--'-_11 InlUvn"""" on , ,.."

a sinB = Jllr = l =* JI = 3, r = 5, and x = .,jr'l - JI'l = 4 (since 0 < 8 < f). Therefore taking x = 4. JI = 3,

r = 5 in the definitions of the trigonometric rati08, we have cod = f, tan 8 = !' CIIC 8 = ~,see 8 = t, and

cotS = t.

30. Since 0 < 0 < f, 0 is in the first quadrant where x and JI are both positive. Therefore, tan 0 = JI I x = f =*

JI = 2, x = I, and r = .,j x'l + y'l = v'5. 'l8king x = I, JI = 2, r = v'5 in the definitions of the trigonometric

ratios, webave8ino = -fa, coso = ~,csco = ~,seeo = v'5, and cot 0 = !.

31. f < tP < 1f =* tP is in the second quadrant, where x is negative and 11 is positM. Therefore

88CtP = rlx = -1.5 = -~ =* r = 3, x = -2, and 11 = ~ = v'5. 'Thking x = -2, JI = vi, and

r = 3 in the definitions of the trigonometric ratios, we have sin tP = ~,cos tP = - f, tan tP = -:II-, CIIC tP = 7a'

and cot 8 = --Jr..

32. Since 1f < x < " x is in the third quadrant where x and 11 are both negative. Therefore cosx = xlr = -i =*

x = -I, r = 3, and JI = -~ = -VB = -2V2. 'Thking x = -I, r = 3,11 = -2V2 in the definitions of""--' ' 'u Lu_- -,-- -- 2.12 '_n n 1ft n- n ~ .. - -~ - , 1

the trigonometric ratios, we have sinz = -~, tanz = 2./2, cscz = -~, secz = -3, and cotz = ~.

33. 11' < fJ < 211' means that fJ is in the third 01' fourth quadrant where 11 is negative. Also since cot fJ = z/1I = 3 which

is positive, z must also be negative. Tbcrefcn cot.fJ = z/1I = ~ => z = -3,11 = -I, and

r = v' z2 + y2 = JiO. Taking z = -3, 11 = -1 and r = JiO in the definitions of the trigonometric ratios, we

have sin II = --L CO6 II = --1... tan II = 1 cae II = - 'iO and see II = _.3Lm/J .;IO' /J .;IO' /J 3' /J VIV, /J 3 .

34. Since' < 9 < 211', 9 is in the fourth quadrant where z is positive and 11 is negative. Therefore cae 9 = r / 11 = - t=> r = 4. II = -3, and z = v'r2 - 112 = ./7. Thking z = ./7, 11 = -3, and r = 4 in the definitions of the

trigonometric ratios, we have sinS = -t, CO8S = 4, tanS = -~, eecS = 77' and cotS = -4.

:r:35. 1lin35° = 10

:r:.. CO8 40" = 25

:r:37. tan't ::a 8

22.cae'=-;-39.

% = 10sin35° ~ 5.73576 em

% = 25cos40° ~ 19.1IHn em

% = 8 tan" ~ 24.62147 em

22% = =ai' ~ 57.48877 em

...

...

..

..;;;1i.

(a) From the diagram we see that sin (J = !! = ~. and, r c.::! asin(-S) = q ,. -c = -sinS.

(b) Again from the diagram we see that coeD:II: = = ~ = COI(-I).r c

1018 0 APPENDIX D TRIGONOMETRY

40. (a) Using (l2a) and (12b). we have

8in x COB 11 COB xain IItan(x+ ) = sin(x+y) = sinxC08I1+~xs~ny = cosxcosfl + ~x~y = tan x + tan II

Y oos(x + y) cosxCOBy - smxsmy cosxcoslI - amxamll 1- tanxtany

cosxCOBy COBXCOBY

(b) From (lOa) and (lOb), we have tan( -8) = - tanS, so (14a) implies that

tan tanx+tan(-II) tanx-tanytan(x-y) = (x+(-y» = I-tanxtan(-II) = 1 + tan x tan y

41. (a) Using (l2a) and (13a), we have

~ [sin (x + y) + sin (x -II)] = ~ (sin x cos II + cosxsiny + sinxCOBII- cosxsiny]

= 4 (2 sin x COSII) = sinxcosy

(b) This time, using (l2b) and (13b), we have

4 [COB (x + y) + C08 (x - II)] = 4 (cosxcosy - sin x sin II + cosxcosy + sinxsiny]

=! (2 cos X cosy) = C08XC08Y

(c) Again using (12b) and (13b), we have

! (cos (x - y) - cos (x + y)] = 4 (C08XCOSY + sinxsiny - cosxCOSII + sin x &iny]

=! (2sinx sin y) = sinxsiny

42. Using (l3b), C08 (f - x) = COB f C08X + sin f sin x = O. coax + 1. sin x = sinx.

43. Using (l2a), we have sin(i + x) = sin f cosx + cos i sin x = 1. COB X + O. sin x = cosx.

44. Using (l3a), we have sin(1I' - x) = sin1l'COSX - 00811' sinx = O. 008 X - (-1) sin x = sinx.

45. Using(6),wehavesin800t8 = sin8. :: =cos8.

46. (sin x + COSX)2 = 8in2 x + 2sinxcosx + COS2 x = (sin2 x + 0082 x) + sin2x [by (15a)]

= 1 + sin 2x [by (7)]

47. secy-cosy= 2- -cosy [by (6)] = l-c0e2y = sin2y [by(7)] = &iny siny = tanysiny [by (6)]cos y COSII COB Y cos Y

sin2 a sin201 - 8in2 OIcos2 a sin2 a (1- C082 a)48. tan2 a - sin2 a = - - sin2 a = . = . = tan2 a sin2 a [by (6),COS2 a 008201 COS201

(7)]

C082 8 1 coe2 ~ 0082 8 + sin2 849. oot2 8 + sec2 8 = sin28 + cos29 [by (6)] = .2800828

- (1 - sin2 9)(1 - sin2 8) + sin2 8 - 1 - sin2 9 + sin' 8- . 28 2 8 [by (7)] - . 28 28am cos 81n cosCOS2 9 + sin' (J 1 . 8in29

= . 28 29 [by (7)] = -:-r8 + ~ 8 = CSC2 9 + tan2 9 [by (6)]sm 008 am COB

? 2 1

~

51. Using (148), we have taD 28 = taD(8 +8) = tan 8 + tan 8 - 2taD81- taD 8 taD 8 - 1- taD28'

&2. 1 + 1. = l+sin8+1-8in8 - 2 2l-1iD8 1+8111' (1-1iD8)(1+8iD8) -1-8iD28 = oos28 [by(7)] =2sec:l8

53. Using (15a) and (100),

.. sint/l = sint/l .1+OO8t/1. sint/l(1+OO8t/1) ==sint/l(1+OO8t/1) [by(7))1-008t/1 1-008t/1 1+008t/1 1-c0e2tj1 sin2tj1

1 +008tj1 1 008tj1== sint/l =sint/l+sint/l=alCt/I+cott/l[by(6»)

II. tan x + tm == sinx + Elf = EX 008 If + 008x8in1f = sin (x + If) [by(118»)If 008 X 008 If 008 X 008 If 008 X 008 If

57. Usins (118).

sin 38 + sin 8 == sin (28 + 8) + sin 8 == sin 28 0088 + CO828sin8+ sin 8

== sin 280088 + (200828 - 1) sin 8 + sin 8 [by (168)]

== sin 28 CO8 8 + 20082 8 sin 8 - sin 8 + sin 8 = sin 28 CO8 8 + sin 28 CO8 8 [by (I5&)J

= 2 sin 280088

.. We we (I2b) with x = 28. If = 8 to get

00838 =008(28 + 8) == 008280086 - sin 28 sin 8 .== (200828 - 1) 0088 - 2ain2 80088 [by (168) and (15&)J

,.. (200828 -1) 0088 - 2(1- 0082 8) 0088 [by (7))

= 20088 8 - 0088 - 20088 + 2008'8 == 4008'8 - 30088

57. Using (12a),

51. We use (12b) with x = 28, If = 8 to get

APPENDIX 0 TRIGONOMETRY 0 1019

[by (12&»)

1iiiII,\

1,1hi.~

I.\

0 CHAPTER t FUNCTIONS AND MODELS2

Example 2: At a certain university, the number of students N on

campus at any time on a particular day is a function of the time t after

midnight. The domain of the function is {t I 0 ~ t ~ 24}, where t is

measured in hours. The range of the function is {N I 0 ~ N ~ k},

where N is an integer and k is the largest number of students on

campus at once.

Example 3: A certain employee is paid $8.00 per hour and works a

maximum of 30 hours per week. The number of hours worked is

rounded down to the nearest quarter of an hour. This employee's

gross weekly pay P is a function of the number of hours worked h.

The domain of the function is [0,30) and the range of the functioo is

to, 2.00, 4.00,.. . ,238.00, 240.oo}.

5. No, the curve is not the graph of a function because a vertical line intersects the curve more than once. Hence, thecurve fails the Vertical Line Test.

I. Yes, the CUI'YC is the graph of a function because it passes the Vertical Line Test. The domain is 1-2,2] and the

ranaeisl-l,2].

7. Yes, the curve is the graph of a function because it passes the Vertical Line Test. The domain is (-3, 2J and therange is [-3, -2) U (-1,3J.

8. No. the curve is not the graph of a function since for x = O. :1::1. and :1::2. there are infinitely many points on the

curve.

t. The person's weight increased to about 160 pounds at age 20 and stayed fairly steady for 10 years. The person's

weight dropped to about 120 pounds for the next 5 years, then increased rapidly to about 170 pounds. The next 30

years saw a gradual increase to 190 pounds. Possible reasons for the drop in weight at 30 years of age: diet,

exercise, health problems.

10. The salesman travels away from home from 8 to 9 A.M. and is then stationary until 10:00. The salesman b'avelsfarther away from 10 until noon. There is no change in his distance from home until 1:00, at which time the distance

from home decreases until 3:00. Then the distance starts increasing again, reaching the maximum distance away

from home at 5:00. There is no change from 5 until 6, and then the distance decreases rapidly until 7:00 P.M., at

which time the salesman reaches home.

11. Tbe water will cool down almost to freezing as the ice melts. 1ben,

when the ice has melted, the water will slowly warm up to room

~

12. The summer solstice (the longest day of the year) is around lune 21,

and the winter solstice (the shortest day) is around December 22.

13. Of course. this graph depends strongly on the geographical

location!

14. 'The temperature of the pie would increase rapidly, level off to

oven temperature, decrease rapidly, and then level off to room

temperature.

15.

~~/

16. (a)

(0)

.'

0 3FOUR WAYS TO A FUNCTIONREPRESENT

Houn of

d8yliJlat

midnight

,

(b)

(d)

r '~ SECTION 1.1 FOUl WAYS TO REPRESENT A RJNCTION C"," s 2+h 2+h z+h"w 22. I(s) ... ;+i'1O 1(2 + h) - 2 + h + 1 = 3 + h ,/(z + h) = z + h + I' and

,,;~.. 1(0+4)-/!0) - ~-~ - (0+4)(0+1)-"\0+4+1) - 1 ," h h h(z + h + l)(z + 1) (z + h + l)(z + 1)

.. f(s) ... s/(3s - 1) is defined for all z except when 0 = 3s - 1 <=> s = i, 10 the domain is

{z e R Is -F i} = (-00,1) u (t, 00).

.. j(s) = (5s + 4) / (z2 + 3s + 2) is defined for all z except when 0 = z2 + 3s + 2 <=> 0 = (z + 2)(z + 1)<=> s = -2 or -I, so the domain is {z e R Iz "" -2, -I} = (-00, -2) U (-2, -1) U (-1,00).

2S. I(t) = .;t + ~ is defined when t ~ O. These values oft give real number results for.;t, whereas any value oft

gives . real number result for ~. The domain is [0, 00).

2&. g(u) = .;u + "';4 - u is defined when u ~ 0 and 4 - u ~ 0 <=> u $ 4. Thus, the domain is

O$u$4=[O,4].

%1. h(z) = 1/ ~z2 - 5z is defined when Z2 - 5z > 0 <=> z(z - 5) > O. Note that z2 - 5z "" 0 since that

would result in division by zero. The expression z(z - 5) is positive if z < 0 or z > 5. (See Appendix A for

methods for solving inequalities.) Thus, the domain is (-00,0) U (5,00).

ah(z)=~.NoWII=~ ~ 1l=4-z2 <=> Z2+y2=4,so Athe gnpb is the top half of a circle of radius 2 with center at the origin. The domain 2

is {z 14 - Z2 ~ O} = {z 14 ~ z2} = {z 12 ~ Izl} = [-2,2]. From the graph,-2 0 2 JC

the range is 0 $ II $ 2, or [0,2].

a f(.) = . ;, "'ul"'" -, .. tho domoin ;, R... (-00,00). 'IThe graph of j is a borizontalline with y-intcn:ept 5. S J - S

~

. F(s) = l(s + 3) is defined for all real numbers. so the domain is R. or

(-CX). CX». The paph of F is a line with x-intercept -3 and y-intercept !.'

31. J(t) = e2 - 6t is defined for all real numbers. so the domain is R. or

(-00,00). The graph of J is a parabola opening upward since the

coefficient of e2 is positi~. To find the t-inten:epts. let 11 = 0 and

soI~ for t. 0 = t2 - 6t = t(t - 6) ~ t = 0 and t = 6. The

t-coordinate of the vertex is halfway between the t-intercepts. that is.- -. "ft' ,,2 "? - _0 th.. V<"rt~1f ill (~. -9),

SECTION 1.1 0 IFWI WAYS TO REPRESENT A FUNCTION

y-intercept 5.

CHAPTER 1

4 - fA (2 + t)(2 - t) 80 for t :F 2, H(t) = 2 + t. The domain is

)"'"'2=t= 2-t .

e :;: 2}. So the graph of H is the same as the graph of the function

= t + 2 (8 line) except for the bole at (2,4).

- ";z - 5 is defined when z - 5 ~ 0 or z ~ 5, so the domain is [5,00).

, 11- ";z - 5 => y'J = Z - 5 => z = y'J + 5, we see that 9 is the, 1/ = ../X - 5 => 1/2 = Xa1f of a parabola.

{ 2x+l I = 12% + 11 =-(2s+ I)

= {~:~I

Iomain is It. or (-00.00).

- { b;Z 3z-z

Z

RJNCTIONS AND MODELS

t

if~+12:0

if~+l<l

if :z: 2: -j

if :z: < -i

ifz~Oif z<O,wchave

:~

If

- { ; if2:>O- ..!..if2:<O

-2:

if x < -1if x ~ -1

Domain is It. or (-00. 00).

. Recall that the slope m of a line between the two points (3:10 Y1) and (3:2, Y2) is m = Y2 - Y1 and an equation of3:2 - 3:1

. . th ., ( ) Th I f thi li . -6 - 1 7the line connecung ose two points 18 Y - Y1 = m 3: - 3:1. e s ope 0 s ne segment 1S 4 - (-2) = -6' so

an equation is y - 1 = -i(x + 2). The function is I(x) = -Ix - f. -2 ~ x ~ 4.

~ The slope of this line segment is : = ~=:~ = ~,so an equation is y + 2 = ~(x + 3). The function is

I(s) >= Iz - t. -3 ~ s S 6.

" We need to solve the given equation for y. x + (y _1)2 = 0 <=> (y - 1)2 = -x <=> y - 1 =:f:yCX <=>

11 = 1 :f: yCX. The expression with the positive radical represents the top half of the parabola. and the one with thenegative radical represents the bottom half. Hence, we want f(x) = 1 - yCX. Note that the domain is x $ O.

~ (:I: -lr~ + 11 = 1 <=> 11 = :';1- (:I: -1)2 = :v'2:z: - :z:2. The top half is given by the function

1(:1:) = v'2:1: - :z:2, 0 ~ :I: ~ 2.

~ For -1 :5 x :5 2, the graph is the line with slope 1 and y-intercept 1, that is, the line 11 = x + 1. For 2 < x :5 4, thegraph is the line with slope - ~ and x-intercept 4 [which corresponds to the point (4, 0)], so

{ X + 1 if -1 < x < 21I-O=-~(x-4)=-~x+6.Sothefunctionisf(x)= 3 . --

-2X + 6 if 2 < x :5 ..

~ For x ::; 0, the graph is the line y = 2. For 0 < x ::; 1, the graph is the line with slope -2 and y-intercept 2, that is,

the line II = -2z + 2. For x > 1, the graph is the line with slope 1 and x-intercept 1, that is, the line

{ 2 if x < 0

II = 1(x -1) = x -1. So the function is j(x) = -2z+2 if 0 < x $1.

x-1 if 1<x

'. Let the length and width of the rectangle be L and W. Then the perimeter is 2L + 2W = 20 and the area is

A = LW. Solving the first equation for W in terms of L gives W = 20,; 2L = 10 - L. Thus,

A(L) = L(10 - L) = IOL - L2. Since lengths are positive, the domain of A is 0 < L < 10. Hwe further restrict

L to be larger than W, then 5 < L < 10 would be the domain.

L Let the length and width of the rectangle be L and W. Then the area is LW = 16, so that W = 16/L. Theperimeter is P = 2L + 2W, so P(L) = 2L + 2(16/L) = 2L + 32/L, and the domain of Pis L > 0, since

if 7SECT10N 1.1 FOUR WAYS TO REPRESENT A RJNCTION

{ -I ifz<-I40. 1(x) = 3x+2 if -I <:c < 1

7-2% ifz~l

Domain is R.

SECTION 1.3 NEW FUNCTIONS FROM OLD FUNCTIONS 0

To obtain the graph ofy = -5/(x) from the graph ofy = I(x), stretch the graph vertically by a factor of 5 and

reflect it about the x-axis.

To obtain the graph of y = 1(5x) from the graph of1l = I(x), shrink the graph horizontally by a factor of 5.

,I. r: To obtain the graph of 11 = 51 (x) - 3 from the graph of 11 = 1 (x), stretch the graph vertically by a factor of 5. ~. and shift it 3 units downward.

r", "'Ii; ,

Ja) (graph 3) The graph of 1 is shifted 4 units to the right and has equation 11 = I(x - 4).

.~(;, ). (graph I) The graph of 1 is shifted 3 units upward and has equation 11 = 1 (x) + 3.~,. (graph 4) The graph of 1 is shrunk vertically by a factor of 3 and has equation 11 = i/(x).

-';(graph 5) The graph of 1 is shifted 4 units to the left and reflected about the x-axis. Its equation is11 = - I(x + 4).

,(graph 2) The graph of 1 is shifted 6 units to the left and stretched vertically by a factor of 2. Its equation is

;"y = 2/(x + 6).

. . 'To graph 11 = I(x + 4) we shift the graph

'. ~fl, 4 units to the left.

point (2, 1) on the graph of 1.. rresponds to the point

f1"4, 1) = (-2,1).

graph y = 2/(x) we stretch the graph of

.yertically by a factor of 2.

,~J!Oint (2, 1) on the graph of 1, nond'l to the point (2.2 . t) = (2. 2).

17

(b) To graph y = f(3:) + 4 we shift the graph of f.4 units upward.

The point (2, 1) on the graph of f corresponds to

the point (2, 1 + 4) = (2,5).

(d) To graph y = -!/(x) + 3, we shrink the graph of

/ vertically by a factor of 2, then reflect the

resulting graph about the x-axis, then shift the

resulting graph 3 units upward.

The point (2, 1) on the graph of / corresponds to

th~f"'P\(2 ,-i .1 + 3' = (2,2~5)

~

4z + 3 = (X2 - 4x + 4) - 1 - (:I: - 2)2 - 1: Start with the graph of 11 = X2, shift 2 units to the right,

bift 1 unit downward.

y=(x-2)Z y=(x-2)z-1

of II = CO8 Z. sttetcb vertically by . factor of 2, and then shift 1 unit upward.

,-zaCO8 x: Start with the grapb

COI.1'

of 11 = ain %, compress horizontally by a factor of 3, and then stretch verticallyIx: Start with the

of4.

'AU 't

graph

,-liD 3x

y = sin(x/2),-sinx

1Y-i

1"z-.

NEW RJNCTIONS FROM WI RJNC11ONS 0 D

SECTION 1.5 EXPONENTIAL AJNCTlONS 0 r1

0 as :& -+ -00, all of them passapproach

increases more quickly than that of 3'" for

8. We start with the graph of y = 4'" (Figure 3) and

then shift 3 units to the right. There is a horizontal

asymptote of y = O.

of y = 4'" (Figure 3)

11 = 48 11 = 4"'-3

-I

EXPONENTiAl FUNCTIONS 0 31SECTION 1.5

pass the Horizontal Line Test.

u,~

INVERSE FUNCTIONS AND LOGARITHMS1.& 0 4tSECTION

model is y = abt, where

60197589 x 10-9 and b = 1.0129334321697.

gives y(1925) ~ 111 million,

million, and y(2020) ~ 375 million.

. ons and Logarithms

1.

I(x) = y for any y in B. The domain of 1-1 is B and the range of 1-1 is A.:*c

: graph of f about the line y = x.

because 2 :F 6, but 1(2) = 2.0 = 1(6).

since for any two different domain values, there are different range values.

ntersects the graph of 1 more than once. Thus, by the Horizontal Line Test, 1 is one-to-one.

y = 0 (the x-axis) intersects the graph of 1 in more than one point. Thus, by the Horizontal Line

>-one.

If = 0 (the x-axis) intersects the graph of f in more than one point. Thus, by the Horizontal Line

,-one.

~tersects the graph of f more than once. Thus, by the Horizontal Line Test, f is one-to-one.

= i (X + 5) is a line with slope i. It passes the Horizontal Line Test, so / is one-to-one.

If Xl ¥ X2. then Xl + 5 ¥ X2 + 5 '* !(Xl + 5) ¥ !(X2 + 5) '* /(Xl) ¥ /(X2). so I

= 1 + 4x - X2 is a parabola with axis of symmetry x = - 2~ = - 2( ~1) = 2. Pick any x-values

:> find two equal function values. For example. 1(1) = 4 and 1(3) = 4, so 1 is not 1-1.

(-1) = 1 = 9(1), so 9 is not one-to-one.

l"I:..,j%2 =* 9(2:1) "" 9(2:2), SO 9 is 1-1.

1 eyery height h up to its maximum height twice: once on the way up. and again on the way

t} does not equal h. f{h) may equal f{t2). so f is not 1-1.

eventually we all stop growing and therefore, there are two times at which we have the same

AJNCTIONS AND MOOB.SCHAPTER 1G 0

15. / does not pass the Horizontal Line Test.

so f is not 1-1.

(b) By the second cancellation equation in (4), we have /(1-1(5») = 5.

ZOo (a) f is 1-1 becau8e it pasaes the Horizontal Line Test.

(b) Domain off == (-3,3) == Range of r1. Range of f == (-2,2) == Domain of r1.

(c) Since f( -2) == I, r1(1) == -2.

21. We solve e == ~(F - 32) for F: fe == F - 32 ~ F == fe + 32. This gives us a formula for the inverse

function, that is, the Fahrenheit temperature F as a function of the Celsius temperature e. F ~ -459.67 ~

~e + 32 > -459.67 ~ Ie> -491.67 ~ e> -273.15. the domain of the inverse function.

mo v2 ~ ,,' ~ . ( mi )22.m= => 1 => -=1-- => ,,2=CA 1-- ...

y'1- "'If? c2 m2 c2 m' m2

D. 1(%) = v10 - 3z ~ If = Vl0 - 3z (I/?: 0) ~ t? = 10 - 3z ....

% = -l~ + ~. Interchange % and V. If = -1%2 + f. So r1(%) = -1%2 +is % ?: O.

4:&-1 4:1:-1Jl /(z) = 2:z: + 3 .. II = 2:z: + 3 ~ 1I(2:z: + 3) = 4:& - 1 ~ 2:z:tI + 3,1 = 4:z: - 1 ~

311+1 3:1:+1311 + 1 = 4:z: - 2:z:tI ~ 3Jt + 1 = (4 - 2,1)z ~ z = 4 - 211' Interchange z and ,I: II = 4 - 2:z:'~+ 1 == (4 - 21/):1: =>3r/+l =43:-22:y ..Sor1(:.:) = ~-+~.

a a11./(:.:) = e8 '* If = e8Sor1(:.:) = ~.

~... 1(:.:) = 22:3 + 3 ~

=* inti = %3 =* % = ~. Interchange % and tI: tI = ~.

:2%3+3 ~ 11-3-2z3 ... 11-3=..2

s r;=3 1 . r;=3X and y: y = V """"2' So r (x) = V """"2'

1/-3=2x3....

11. f passes the Horizontal Line Test,

so f is 1-1.

3x = 10 - y'l

Jf. Note that I

..Note

z= f/"-32 ....