If both equations are put in the slope-intercept form, we have

a:?x-4

and

-tx*2

Eliminating y gives 3x-4:-gx*2 2x-L2:-4x#6 x:3 SoF i g u r e1 . 5 . 7

y:BG)_4Therefore, the point of intersection is (3, -2). 1'5'3 Theorem If /r and l, are two distinct nonvertical lines having slopes m, and mr, respectively, then /, and l, are parallel if and only if mr: //t2. PRooF: Let an equation of /, be : tftrx * br, and let an equation of l, A be y : rnzx * br. Because there is an "if and only if " qualiiication, the proof consists of two parts. panr 1: Prove that l, and I, are parallel if ftir: tnz.

OF 1.5E QU A TION S A LIN E Assume that I, and I, are not parallel. Let us show that this assumPtion leads to a contradiction. If I, and I, are not parallel, then they intersect. Catl this point of intersection P(*o, /o).The coordinates of P must satisfy the equations of 11and Ir, and so we have ls: But mr: ls: rnlxs* b1 and rrt2r which gives mfis * b1 and !o: nttfio* b2 !o: fti2xo b2 *

b2,both from which it follows thatbl: bz.Thus, becauserrtr: mrandbl: lines 11and /, have the same equation , A : mrx * br, and so the lines are the same. But this contradicts the hypothesis that l, and I, are distinct lines. Therefore, our assumption is false. So we conclude that /r and 12 are parallel. pARr 2: Prove that l, and I, are parallel only if mt: lftz. Here we must show that if Ir and l, are parallel, then rflt: rtrz. Assume that m, * mr. Solving the equations for \ and I, simultaneously, we get, upon eliminating V, m t x * b r : m 2 x* b 2 from which it follows that ( m , - m r ) x : b z- b t Because we have assumed that m, * *r, this givesL _

- - _ _b r - b ,t7\ffiz

Hence, 11and l, have a point of intersection, which contradicts the hypothesis that l, and 12are parallel. So our assumption is false, and thereI fore mr: nflz. In Fig. 1.5.8, the two lines Ir and l, are perpendicular. We state and prove the following theorem on the slopes of two perpendicular lines. 1.5.4 Theorem If neither line /, nor line l, is vertical, then Ir and 12are perpendicular if and only if the product of their slopes is -1.. That is, if m, is the slope of 11and m, is the slope of. Ir, then Ir and I, are perpendicular if and only if n tfl tz - -1 .PROOF:

- -1' Panr 1: Prove that l, and 12are perPendicular only if mtm2 Let L, be the line through the origin parallel to l, and let L, be the line through the origin parallel to Ir. See Fig. 1.5.8. Therefore,by Theorem 1..5.3,the slope of line L, ts m, and the slope of line L, is m2.Because neither

REAL NUMBERS, INTRODUCTION ANALYTICGEOMETRY, TO AND FUNCTIONS

L, nor L, rs vertical, these two lines intersect the line x: ! at points p, and Pr, respectively. The abscissa both P, and P, is L. Let / be the ordiof nate of Pt. Since Lr contains the points (0,0) and (1., and its slopeis mr, y) we have from Definition 1.5.1ttL: a-0 i=

and so y = mr.Similarly, the ordinate of P, is shown to be mr. Applying the Pythagorean theorem to right triangle prop2, we get

l o t r ,+z @ f : l P , k l , ;lF i g u r e1 . 5 . 8

(1 3 )

By applyingthe distance formula,we obtain l o - 4 l r : ( 1- 0 ) ,f - ( m r - 0 ) ,: 7 * m ] l O t r l r : ( 1- o ) , * ( m r - o ) , : r * m z 2l F J r P : ( 1 - 1 ) , * ( m r - m r ) ,- n t r z 2 m r m r t m r , Substituting into Eq. (13)gives (l + mrz)+ (1 + m22) ftizz 2mrm,I.mrt2: -2mrffiz tltinz - -1 penr 2: Prove that I, and l, are perpendicular if.mrm" - -1. Starting with lt|fitz - -'1, we can reverse the steps of the proof of Part 1 in order to obtain

lO-&l'+ l1trl': lPEl'zfrom which it follows, from the converse of the Pythagorean theorem, that Lt andL, are pe{pendicular; hence, /, and l, are perpendicular. I Theorem I.5.4 states that if two lines are perpendicular and neither one is vertical, the slope of one of the lines is the negative reciprocal of the slope of the other line. o rLLUsrRArroN 4: If line /t is perpendicular to line I, and the slope of /, is 3, then the slope of /, must be -9. o

sxarvrpr,n Prove by meahs of 4: slopes that the four points A(5,2), B(9,6), C(4,g), and DQ,4) are the verticesof a rectangle.

solurroN:

See Fig. 1.5.9. Let

6-2 ntr: the slope of AB - 8 - 6 8-6 m, J the slope of BC -

4- I

1.5 EQUATIONS A LINE OF

41

C(4,8)

/fi.:the sloPe DC:ft: ofB(8,6)

''

rn.:the slope AD:=:-+. of.Because

D

Tnr: mr, AB ll DC. /n2: ma,BC ll AD. ntrttrz--'1, AB L BC.

F i g u r e1 . 5 . 9

Therefore, quadrilateral ABCD has opposite sides parallel and two adjacent sides perpendicular, and we conclude that ABCD is a rectangle.

5: Given the line / EXAMPLE having the equation 2x*3Y-5:0

Because the required line is perpendicular to line l, its slope solurroN: must be the negative reciprocal of the slope of l. We find the slope of lby putting its equation into the slope-intercept form. Solving the given' equation for y, we obtain

find an equation of the line perpendicular to line I and passing through the point

y--tx**Therefore, the slope of I is -3, and the slope of the required line is 9. gecausewe also know that the required line contains the point (-L, 3), we use the point-slope form, which gives y-3:$(x*t) 2y-6:3x*3 3x-2Y+9:0

Aet,3).

1-.5 ExercisesIn Exercises L through 4, find the slope of the line through the given points.

1 . ( 2 ,- 3 ) , ( - 4 , 3 )

2 . ( 5 , 2 ), ( - 2 , - 3 ) 4. (-2.T,0.3) (2.3,1.4) ,

3.(+, eE, +), &)5. The slope is 4 and through the point (2,-3). 6. Through the two points (3, L) and (-5,4). 7. Through the point (-3, -4) and parallel to the y axis. 8. Through the point (I,-7) and parallelto the x axis. 9. The x intercept is -3, and the y intercept is 4.

In Exercises5 through14, find an equation of the line satisfying the given conditions.

INTRODUCTION ANALYTIC TO REAL NUMBERS, GEOMETRY. AND FUNCTIONS

10. Through (1, 4) and parallel to the line whose equation is 2r - 5y + 7:0. 11. Through (-2, -5) and having a slope of l/5. 12. Through the origin and bisecting the angle between the axes in the first and third quadranb, 13. Through the origin and bisecting the angle between the axes in the second and fou h quadrants. 14. The slope is -2, and the r intercept is 4. the points (1, 3) and (2, -2), and put the equation in the intercept form. equation of the line through the points (3, -5) and (1, -2), and put the equation in the slope-interceptform. dg. Fitrd "tr 17. Show by means of slopes that the points (-4, -1) , (3, +) , (8, -4) , and (2, -9) are the vertices of a trapezoid. 15. Find an equation of the line thouth 18. Thrce consecutivevertices of a parallelogramare (-4, 1), (2,3),and (8,9). Find the coordinatesof the fou h vertex. 19. For each of the following sets of three point6, detemine by means of slopes if the points are on a line: (a) (2, 3), (-4,-7), (s,8);(b) (-3,6), (3,2), (e,-2); (cl (2,-7), (7,7), (3,t); and (d) (1,6), (1,2), (-s,-4). 20. Prove by means of slopes that the three points A(3, 1), B(6,0), and C(4, ) are the vertices of a right triangle, and find the area of the triangle. /\2T. Given the line I having the equation 2y - 3r : 4 and the point P(1, -3), find (a) an equation of the line through P " and perpendicul to l/(b) the shortest distance from P to line l. \ 22. lf A,B, C, and D arc cdhstarts, show that (a) the lines Ax + By+ C:0 a dAx+By+D:0 are parallel and (b) the ljmes Ar+ By + C:0 and Bx- Ay * D:0 areperpendicular. 23. Given the line l, having the equation Ar + By + C: O, B + 0, find (a) the slope, (b) the y intercept, (c) the x intercept, (d) an equation of the line through the origin perpendicular to L 24. Find an equation of the line which has equal intercepts and which passesthrouth the point (8, -6). 25. Find equations o{ the three medians of the triangle having vertices A(3, -2) , BG, a), and C(-l,1), they met in a point. and prove that

26. Find equations of the perpendicular bisectors of the sides of the triande having vertices A(-l , -3) , B (5, -3), and C(5,5), and prove that they meet in a point. 27. Find an equation of each of the lines through the point (3,2), which forms with the coordinateaxesa triangle of area12. 28. Let il be the line having the equation.r4'r Bly+ C1:0,andlet I'be the line having the equationArx+ 82! + Ca:0. * If ll is not parallel to L and iJ & is any constant,the equation A'x 'l Bry I C' * k(A"x * Bry * C") :0 represents an unlimited number of lines. Prove that eadr of these lines contains the point of intersection df lr and t. 29. Given an equation ot It is 2x 'f 3y - 5: 0 and an equation of l, is 3x + 5y - 8 : 0, by using Exercise 28 and without finding the coordinates of the point of intersection of ,r and lr, find an equation of the line through this point of intersection and (a) passing through the point (1, 3); (b) parallel to the r axis; (c) parallel to the y axis; (d) having slope-2; (e) perpendicular to the line having the eqtatio\ 2x + y :7; (f) forming an isoscelestriangle with the coordinate axes. 30. Find an equation of each straight line that is perpendicular to the line having the equation 5r - y: I and that forms with the coordinate axes a triantle having an area of measure 5. 31. Prove analytically that the diatonals of a rhombus are perpendicular. 32. Prove analytically that the line segments joining conEecutivemidpoints of the 6ide8 of any quadrilateral form a parallelogram.

1.6THECIBCLE 43 33. Prove analytically that the diagonals of a parallelogram bisect each other. 34. Prove analytically that if the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram.

1.5 THE CIRCLE The simplest curve that is the graph of a quadratic equation in two variables is the "citcle," which we now define.

'1,.6.1 Definition

A circle is the set of all points in a plane equidistant from a fixed point. The fixed point is called the center of the circle, and the measure of the constant equal distance is called the radius of the circle. The circle with center at the point C(h, k) and radius r has as an equation ;,(x *'h}P, * (g1 k)2,* 7zi'1 = pRooF: The point P(x, y) lies on the circle if and only if'; l -,: i :

1.6.2 Theorem

(1)

l n c l :,that is, if and only if

@:rThis is true if and only if (x-h)'+ (y-k)z--tz

which is Eq. (1). Equation (1) is satisfied by the coordinates of those and only those points which lie on the given circle. Hence, (1) is an equation I of the circle. From Definition L.3.2, it follows that the graph of Eq. (1) is the circle with center at (h, k) and radius r. If the center of the circle is at the origin, then h: k: 0; therefore, its equation is x' + y' : 12. o rLLUsrRArroN 1: Figure 1.6.1 shows the circle with center at (2, -3) and radius equal to 4. For this circle h:2,k:-3, and r:4. We obtain an equation of the circle by substituting these values into Eq. (1) and we obtain (x-2)'+ly - (-3)12:42

' :-LC(2, - 3)J-'

(x-2)'+ (y+3)':t6Squaring and then combining terms, we have x2-4x+4+y'*5y*9:t6

F i g u r e1 . 6 . 1

)c'+y'-4x+6y-3:0

REALNUMBERS, INTRODUCTION ANALYTIC TO GEOMETRY, AND FUNCTIONS

EXAMPLE 1:

Given the equation

solurroN:

The given equation may be written as

x'+ y'* 6x-2y - 15: oprove that the graph of this equation is a circle, and find its center and radius.

(xr+6x)t(yr-2y):tsCompleting the squares of the terms in parentheses by adding 9 and 1 on both sides of the equation, we have

(x' * 6x*9) + (y' - 2y I1) : 15+ 9 + 1, (r*3)2+(V-1)':25Comparing this equation with Eq. (1), we see that this is an equation of a circle with its center at (-3, I) and a radius of 5.

In Eq. (L), removing parentheses and combining terms gives x ' + y ' - Z h x- Z k y+ ( h , + k 2- r r ) : 0 Equation (2) is of the form (3) : -2h, E : -2k, and F : h2* k2- f. where D Equation (3) is called the general form of an equation of a circle, whereas Eq. (1) is called the center-rsdius form. Becauseevery circle has a center and a radius, its equation can be put in the center-radius form, and hence into the general form. We now consider the question of whether or not the graph of every equation of the form x'+y,*Dx*Ey*F:0 is a circle. To determine this, we shall attempt to write this equation in the center-iadius form. We rewrite the equation as x'+y,*Dx*Ey--F and completethe squaresof the terms in parentheses adding]nDzand by iE' on both sides, thus giving us )e+y'*Dx+Ey*F:0 Q)

--F + +Dz iE, (x' * Dxt +D')+ (y' * Ey++E'z) *or, guivalently, (x + LD)2 (y + +E)2: i(D' + Ez- 4F) +Equation (a) is in the form of Eq. (1) if and only if ( 4)

+(D,+E2-4F):rz We now consider three cases,namely, (D'+ E2- 4F) as positive, zero, and negative. Case 7: (D, + E2- 4F) > 0. Then r': *(D' + E2- 4F), and so Eq. (a) is an equation of a circle having a radius equal to it/FE= 4F and its center at (-*D,-+E).

1.6 THE CIRCLE

2 C a s e : D z + E z- 4 F : 0 . Equation ( ) is then of the form (x + tD)z + (Y * iE)z :0

(5)

the only real values of r and y satisfying Eq. (5) are v -L2D and Because -+E). Comparing Eq. (5) with y:-+E, the graph is the point (-+D, and r - 0. Thus, this point can be Eq. (1),we see that h:-+D, k:-iE, called a point-circle. 3: (Dt + E2- 4F) < 0. Case Then Eq. (4) has a negative number on the right side and the sum of the squaresof two real numbers on the left side. There are no real values of r and y that satisfy such an equation; consequently,we say the graph is the empty set. Before stating the results of these three casesas a theorem, we observe that an equation of the form Axz*Ay'+Dx*Ey* F:0 w h e r eA + 0(6)

can be written in the form of Eq. (3) by dividing by A, thereby obtaining D E F r c ' + y ' + t r *, * = A,U + i : 0Equation (6) is a special case of the general equation of the second degree: Axz * Bxy * Cy' t Dx * Ey * F : 0 in which the coefficients of x2 and yz are equal and which has no xy term. We have, then, the following theorem. 1.6.3 Theorem The graph of any second-degree equation in R2 in r and U, in which the coefficients of x2 and y2 are equal and in which there is no xy term, is either a circle, a point-circle, or the empty set. o rLLUSrRArroN 2: The equation 2x2*2y'*LZx-By+31:0 is of the form (6), and therefore its graph is either a circle, a point-circle, or the empty set. If the equation is put in the form of Eq. (1), we have

x'+y't6x-4y-F#:o ( x ' + 6 x )+ ( y ' - 4 y ) : - T ( x ' * 6 x* 9 ) + ( y ' - 4 y + 4 ) : - # + (r*3)2 + (Y-2)':-E the Therefore, graphis the empty set. 9+ 4


ExAMPLE2: Find an equation of the circle through the three points A(4,5), B(3, -2), and

soLUrIoN:

The general form of an equation of the circle is

x'+y,tDx-fEy*F:0 Because threepoints A, B,and C must lie on the circle,the coordithenates of these points must satisfy the equation. So we have

-4). c(1,,

L6+25+4D+5E+F:0 9+ 4+3D-2E*F:0 1+16+ D-4E*F:0 or, guivalently, 4D+ 5E+F:-47 3D-2E+F:-L3 D - 4 E* F : - 1 7Solving these three equations simultaneously, we get D:7 E:-5 F:-44

Thus, an equation of the circle is

x'+y'*7x-5y-44:0

In the following example we have a line which is tangent to a circle. The definition of the tangent line to a general curve at a specific point is given in Sec. 3.L. However, for a circle we use the definition from plane geometry which states that a tangent line at a point P on the circle is the line intersecting the circle at only the point P.

nxeuprr 3: Find an equation of the circle with its center at the point C(1,6) and tangent to the line / having the equation x-v-1:0.

soLUrIoN: See Fig. L.6.2. Given that h: L and k: 6, if we find r, we can obtain an equation of the circle by using the center-radius form. Let I, be the line through C and the point P, which is the point of tangency of line I with the circle.

r: lPClHence, we must find the coordinates of P. We do this by finding an equation of /, and then finding the point of intersection of /1 with /. Since 11 is along a diameter of the circle, and / is tangent to the circle , I, is perpendicular to l. Because the slope of / is L, the slope of 11is -L. Therefore, using the point-slope form of an equation of a line, we obtain as an equation of I,

y-5--1(r-1)

1,6THE CIRCLE

or, equivalently,

x+Y-7:0Solving this equation simultaneouslywith the given equation of l, namely, X- A- L:0 we get x: 4 and y: 3. Thus, P is the point (4, 3). Therefore,

,:l-PCl:ffior, equivalently,

r: !18So, an equation of the circle is

(x-l)'+F i g u r e1 . 6 . 2

(Y-6)':

(\ft)'

or, equivalently, x' + y'- 2x- 12y L9:0 *

1.6 /- )ExercisesIn Exercises 1 through 4, find an equation of the circle with center at C and radius r. Write the equation in both the centerradius forrr and the gmeral form. ' , . c ( Q ,0 ) , r : 8 r . c 1 4 , - 3 ) ,r : S3. C(-5, -12), f :3

4. C(-l,l),r:2

In Exercises5 through 10, find an equation of the circle satisfying the given condition$. 5. Center is at (1, 2\ and through the point (3, -1). 5. Center is at (-2,5) and tangent to the line x:7. is at (-3, -5) and tangent to the line l}x t 5y - 4:0. /Z) C"n w "r 8. Through the three points (2, 8) , (7, 3), and (-2,0). 9. Tangent to the line 3x * y * 2:0 at (-L, 1) and through the point (3, 5). 10. Tangent to the line 3x * 4y - 16:0 at (4,1.) and with a radius of 5. (Two possiblecircles.) In ExercisesL1 through 14, find the center and radius of each circle, and draw a sketch of the graph. ll. f*y'-6x-8y*9:0 13. 3f * 3y' * 4y - 7 :0 12.2f*2y'-2x*2y*7:0 1 4 .* * y ' 10r-lOy+25:0

In ExercisesL5 through 20, determine whether the graph is a circle, a point-circle, or the empty set. 16. 4* l- 4y' * 24x - 4y t L : 0 * 1 5. 12* y'- 2x * LO y L9:0 1 7 .f * y ' 10r* 6y*36:0 1 8 .x 2 * y ' * 2 x - 4 y *5:0

48

REALNUMBERS, INTRODUCTION ANALYTIC TO GEOMETRY, FUNCTIONS AND

19. *+ f -{f.x+36v-719:0 20. W 1 9 f I 6 x - 6 y + 5 - 0 21. Find an equation of the comrnon chord_of the two cirdes f*- 6y - 12:0 and f* 2yrg:o. 3F+rlt l+gr(rrrrvr: If the coordinates of a Point gatisfy two different equations, then thjcoordinates also satisfy the difierence of the two equatione.)

22. Find the points of intersection of the two cirdes in Exercise 21.Find an equation of the line which is tangent to the circle .d + y, - 4t + 6y - 12: O at the point (5, 1). 24. Findanequationofeachofthetwolineshavingslope-jgwhicharetangenttothecirclef*l*2x-W-g:0.I i\

25. From the origin, dbrda of the cirde t' * I * 4r: 0 are drawn. Prve that the set of midpoinb of these chords is a circle. 26. Prove analytically that a line from th center of any circle biEecting any drord is perpendicular to the chord. 27. Prcve analytically that an angle inscribed in a semicircle is a right angle. 28. Given the line y: flr + b tangent to the circle rP * y, = r2, find an equation involving m, b, and t

1.7 FUNCTIONS AND We intuitively consider y to be a function of r if there is some rule by THEIR GRAPHS which a unique value is assignedto y by a correspondingvalue of x. Familiar examples of such relationships are given by equations such as y:2f+5 and (1)(2)

Y:tP-g

It is not necessary that r and y be related by an equation in order for a functional relationship to exist between them, as shown in the following illustration. o ILLUsTRATIoN lf y is the number of cents in the postage of a domestic 1,: first class letter, and if r is the number of ounces in the weight of the letter, then y is a function of x. For this functional relationship, there is no equation involving r and /; however, the relationship between r and y may be given by means of a table, such as Table 1.7.L. .T a b l e1 . 7 . 1

r: number of ounces in the weight of the letter y: number of cents in first classpostageThe formal definition 1.7.1 Definition

x 0. That is, x>0 and x>2

solurroN: Because\RFT is not a rcal number when x(x - 2) < 0, the d-omainof g consistsof the values of x for which x(x - 2) > 0. This in_two cases holds:

Both inequalities hold if x > 2, which is the interval [2, +*;. Case2: x 0-

2 ss : f f i - z u f , f .{2?. s: $Ssrz 2trt2,f > 0 +

>o

2 6 .s : g t t + 2 \ / 2 t + l , t > 0 (x'

ls I n E xe rci ses 28t hr ough31 .,fi n d fo rm u l ao rf' (r) a n d f" (x ) andstatethedomai nsof f' and.f"

x zB. IFTt.r # o f(x): i f x : 0 L03 0 .f ( x ) : l r l '

ze. : ff f(x)31.

lf; : 3

32. For the function of Exercise30 find f "' (x) when it exists. 31 33. For the function of Exercise find f "'(x) when it exists. 34. Show that if xA:1, then D,zy ' Dozx: 4. expressh"(r) in terms of the derivatives offand g. 35. fi f', g', f", and8" exist and if tt:/"9, 36. If / and g are two functions such that their first and secondderivatives exist and if h is the function definedby h(x): ' f(x) sG) ' Prove that

n" G\ : f(x) . g" (x) + 2f'(x) ' g'(x) + f" (x) ' g(x)

37. lI V = t, 38. If-

whete z is any positive integer, Prove by mathematical induction that D'?: 1,L2X

n!

prcve by mathematical induction thatD nt : t,rl -------:-i::-

(l - 2x't"*l

(Chapter3) Reaiew ExercisesIn Exercises1.through 14, find D"y.+2

l.y:fra

2.5.

3 . 4 x z* 4 y ' - U 3 : 0 6.

4.y:\GV+\Fx

v:

x2 (x*2)2(4r-5)

162

THE DERIVATIVEt4

7.y:{tr10. XztB* yzts: g2t3

8.*y'*2y": x-2UlL. y: (x'l)stz(xz 4)rrz

13.Y:f*

lrt* (xn*x)'ft

xffi t4-y: -7*.,

15. A parttcle is movint in a straight line according to the equation of motion s: r3- llP + 24t + 100 where s ft i8 the directd distance of the particle fiom the starting point at , sec. (a) The partide is at the starting poht when t: 0. For what other values of t is the particle at the starting point? (b) Determine the velocity of the partide at each instant that it is at the startint point, and interpret the sign of the velocity in each case. + 4.C - r that have the slope l. 17. Find equations of the tangent line and normal line to the curve 2/ * 2y3 - 9xy : g at the point (2, 1). 16. Find equations of the tangent lines to the curve y:2f 18. An obiect is sliding down an hclined plane according to the equation of motion s : 121,* 5t wherc s ft is the dirccted distance of the obiect ftom the top I sec after stafting. (a) Find the velocity 3 sec after the start. (b) Find the initial velocity. 19. Using only the definition of a derivative find l'(r) it f (x') : \/-4x - 3. 20. Using only the definition of a derivative find /'(5) it f(x) : 19iiT1. 2 7 - F i n df ' ( x ) i f l ( r ) : ( l r + 1 1 - l r l ) , . , 2 2 . F i n df , ( x \ i I f ( x ) : x - [ x \ . : lrl3. 1a)Draw a sketch of the graph of (b) Find lirn 23. Given /(r) l(r) if ir existe. (c) Find l,(0) if it exists. f 24. Given l(r) : t' sgn x. (a) Discuss the differentiability of. f. (b) Is f' continuous on its domain?

25.Findl'(-3) if /(r): [r + ]lV9r.27- Ptove tha! the line tantent to the curve y:-y' point, and find this point

26.Find t,(-3) iff(r):(lrl

- x)fl-sx.

+ Zf + r at the point (1, 2) is also tangent to the curve at another

at the point 13, l). fETl 29. A ball is thrown vertically uPward from the top of a houee 112 ft high. Its equation of motion is s : -15f + l)5t wherc s ft is the dirccted distance of the ball from the starting point at t sec. Find: (a) the in8tantaneous velocity of the batl at 2_ sec; O) how high the ball will go; (c) how long it takes for the ball to reach the ground; (d) the instantaneous velocity of the ball when it reaches the ground. 30. Prove that the tangent lines to the curves 4Vs- t1y - x + sy:g 31. Givn (qf+b((t\ :1 1

28. Find an equation of the normal line to the eurve r - y:

and,/ - 4y3+'Sxt y:0

at the origin are perpmdicular.

if r =l

Ei32. Suppose IVJ:

ifx>1exists.

Find the values of c and, 8o that/'(1) (f lot"+bt +,

if x 0, according to Theorem 4.3.1 there is an open interval containing 3 such that /(r) ) 0 for every x * 3 in the interval. such an interval is (2, 4). Actually, any open interval (a,b)for which i = a< 3 and b > gwill do. o pRooFoF rHEoRa''4.3.7: Let L: tjT f@).By hypothesis,L > 0. Apply_ ing Definition z.r.r and taking e:!L, we know thereis a 6 > 0 such that

lf (*) - Ll < +f whenever 0 < l* - ol < a (1) Also, lf (x)- LI < lL is equivalent -*L < f (x)- L < tf (refer to Theotorcm'/-..2.2),which in turn is equivalent to

Figure4,3.1

tt 0 by hypothesis. Theretore f(x) - f(rr) > 0, and so f(xr) > f(xr). We have shown, then, that f(xr) < f(h) whenever x, < xr, where x., and r, are any numbers in the interval- [4, b]. Therefore, by Definition 5.1.1, it follows ti.rat is increasing / on fa, bl. The proof of part (ii) is similar and is left as an exercise(see Exer_ cise 34). I An immediate application of Theorem 5.1.1,is in the Droof of what is known as the first-deriaatioetestfor relafiaeextremaof a funcuon. 5.1'4 Theorem Let the function f be continuous at all points of the opm interval (a, b) Fitst-Derioatioelor Relatioe Test Extrena containing the number c, and suppose that l' exists at'aii points of (a, b) except possibly at c: (1) if f' (x) > 0 for all values of r in some open intewal having c as its right endpoint, and if f'(x) < O for all values of r in iome open interval having c as its left endpoint, then f has a relative maximum value at c;

TEST AND FUNCTIONS THEFIRST-DERIVATIVE AND 5 . 1 INCREASING DECREASING

207

(ii) if f '(x) < 0 for all values of r in some oPen interval having c as its right endpoint, and if f ' (x) ) 0 for all values of x in some open interval having c as its left endpoint, then /"has a relative minimum value at c. pRooF or (i): Let (d, c) be the interval having c as its right endpoint for which /'(r) > 0 for all r in the interval. It follows from Theorem 5.1.3(i) that / is increasing on ld, c). Let (c, e) be the interval having c as '(x) < 0 for all x in the interval. By Theorem its left endpoint for which f 5.1.3(ii) / is decreasing on lc, ef. Because/is increasing onld, c], it follows from DefinitionS.l..L that if x, is tnld, c] and xr* c, then f(xt) < f (c) .Also, because / is decreasing on fc, ef , it follows from Definition1.'1..2 that if x, is in [c, e] and xr* c, then f(c) > f(rr). Therefore, from Definition 4.5.L, / has a relative maximum value at c. The proof of part (ii) is similar to the proof of part (i), and it is left I as an exercise (see Exercise 35). The first-derivative test for relative extrema essentially states that if '(x) changes algebraic sign from positive to is continuous at c and f / negative as r increases through the number c, then / has a relative maxi'(x) changes algebraic sign from negative to posimum value at c; and tf f tive as r increases through c, then / has a relative minimum value at c. Figures 5.1.2 and 5.1.3 illustrate parts (i) and (ii), respectively, of Theorem 5.1,.4when /'(c) exists. Figure 5.1,.4shows a sketch of the graph '(c) of a function / that has a relative maximum value at a numbet c,but f d o e s n o te x i s t; how ever,f' (x) ) 0w hen x l c,andf' (r) < 0w hen x ) c . In Fig. 5.1.5, we show a sketch of the graph of a function / for which c is (r) < 0 or, equivalently, 0, and it follows from (1) that lf f < f'(r). f'(x) '(r) > 0, and if But because '(c) : 0, we concludethat if x is in l' , f f '(r) < 0. Therefore, '(r) changesalgebraicsign from positive r is in 1" , f f through c, and so, by Theorem5.L.4,/hasa relato negativeas r increases tive maximum value at c. The proof of part (ii) is similar and is left as an exercise (see ExerI cise L8).

212

ADDITIONAL APPLICATIONS THEDERIVATIVE OF Given

EXAMPLE

SOLUTION:

f(x):xa*$x3-4x2 find the relative maxima and the relative minima of f by applying the second-derivativetest.

f'(x) :4xs * 4xz 8x f"(x):12x2*8r-8 Setting '(x):0, we have f

4 x ( x+ 2 )( x - 1 )which givesff:0

x:L

Thus, the critical numbers of f are-2,0, and L. We determine whether or not there is a relative extremum at any of these critical numbers by finding the sign of the second derivative there. The results are summ arized. in Table 5.2.1. T a b l e .2 .1 5

f(x)

f'(x)0 0 0

f" (*)T

Conclusion t' has a relative minimum value

-+0_5 3

/ has a relative maximum value / has a relative minimum value

x

rf f " (c) : 0, as well as f '(r) :0, nothing can be concluded regarding a relative extremum of f at c. The following three illustrations lustify thii statement. .IL L U ST R A TToN If f(x)- x4, then f ' (x):4x3 and ,,(x):' !,2x2. Thus, 7: f (0), f '(0), and f ''(0) all have the value zero. By applying the first-derivative f test we see that / has a relative minimum value at 0. A sketch of the graph of / is shown in Fig. 5.2.1. o o IL L U S TR A TION If 8(x):-xn, 2: then g' (r) :-4x3 and g,,(r): - 12x2. Hence, g(0) : g'(0) : g"(0) : 0. In this case,g has a rerative maximum value at 0, as can be seen by applying the first-derivative test. A sketch of the graph of g is shown in Fig. 5.2.2. o . rLLUsrRArroN 3: If h(x) : r3, then h'(*) : 3x2 and,h" (*) : 6x; so h(0) : h'(0) : h" (0): 0. The function h does not have a relative extremum at 0 because if r < 0, h(x) < h(0); and if x ) o, h(x) > h(0). A sketch of the graph of h rs shown in Fig. 5.2.9. o In Illustrations 7,2, and 3 we have examples of three functions, each of which has zero for its second derivative at a number for which its first derivative is zero; yet one function has a relative minimum value at that

Figure5.2.1

Figure5.2.2

Figure5.2,3

5.3 ADDITIONAL PROBLEMS INVOLVING ABSOLUTE EXTREMA

213

number, another function has a relative maximum value at that number, and the third function has neither a relative maximum value nor a relative minimum value at that number.

5.2 ExercisesIn Exercises 1 through 14, find the relative o(trema of the given function by using the second-derivative test, if it can be applied. If the second-derivative test cannot be applied, us the first-derivative test. 1.f(x):3x2-2x*l a. h@) :2xs - %c2 27 + 7. G(x) : ( r - 3) a 1 0.f(x): x\/R 2.s@):13-sx*6 5. f(x) : (x - 4)2 8 . f(* ) : x (x - 1)' 11'.f(x) - lx6rtz x-rtz + M. G(x) - xszts(a4)'

3.f(x):-4x3+3f *'8r 6. G(x): (x * 2)3 9. h(x): xfr$ 1 2 .S U ) : ; * ;Ov2

13.F ( r ) : 6 x r t 3 - x z t l

15.G i v e n f ( x ) : x 3 + s r x * 5 , p r o v e t h a t ( a ) i f r ) 0 , / h a s n o r e l a t i v e e x t r e m a ; ( b ) i f r / - 0 , / h a s b o t h a r e l a t i v e m a x i m u mvalue and a relative minimum value. 16. Givenf(r):x'-1.r+ k, where r > 0 and r + 1, prove that (a) if 0 < r< 1,/hasa relativemaximumvalue at 1; (b) if r > 1.,f ha8 a relative minimwn valu at l. 77. Given f(x) : f * rr-1, prove that regardless of the value of r, f has a relative qrinimum value and no rclative maximurn value. 18. Prove Theorem 5.2.1(ii).

5.3 ADDITIONAL

PROBLEMS INVOLVING ABSOLUTE EXTREMA

The extreme-value thmrem (Iheorem 4.5.9) guarantees an absolute maximum value and an absolute minimum value for a function which is continuous on a dosed interval. We now consider some functions defned on intervals for which the extreme-value theorem does not apply and which may or may not have absolute extrema.

EXAMpLE l,:t\re,

Given

f(x\:

yz_27x_5

solurroN: f is continuous on the interval [0, 6) becausethe only discontinuity of f is at 6, which is not in the interval. t, t ^"\_zx(ffi _ xz L2x* 27 _ (r 3) (r 9)

find the absolute extrema of f on the interval [0, 5) if there are any.

I\x)

@:

1r_5y::-ffiy

' '(r) : 0 when x: 3 or 9; so f (*) exists for all values of r in 10,6), and f the only critical number of / in the interval [0, 5) is 3. The first-derivative test is applied to determine if f has a relative extremum at 3. The results are summarizedin Table 5.3.1. Because/ has a relative maximum value at 3, and f is increasing on the interval [0, 3) and decreasingon the interval (3,6), we conclude that

214

ADDITIONAL APPLICATIONS THE DERIVATIVE OF

on [0, 6) f has an absolute maximum value at 3, and it is f(3), which is 6. Noting that lim f (x) : -m, w conclude that there is no absolute mini-

mum value of / on [0, 6).T a b l e .3 . 1 5

Conclusion 0 0 if x ( c and f" (x) < 0 if x ) c o rLLUsrRArroN 2: Figure 5.4.5 illustrates a point of inflection where condition (i) of Definition 5.4.4 holds; in this case,the graph is concave downward at points immediately to the left of the point of inflection, and the graph is concave upward at points immediately to the right of the point of inflection. Condition (ii) is illustrated in Fig. 5.4.6, where the sense of concavity changes from upward to downward at the point of inflection. Figure 5.4.7 is another illustration of condition (i), where the sense of concavity changes from downward to upward at the point of inflection. Note that in Fig. 5.4.7 there is a horizontal tangent line at the point of inflection. .

T -t{ i

{

Figure5.4.6

5.4.4 Definition

OF APPLICATIONS THEDERIVATIVE ADDITIONAL For the graph in Fig. 5.4.1, there are points of inflection at C, E, and F. Definition 5.4.4 indicates nothing about the value of the second derivative of f at a point of inflection. The following theorem states that if the second derivative exists al a point of inflection, it must be zero there. 5.4.5 Theorem If the function / is differentiable on some open interval containing c, and fi (c, f(c)) is a point of inflection of the graph of f, then if f" (c) exists,

/" (c)- o'

pRooF: Let g be the function such that g(r) - 1' k); then g'(x) : f " (x). Because(c, f (c)) is a point of inflection of the graph of /, then f " (x) changessign at c and so 8'(r) changessign at c. Therefore,by the firstderivative,test(Theorem5.1.4),g has a relative extremum at c, and c is a critical number of g. Because G) : f " (c) and sinceby hypothesisf " (c) g' exists, it follows that g'k) exists.Therefore,by Theorem4.5.3 g'G) :0 and f" (c) : 0, which is what we wanted to prove. IThe converse of Theorem 5.4.5 is not true. That is, if the second derivative of a function is zero at a number c, it is not necessarily true that the graph of the function has a point of inflection at the point where x: c. This fact is shown in the following illustration. o rLLUsrRArroN 3: Consider the function / defined by f (x) : x4.f '(x) : 4x3 a n d f" (x):L2x2. Further, f " (0):0; but becausef" (x) > 0 i f x ( 0 and f"(x) > 0 if x) 0, the graph is concave upward at points on the graph immediately to the left of (0, 0) and at points immediately to the right of (0, 0). Consequently, (0, 0) is not a point of inflection. In Illustration 1 of Section 5.2 we showed that this function / has a relative minimum value at zero. Furthermore, the graph is concave upward at the point (0, 0) (see F i g . 5 .2 .1). o The graph of a function may have a point of inflection at a point, and the second derivative may fail to exist there, as shown in the next illustration. . rLLUSrRerrox 4: If / is the function defined by f (x) : tr1l3, then * $y-l tl f' (x ) : !x6-zt? and f" (x) :

r ,rr6,r* rc/f\

4--

-/l

(r) < 0. f " ( 0 ) d o e s n o t e x i s t ;b u t i f x < 0 , f " ( x ) > 0 , a n d t f x ) 0 , f " Hence, / has a point of inflection at (0, 0). A sketch of the graph of this function is shown in Fig. 5.4.8. Note that for this function /'(0) also fails to exist. The tangentline to the graph at(0,0) is the y axis. . In drawing a sketch of a graph having points of inflection, it is helpful to draw a segment of the tangent line at a point of inflection. Such a tangent line is called an inflectional tangent.

I

Figure5.4.8

5.4 CONCAVITY AND POINTSOF INFLECTION

EXAMPLE For the function in 1: Example1 of Sec.5.1, find the points of inflection of the graph of the function, and determine where the graph is concaveupward and where it is concave downward.

SOLUTION:

f(x):x3-6x2*9x*1' f ' ( x ) : 3 x 2 r 2 x* 9 f " ( x ): 5 x - L 2 point of inflecf " (x) existsfor all valuesof x; so the only possibletion is where f " (x): 0, which occurs at x:2. To determine whether there is a point of inflection at x:2, we must check to see if f " (x) changes sign; at the same time, we determine the concavity of the graph for the respective intervals. The results are summanzed in Table 5.4.1. Table 5.4,1

f (x)-oo< x12

f

'(x)

f " (x)

Conclusiongraph is concave downward

x:2Figure5.4.9 2{x(foo

3

-3

0

graph has a point of inflection graph is concave upward

+

In Example L of Sec.5.1 we showed that / has a relative maximum value at L and a relative minimum value at 3. A sketch of the graph showing a segmentof the inflectional tangent is shown in Fig. 5.4.9.

Exruprn 2: If f (x) : (1 - 2x)t, find the points of inflection of the graph of f and determine where the graph is concave upward and where it is concave downward. Draw a sketch of the graph of f. v

SOLUTION:

f

'(x) - -6(1. * 2x)2

- 2x) f " (x) :24(l Because " (x) exists for all values of x, the only possible point of f inflection is where f " (x) : 0, that is, at x: *.By using the results summarized in Table 5.4.2,we see that /"(x) changessign from rt+" to "-" at x: L, and so the graph has a point of inflection there. Note also that becausef '(+): 0, the graph has a horizontal tangent line at the point of inflection. A sketchof the graph is shown in Fig. 5.4.10.Table 5.4.2

f (x) -m< x 0 i t x < c ; f , , ( t ) < 0 i f r > c . 7 7 .f " ( c ) : o rf t ( c ) : o ; 1 " ( x ) > 0 i f r < c ; 1 " ( x ) > o i I x > c . 1 8 .f ' ( c ) : 0 ; f ' ( x ) > 0 i f r < c;f"(x) > 0ifr>c. - 0; f' (x) < 0 if r < c; f" (x) > 0 iI x > c. 79. f' (c)

m . f " ( c ): 0 ;f ' ( c ): * ; 1 " ( r ) > 0 i f r < c ; f , , ( x ) < o i t r > c . 21.l'(c) does exist; (r\ >0itr < c;f,(x\ > 0 ifr > c. not f" 22. l'(c) doesnot exisf /" (c) doesnot eist f" (x) < 0 ifr < c; f,'(r) > 0 ifr > c. 23. lim f'(x) - +co, l'(t) : o; f" (x\ > 0 ifr < c;f"(x) < 0 ifr > c. lim24. lim f'(x):*@j limf' (r): -*;f" (x) > 0 if r < c;f" (x) > 0 if x > c.

E. Draw a Bketchof the graph of a functionf for which l(x), f'(r), andf,,(r) exist and are positive for all r. 26. II f(t\ :3f + xlxl, prove that l" (0) does not exist but the graph of f is concaveupward everyrvhere. 27. Prove Theorem 5.4.3(ii). 28. SuPPos that I is a function for which l" (x) exists for all values of r in some open interval I and that at a number c in I, f" (r) : O and'l"' (c) xists and is not zero. Prove that the point (c, l(c) ) is a point of inflection oI the graph of /. (HrNr; The proof is similar to the prcof of the second-derivative test (Theorem 5.2.1.).)

TO A OF 5.5 APPLICATIONS DRAWING SKETCH THE GRAPH A FUNCTION OF

5.5 APPLICATIONS TO DRAWING A SKETCH OF THE GRAPH OF A FUNCTION

We now apply the discussionsin Secs.5.'1.,5.2,and 5.4 to drawing a sketch of the graph of a function. If we are given f (x) and wish to draw a sketchof the graph of.f ,we proceedas follows. First, findf' (r) and f"(x). Then the critical numbers of f arc the values of r in the domain of f for which either f '(x) does not exist or f ' (r):0. Next, apply the first-derivative test (Theorem 5.1.4)or the second-derivative test (Theorem 5.2.1)to determine whether we have at a critical number a-relativemaximum value, a relative minimum v3lue, or neither. To determine the intervals on which '(x) is positive; to de/ is increasing, we find the values.of.x for which f termine the intervals on which / is decreasing,we find the values of r for which f '(x) is negative. In determining the intervals on which / is monotonic, we also check the critical numbers at which / does not have a relative extremum. The values of r forwhich f"(*):0 or f"(x) does not exist give us the possible points of inflection, and we check to seeif f " (x) changessign at eachof these values of r to determine whether we actually have a point of inflectioh. The values of r for which f " (x) is positive and those for which f" (r) is negative will give us points at which the graph is concaveupward and points at which the graph is concavedownward. It is also helpful to find the slope of each inflectional tangent. It is suggestedthat all the information so obtained be incorporated into a table, as illustrated in the following examples.

nxanaptnL: Given f(x):x3-3**3 find: the relative extrema of f the points of inflection of the graph of f the intervals on which f is increasing;the intervals on which / is decreasing;where the graph is concaveupward; where the graph is concavedownward; and the slope of any inflectional tangent. Draw a sketch of the graph.

S e t t i n gf ' ( x ) : 0 , w e o b sol,urroN f' (r)-3x2-5x; f" (r):5x-5. tain r : 0 and x: 2. Setting f " (x) : 0, we obtain x: t. In making the table,we considerthe points at which x:0, x:L, and r:2, and the intervals excluding these values of.x:-oo tF(t) F(, 1)l F(n) F(0) i:1

PROOF:

n - F ( 1 ) ] + [ r ( 3 )- F ( 2 ) ] + ' t trtil -F(i-1)l : [F(1)F(0)] + [F(2) Z-/ -

i:1

+ l F ( n - 1 ) - F ( n - 2 ) l + l F ( n ) - F ( n - 1 )l

THE DEFINITE INTEGRAL

: - F ( 0 ) + [ F ( 1 )_ r ( r ) ] + [ F ( 2 )_ r 1 z ; 1 + . . . + l F ( n - 1 ) - F ( n- 1 ) l + F ( n ) :-F(0) +0+0+ . . . + 0+ F(n): F(n) - F(0) I

The following formulas, which are numbered for future reference, are also useful.n

7.'l.,.5 Formula Li:r n

. n ,:T ( n * 7 1 . r-_ n ( n * 1 ) ( 2 n + 1 ) b

7.1.6 Formula 2i:rn

7.'1,.7Formula 3i:r 'n

, r - - l '-(4 -+ 7 ) ' , f- n

7.'1,.8 Formula 4i:1

n(n r t) (6nB 9n2 n - 1) + t ,+_ 30

Formulas 1 through 4 can be proved with or without using mathematical induction. The next illustration shows how Formula L can be proved without using mathematical induction. The proof of Form ula 1,by mathematical induction is left as an exercise (see Exercise 11). . ILLUSTRATToN We prove Formula 1. 4:n

) l:1,+2+3+andn

. . . -t (n-t) *n

)i:n*(n-1)+

(n-2)+..

+2+t

,it*" add thesetwo equations term by term, the left side is

2>i""0 .;;hen

n

right side are n terms, each having the value (n + 1). Hence, '+(n+I) nterms

2 > i : ( n + 1 ) + ( n * L ) + ( n+ 1 ) + 'i:r :n(n*t) Therefore,

z.,l i=l

S: , _ n ( n,)+ t ) \'

7.1 THE SIGMA NOTATION

EXAMrLE1: Prove Formula 2bY mathematical induction.

solurroN:

We wish to Prove that

*,, _n(n+L)(Zn+I)

>r:

6

'L.The sideis then ,': l'. When left ,rrr,ln" formulais verifiedfor n: 2 '2'3-li6- l. Theren : L , t h er i g h ts i d ei s [ 1 ( 1+ 1 ) ( 2+ D ) t e - ( 1fore, the formula is true when n: 1. Now we assume that the formula is true for n: k, where k is any positive integer; and with this assumption we wish to prove that the formula is also true for n:k + 1. If the formula is true for tt: k, we havek

S i,: ./-/i:1

(2k k(k + 1,) + 1,) 6k * L , w e have

(2)

Wh e n n :k+l

f

i r : 1 2 * 2 2 + 9 2 + . . . + k 2+ ( k + 1 ) '

i:l

:)i2+(k+1)' + _ k(k + L)_(zk L) + (k + 1), (by applying (2)) Eq.6

k

_ k ( k+ L )( 2 k+ L ) + 6 ( k+ L ) ' z 6 + _ - - ( k + 1 )[ k ( 2 k 1 ) + 6 ( k+ 1 ) ] _ (k+L)(zk'z+7k+6) 5 _ (k+1)(k+2)(2k+3) 6 - ( / c +1 ) [ ( k + 1 ) + 1 ] [ 2 ( k +1 ) + 1 ] 6Therefore, the formula is true f.ot n: k * L. We have proved that the formula holds f.or n: 1, and we have also proved that when the formula holds f.or n: k, the formula also holds for n: k * 1. Therefore, it follows that the formula holds when n is any positive integer.

A proof of Formula 2, without using mathematical induction, is left as an exercise. The proofs of Formulas 3 and 4 are also left as exercises (see Exercises L2 to 76).

T H E D E F I N I T EN T E G R A L I

ExAMpLE 2:n.{-J

Evaluate

soLUTroN: From Property 4, where F(i) : 4i,1t follows that -4-40 >r (4' 4i-') i: :4n-In

t

(4' - 4i-')

i:r

nxaivrPrn3:n

Evaluate

SOLUTION:

>r i(3i 2) i:by using Properties 1-4 and Formulas 1-4.

> r(3t 2) i:r

n

(3i2 - 2i)

(3i')

+> (-2i)

(by Property 3)

:3

> i'- Z2 i i:7

(by Property 2)n(n+ 7) 2 (by Formulas2 and 1 )

2nBl3nzIn-2n2-2n

22n3*n2-n

7.1 ExercisesIn Exercises 1 through 8, find the given sum.6"

1 .> ( s i - 2 )i=l

2.>(i+1.)'i:r

3.

?

it

i-1,

t $

2

KFz

U ,D-,''3 b 8'. y t I " ,ro-rk+3

3

6. z ' - l + i 2

s1

-. + , -( - t ; u * r i ) , k F 9. Prove Property 1 (7.1,.1,). 10. Prove Property 3 (7.1.3).

11. Prove Formula 7 (7.1.5) mathematicalinduction. by L2. Prove Formula 2 (7.1.6\without using mathematicalinduction. (HrNr: iB- (i - 7)":3i ' - 3i + L, so thatnn

)

t ; '- ( i - 1 ) ' l : )

tgt,- 3t+ 1)

On the left side of the above equation, use Property 4; on the right side, use Properties '1,, and 3 and Formula Z, 1.)

7.2 AREA 281 13. Prove Formula 3 (7.1.7)without using mathematicalinduction. (HrNr: ia- (i - qa:4p - rO * 4i- 1, and use a method similar to the one for Exercise12.) 14. Prove Formula 4 (7.1.8)without using mathematical induction (seehints for Exercises12 and 13 above). 15. Prove Formula 3 (7,1.4 by mathematical induction. 16, Prove Formula 4 (7.1.8)by mathematicalinduction. In Exercises17 through 25, evaluatethe indicated sum by using Properties 1 thlough 4 and Formulas 1 through 4.20

t

2zi(i-1,)n

1 8 .> 3 i ( i 2 + 2 )i:rn

19.>(10i*'-10,)

20. > (2k-t-zk)k:r 40 22.>3/T+r -\Dl-l

--'f:,Lk k+L) ,r.''flff-=+-l ,t 24iz(in

i:rn

- z\

24.>\i(L+i2)i--l lr f /i\21r12

25. > [(3-o - 3o)'- (3t'-t+ 3-k-r)2] ,.i;UseLli:1

2 6 .P r o v e : l t - { - } ) | \nl J ,?nL

:2}

n

[ ?=, '-

(*)']''' *,

Prope rty4 toshowthat

) > t t t , + 1 ) - F ( i - 1 ) l : F ( n r - 1 ' + F ( n )- F ( 1 )- r ( 0 )

(b) Prove thatn,1J

t

t(t+1)'-(i-'

1 ) ' l: ( n * 1 , ) 3 * n 3 - L

i:r

(c) Prove thatnnn

)

t t ; + 1 ) ' - ( i - 1 ) ' l : 2 ( 0 i '+ 2 ) : 2 n* 6 )

i2

i:r

(d) Using the results of parts (b) and (c), prove Formula 2. 28. Use the method of Exercise 27 to prove Formula 3.n )t' n

29. If X:El - , pr ov en

t ha t )i:r

(x , - X)' :2

n

i=l

r,' - X 2 * ,i--L

r.

7.2 AREA

We use the word mensure extensively throughout the book. A measure refers to a number (no units are included). For example, if the area of a triangle is L0 irt.z, we say that the measure of the area of the triangle is 10. You probably have an intuitive idea of what is meant by the measure of the area of certain geometrical figures; it is a number that in some way

INTEGRAL THE DEFINITE

Figure7.2.1

Figure7.2.2

gives the size of the region enclosed by the figure. The area of a rectangle is the product of its length and width, and the area of a triangle is half the product of the lengths of the base and the altitude. The area of a polygon can be defined as the sum of the areas of triangles into which it is decomposed, and it can be proved that the area thus obtained is independent of how the polygon is decomposed into triangles (see Fig. 7.2.1). However, how do we define the measure of the area of a region in a plane if the region is bounded by a curve? Are we even certain that such a region has an area? Let us consider a region R in the plane as shown in Fig. 7.2.2. The region R is bounded by the r axis, the lines x: a and x: b, and the curve having the equation y -- f(x), where f is a function continuous on the closed interval [a,b]. For simplicity, we take f (x) > 0 for all x in [a,bl.We wish to assign a number A to be the measure of the area of R. We use a limiting process similar to the one used in defining the area of a circle: The area of a circle is defined as the limit of the areas of inscribed regular polygons as the number of sides increases without bound. We realize intuitively that, whatever number is chosen to represent A, that number must be at least as great as the measure of the area of any polygonal region contained in R, and must be no greater than the measure of the area of any polygonal region containing R. We first define a polygonal region contained in R. Divide the closed interval [a, b] into resubintervals. For simplicity, we shall now take each of these subintervals as being of equal length, for instance, Ar. Therefore, Ar: (b - a)ln. Denote the endpoints of these subintervals by xo, xt, x 2 , . . , x n - r ,r r , w h e r f f o : f l , x t : a i L x , . . . , x i : A * i L , x , . . , xn-r:

_tl9P53EINhZ

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negative number x

positive number x

o x sr

x units

rand x

r r o1

number plane

u s r r