Pre Calculus qs

Prévia do material em texto

FUNCTIONS 
A. A function is a relation in wh ich each el ement of the d oma in (x value 
- independen t va ria bl e) is pa ired w ith only one e lement of the range 
(y value ­ d epen de nt va riable). 
B. A re lation can be tested to see if it is a funct ion by the vertical line test. 
Draw a vertical line through any g raph, and if it hi t an x-value more than 
once, it is not a function . (1-4) 
b. b2 ­ 4ac = 0, exact ly one rea l r ot 
c. b2 - 4ac < 0, no real roots (two disti nct imaginary roots) 
I) Example l:f(x) =x2_4x+ I use 
-b ± /b 2 _ 4ac 
f( x)= 2a 
x = -(-4) + f< - 4 2 [!]E8rnrn
A function Not a function Afunction A fu nction 
C. Linear funct ions take the form: f(x) = mx+b, or y = FE
m x+b where m = the. lope, and b = the y-intercept. 
Example: f(x) = 4x-l , the s lope i 4/1 (ri e over run), and 
the y-intercept is -I. 
D. he dis tance between two points on a lin can be found 
u ing the distance form ula, d = j(X 2 - XI ) 2+ (Y2 - YI )2 X;~~5 
E. The mid-point of a line segment can be found us ing the 
.. ( X2 + X I) (Y 2+ Vd )
m id-po lllt form ula, 2 • 2 ' . 
F. The s tandard form ofa linear function is 0 =Ax + By + 
C. The lope is m = -A/B, and the y-intercept is -C/B. 
G. The zeros of a funct ion a re found by setting y to O. and ffi
No zeros solving for x. 
I. E xample .1: f(x) = 4x-1 (5) 
2. 	Exam p le 2: f(x) = 6, this func tio n has no zero, and is 
a h ri z ntal line th rough +6 on the y-axi . (6) 
3, Exam ple 3: x = 4, th i i not a function, because the re 
is a vertica l line through +4 on the x-axi , g ivi ng an 
in fi n ite set of va lues for y. (7) 
H. Polynomial function s take the form : f(x) = axn + bxn- I 
+ cxn-2 ••• + d x + e 
I . When the highe t power o f the func ti on is an odd 
integer, there is at least one real zero. 
2 . When the highe t power is an even integer, there may 
be no real zeros . 
 EE3. Both type can have imaginary roots of the form a + bi. 
x=-1.170213 
4. T h 	 highest p wer of a po lynomi a l w ith one vari able is y=.0100806 
called it degree. 
Example l: f(x) = 2x4 + x2 + X + 10, has a degree of tE
4, there are four roots (solutions) to th is polynomia l. 9 
Example 2: f(x) = 2xJ + x2 - 2x + 3, thi function has 
one real zero at x = -1.17, and two non-real roots. (8) 
Example 3 : f(x) = x2 + I. this fUllction has'two non-
real roots, (9) x=.10638298 
l. Quadratic funct ions take the form : f(x) = ax2 + bx + c. y=1.0113173 
I. 111e graph ofa quadrati function i called a parabola. (10) 
2 . ome parabolas are quadrat ic eq uation , but onot tE
q uadratic fu nctions. (II) 
3. 	Quadratic functions or equat ions can have one real 
so lution. two rea l solutions, or no real solution. (12-14) 
4 . Th vertex of a parabola i ca lled its critical p oint. 
- b ± jb2 - 4ac x=-1.010638 ). The quadratic equat ion f(x) = 2a can y=2.9737903 
be u ed to f ind the roots of a ll quadratic eq uations. 
6 . The a lue under th square root symbol is ca lled the 
d iscrim inan t. It te ll s u the type of roots o f a quadratic 
equat ion . 
a . b2 - 4ac > 0, t lVO dist inct rea l roots 
_ 	 4)/2 = 3.732, and 
x = -(-4) - f< - 4 - 4) /2 = .267. s ince the 
discriminant is > 0, there are two rea l roots. (15) 
2) Example 2: f(x) = 2x2 + 2x + I us ing b2 - 4ac = 
-4, since the d iscriminant is < 0, there are two 
imaginary root . (16) 
3) Example 3: f(x) = x2 + 2x + I us ing b2 - 4ac = 0, 
1 	d· .. . 0 th ' I 17since t le ISCnl11l11ant IS = ere IS one rea I' \. ( ) 
.. 	 g(x)
J . 	RatIOnal function s take the fori: f(x) = b (x) . 
I . The parent fu nction i f(x) = X · 
2. The g raph of these functi ons cons ist of two pan s, one 
in quadrant I, and one in quadrant 3. 
3. The bran hes of rational functi on approac h line 
called asymptotes. (18) 
x 
4. Example I: f(x) = x + 3 (19) 
3 
5. Exam ple 2: f(x) = x (20) 
6 . Example 3: f(x) = ~2 + 3 (21)
x -
K. Oper ations of fu nctions: 
I. Sum: (f + g)(x) = f(x) + g(x) 
2. Difference: (f - g)(x) = f(x) - g(x) 
3. P roduct: (f x g)(x) = f(x) x g(x) 
. (f) f(x) 4. QuotIent: g (x) = g (x )' g( x) f. 0 
5. Example I: Given f(x) = x + 2, g (x) == x:. 4 
a . Find the sum : (f + g)(x), x + 2 + ~ 
( x + 2 ) (x - 4) + x X 2 _ X _ 8 x ­
x - 4 = x - 4 ' and x f. 4. 
b. Find the di fference: (f - g)(x), x + 2 -~ (x + 2) (x - 4 ) - x x _ 3x _ 8 x ­
x - 4 = x _ 4 • and x f. 4. 
6. Example 2: Given f(x) = x + 2, g(x) = x:. 4 
a. Find the product: (fxg)(x), (x + 2)( x :. 4 ) = 
x 2 + 2x 
---x=-;t. an d x f. 4. 
b. Find the quot ient: ( t )(X), x;x+_24 = 
( X - 4) x2 -2x - S(x+2) .-x- = x , andx f. O. 
L. c omP.osition of functions:lfog l(x) = f(g(x» E!J8 
Example: Gi en f(x)=x+2, g(x)= :. 4 + 2. 
Find IfogJ(x): f ( x :. 4 + 2 ) = 
2 ( 4 ) x-- 0265958( _ x_ + 2 ) + 2 = x + x - + 2 y~.OlO0806 
x - 4 (x - 4) A rationalfunction 
with asymptotes5x - 16 at the x& y axes. 
, and x f. 4.4x-
M .lnverse function s : Ifog l(x) = Igofj( x) OFE
· +4 Example: Given f(x)=2x - 4, g(x) = ~, 
_ ('X+4)_ (X +4) _Ifog l(x) - f -2- - 2 -2- - 4 - x, 
x=.02659573( 2 4 ) 4X - + y=-.0100806 
and I gO fj( x) = 2 = x. The asymptotes 
are the axes 
y=-.0100807 
One real solullOn 
EE/3 
x=1.2765957 
y=-2. 1 06855 
Two real solutions 
EE4 
x=-.9574468 
y=7.9737903 
No real solutions 
EE
x=1.3829787 
y=-2.671371 
Two real roots, 
(.26,0) &(3.73,0) 
ra
x=-.5053192 
y=.47379034 
Two imaginal)! 
ra
x~930B511 
y=.00478157 
y1=x~2x+1 
One real rOO! (- .93,0) 
1 
22 
N. Families of functions: Graphs of fu nct ion fami lies. Changes in values of 
the parent affect the appearance of the paren t g raph. A par ent g raph is 
the basic graph in a fam ily. II the other family m mbers move up, down, 
left. right, o r turn based on changes in values . 
I. Polynomial functions 1: 
a. f(x) = x2 (22) 
b. f(x) = 2x2 (23) 
c. f(x) = .5x2 (24) 
_ 2 
d.f(x) - -x (25) 
e . f(x) = x2 + 2 (26) 
x2 ­f f(x) = 2 (27) 
g. r(x) = (x + 2)2 (28) 
h.r(x) = (x - 2)2 (29) 
2. I>olynomial functions 2: 
a. f(x) = x3 (30) 
b. f(x) = _x3 (31) 
c. f{ x) = x3 + 2 (32) 
d.f(x) = x3 - 2 (33) 
e. f(x) = 2x3 (34) 
f. f( x) = .5x3 (35) 
g. f( x) = (x + 2)3 (36) 
h.f(x) = (x - 2)3 (37) 
3. A bsolute value functions: 
a. f(x) = Ixl (38) 
b. f(x) = -Ixl (39) 
c. f{x) = 12xl (40) 
d.( f(x) = 1.5xl (41) 
e .f(x) = Ix + 21 (42) 
f. rex) = Ix - 21 (43) 
g . f(x) = Ixl + 2 (44) 
h. f(x) = Ixl- 2 (45) 
A. Rectangula r coordinates arc o f the form (x,y), and arc pl otted on the 
Cartes ian coord inate system. 
B. Poin ts are p lotted 	with two va lues, one the absci sa and the other th 
ordinate. 
C. The absc issa i the x-va lue, ca ll ed the domain. and the ord inate is the 
y-value, called the range. 
D. Many di ffe rent shape and func tion can be drawn on the 
Carte ian system. 
E. Here 	 is a g iven ang le, orig inating fr III the -axi and 
rotating counter-clockwise. This angle is represented by a 
line segment ori ginati ng at t.he o rigin, and extend ing to a 
given point (P). (46) 
R 	 Polar coordinates are of the fo rm P( r, 9), where r = the 
radiu , the di stance from the orig in (0,0) to I) (a g iven 
point), and El = the magnitude o f an angle. 
I. If r is pos itive, e is the meas ure o f any a ng le in 
sta ndard pos ition th at has segment 0 ,1' a its te rmina l 
s ide. 
2. If r is negat ive, El is the measure of any angle that has 
th ray oppos ite segment O,P as its termina l side. 
(47&48) 
G. Graphing w ith polar coordinates : 
I. Example I: 1'(4, 120 d egrees) (49) 
2. E xample 2: P ( 4. ~ ) (50) 
H. One angle graphed with polar coordinates can I' present 
severa l angle . 
I. If Pis a po int w ith polar coordinate (r, 9), then l' can 
Ef]S /P 
P(4.6) 
EB7, p 
I ( . 
P(H,O) 
EEJ
P(-r,9) 
ffi
P(4, 120) 
2 
also be g raphed by the po lar coordi nates (-r, e+ (2x + 1)1t) or 
(r, e+ 2x1t), where x i any in teger. 
2. E xa mlJle: Show fo ur differ nt pa ir 	 aofp la r coo rd i- rn. 
nate that can be re presented b the po int 1'(3, 60 I. 
degrees). 
3. (-r, e+ (2x + 1)1 80 degrees) ~ (- (3), 60 + (1)180) 
(- (3), 60 - (1)1 80), = 1'(-3, 240) or 1'(-3,120) P(4'3) 
4. (r, e+ 360x) ~ P(3, 60 + (1)360) or 1'(3, 60 + (2(360) = P(3, 780) 
hanging from rectangula r to pola r coordinates : The fo llowing 
fo rmulas are used to make this change: 
I. r = j( x 2 + y2) , e=A rctan f , x > O. 
2. 9 =Arc tan f + 11:, X < 0, a nd fj = radians. 
3. E xample I: Find the polar coordinates f, r 1'(-2,4). r = j _ 2)2 + (4 )2 
= j20 = 4.47. e= Arctan _42 + 1t = 2.03, P(4.47, 2.03). ~_~ 
4. Exam ple 2: Find the polar co rd inates for P(3 ,5). r = j()2 + 5 2) = 
34 = 5.83, El = A rctan "35 = 1.03. P(5.83, 1.03). 
hanging from polar to rectangular coordinates: The formulas used to 
make this change are : 
I. x = r cos El 
2. v = r sin e 
3. ~~a~Ple 1: P ( 4. ~), x = 4 cos ( ~) = 2, and 4 s in (~) = 
3.46 - P(2, 3.4(h. 
4. Example 2: P(5, 60°), x = 5 cos (60) = -4.76. Y = 5 sin (60°) = -1.52 = 
P( -4. 76, -1.52). 
K. Grapbing imaginary numbers with po lar coord inate: The polar 
form of a complex number i x + yi = r(cos El + i s in e). 
Example: Graph the complex number -4, + 2i, and change to polar form. 
r = ;;r+yL = j( _ 4 2 + 2i 2) = /16 + 4 = ,fiO = 4.47, El = rctan 
( _24 ) + IT = 2.68,1)0Ia r fo rm = 1'( -4, 2 i) = 4.47(cos 2.68 + i s in 2.68) 
A. Exponential properties: 
I. Multiplication: x· xh = x. + b 
Example: x2x4 = x6 
2. Division : (;: ) = x·-b 
Example: (::..64)= x lO 
3 . Distribution with multiplication : (xy)· = x·y· 
Example: (xy)s = xSy5 
4. Distribu tion with d ivision: ( yr= x: 
Exam ple: ( y / = ( ;: ) y 
5. Power of a power: (X")b = x· b 
Example: (x2)3 = x6 
6. Inverse power: x-I = { 
Exa mple: x"" = ~ 
x 
7. Root power: xI I. = .j; 
Example: XII2 = j; 
8. Rational power: x· /b = bR 
Example: x312 = R 
B. Logarithmic Properties and Logarithmic Form: 
11. 	Logarithmic Form: log.x = y, thi s is read a "the 
exponent of a to get the result x is y." 
 EE
Example: log, 100 = 10, the exponent of x to get the 
result .100 = 10 or xlO = 100. 
2. Loga rithmic I)roperties : 
a. M ultiplication: log.xy = log.x + log.y 
b. Division: log. y= logax - log.y FE
c. Power p roperty: log.xb = b * log.x 
d. rdentity property: If log.x = logaY' then x = y 
3. Change of Base property: I Lx, y and z are + numbers. 
log ) z 
and x and }' are n t = I , then, log,z = -I -- . EBog)x 
C. Solving logarithmic equations: 
Example I: Write log 1000 = 3 in exponential form: 
10" = 1000 
Example 2: olve, log, Ii = i ~ xl/4 =Ii ~ 
EfJ 
(xI /4)4 = (1i)4 ~ x = 4. 5 
Example 3: logi2x - 6) = log4(24 - 3x). 2x - 6 = 
t§24 - 3x. x = 6. 
Example 4: log (2x + 8) - log (x + 2) = I log 
2 + 8) (2x + 8) 
(x +2 ) = I ~ = (x + 2) ~ 6101 IOx +20= 2x EEJ
+ 8 ~ 8x = -12 ~ x = - L5. 
x- 1/3 Exam ple 5: log, 5 = -* ~ = 5 ~ x = - 1~5. 
D. Graphing Exponent ial and Loga rithmic Equations: 
Exam ple I: y = 3 ', and (1/3)' on the same graph. (51) 
Example 2: y = 92+1 (52) 
Example 3: y = 22. - 1 - I (53) 
 BtlExample 4: y = 22>-1 + I (54) 
Example 5: y = logz<x + 2) (55) 
Example 6 : y = log2(x - 2) (56) 
Example 7: y = log2(x - 2) + 2 (57) ttJ 
A. The notation P(n,n) = the number of permutations of n objects taken all 
at one time. 
B. The notat ion P(n,r) represents the number of perm utation of n obj cts 
I 
taken r at a time P(n,r) = !, P(n,r) = ( 11.) I'Il -r. 
C. The notation n! is read as n - factorial. 
Example I: How many ways can fi ve cartons of cereal be arranged? 
P(5,5) ~ 5! = 5432 I = 120 
Example 2: If 18 people show up to serve on ajury, hO\ many 12 person 
18!juries can be chosen? P(1 8, 12) ( 18 -12 ) ! 
18·17·16·15·14·13·12·11·10·9·8·7·6·5·4 ·3·2·1 . h 
6.5.4.3.2.1 	 , not!cetatyou 
can cancel 6!, leaving 18 ~ 7 = 8.89 X 10 12 choices. 
Example 3: A combination lock has four tumblers, and i num bered I ­
20 on the dial. How many combinations are possible . P(20,4) ~ ;gi 
_ 20·19·18·17·1 6 ·15·1 4 ·13·12·11·10· 9 · 8 · 7 · 6·5 · 4 ·3· 2 ·1 _ 
- 16·15· 14 · 13 ·12 ·11·1O·9· 8 ·7·6·5·4.J.2 ·1 ­
116,280. 
Example 4: If26 people enter a 16-mile race and all of them finish, h w many 
possible orders of fin ishing are there? P(26,26) = 26! 26! = 4.032 x 1026 
D. Permutations that contain repetitions or are placed in a Circular Pattern 
Repetitions: The number of n objects of which x & yare alike = ~. 
.y. 
C ircular Patterns = (n-I)! 
Example 1: How many word arrangements can be made from the word 
5!
radar? n = 5, x&y = a&r = 2. 2!2! = 30. 
Example 2: How many word pattern can be form ed from Mi issippi? 
ll! 
2!2!2!4! = 207,900. 
Example 3: There are 24 ch ildren who are going to play dodge ball. 10 
of them w ill start out in the circle. Ilow many ways cou ld the 
remai ning children fo rm ci rcular combinations? (n-I)! = 13! = 
6,227,020,800. 
E. Combinations: Di ffer from permutations in that order i not a cons ider ­
ation. The number of n objects taken r at a time is C(n,r) = ( II! ) . 
11- r !r! 
Example 1: A book c lub ha selected e ight books to read, how many four 
group com binations are pos ible? ~ qn,r) = ( 8! ) = 708 - 4 !4! 
Example 2: How many five card hands an be dealt from a regu lar deck 
52' 
of cards? qn,r) ~ = (52 _ 5) !5! = 2598960 
A. Synthetic division is a method used to make long d ivision of p Iynomial 
les cumbersome. 
I . It is mainly used when you have a very long numeral r. 
2. It can only be lIsed wi th a di i or in the form (x - n). 
3. 	You can convert the form (3 x + n) to (x + 1/3n) and then use synthetic 
divis ion. 
B. Synthetic Division of the fo rm x2 - 3x - 54.;- x - 9. 
I. Example I: The standard way of d ividing polynomials 
x - 9 ) X 2 - 3x - 54 
x + 6 
x - 9 ) X 2 - 3x - 54 
_(x2 - 9x) 
6x - 54 
-(6x - 54) 
o Answer = x + 6 r. 0 
2. 	Using synthetic division, we use the econd part of the divi or, but change 
the sign (+9, trom the above example). We then Ii I the coefficients of the 
div idend. From the example above, th y \\ould be I , -3, -54. 
Then bring down the coefficients one at a time, mUltiply them b} the 
divisor (+9), and add them to the next coefficient. This gives the same 
result. From the example on pg. 3: 
2..1 1 -3 -54 
9 54 
1 6 0 An wer = x + 6 r. 0 
Note: The power of x a lways begins one lower in the answer. 
Exam ple 2: 3x3 + x2 + 5x - 2 -;.- x + 1 -? 
::.113 1 5 -2 
-3 2 -7 
3 -2 7 -9 nswer = 3x2 - 2x + 7 r. -9 
Example 3: In the following example, notice that there mu t be a 
place-ho lder fo r missing powers of the variable. Z4 + Oz3 + Oz2 + 
Oz + 16 -;.- z + 2 
-21 0 0 0 16 
-2 4 -8 16 
I -2 4 -8 32 Answer = z3 _2z2 + 4z -8 r.32 
C. S nthetic divi ion of the form 3x2 + lOx - 9 + 3x~2. In thi example, all 
number are first divided by 3, then at the end of the process a ll fractions 
are changed by mul tiplication. 
Exam pie 1: 3x - 2 ) ( 3x 2 + lOx - 9) become x - t ) ( X 2 + lj) x - t 
10 - 9tl "3 3 
2 8 
3 3 
1
4 
- 3 
Answer = x + 4 r. - t ' now mult iply a ll fract ions by 3. 
Answer = x + 4 r. -1. When th i same problem is performed with trad i­
tional divi sion, the an wer i the same. 
Exam ple 2: 2x - I )( 6x 4 - x3- Ilx 2+ 9x - 2), now div ide all parts 
by 2, and use s nthetic divis ion . 
1 I I 9
t l 3 - 2 - 2 2 -I 
3 I 5 
2 2 - 2 
3 -5 2 0 Answer = 3x3 + x2 -5x + 2 r. 0 
Aga in, when standard long division is perform d, the same answer is 
atta ined . Note:I I' there w re any frac ti ons in a ny part of the an wer, 
they wou ld be removed by mul tip licat ion. 
MATRICES 
A. Matrix: rectangu lar array of elements in columns and rows. Rows are 
named before columns, therefore a 2x4matrix has two rows and four colwnn . 
B. Proper ties of Matrices: 
I. A matrix with on l one row is called a row matri x. 
2. matri that ha onl. one olum ll i ca lled a colum n matrix . 
3. Two matrices are equal if and only if they have the same dimensions 
and contai n the same identical e lements. 
4. The 	sum or a 2x3 and a 2x3 matrix is a 2x3 matrix in which the 
e lements are added to the correspond ing e lements. 
9 21
Exam ple: Find.1 + K ifJ = 1 2 4 -61 and K = f3 
1-3 0 2 1 15 
 -3 -'i' 
.J +KJ5 13 - 4 1. 
12 - 3 I I 
5. The ditference of two matrices.J - K is equa l t adding J to the add itive 
2 4 -61 1-
-9 2 1 
J ­inverse of K. .I = 1 and K = 
1- 3 0 21 
 I 3 II 
J-K J -I -5 - 81 
1-8 3 31 
6. The product ofa scalar (x), (a va lue with magn it ude, but no direct ion), 
and a matrix (.I) is d, with each e lement f . .1 mu ltip lied time the 
S·· . I 12 24 - 36 1
I
va ue x. IX tImes the matrIX .I = . 
1-1 8 0 12 1 
7. The product of 2 - two by t\\'o matrices i a two b two matrix. The 
procedure isa fo llow: .J = 12 41 ,1 6 9 1. 
16 I I 13 21 
JxK='2 ( 6 ) + 4( 3 ) 2 (9) + 4( 2 )IJ 24 26 1 
[6 ( 6 ) + t( 3 ) 6 ( 9 ) + l( 2 ) 1 139 56 1 
C. 	 sing matnce to solve systems of eq uat ions: If you have thr e systems 
of eq uation you can use an augmented matrix to find the solu tion et of 
the va riable. You mu t fo llow the e guide li nes: 
I. Any two rows Illa be interc hanged. 
2. ny row may be replaced by a non-zero multiple of that row. 
3. Any row may 	be replaced by the um of that row and the multiple of 
another. The goa l is to achieve an augmented matrix of the form; 
110 0 : x,1 
10 I 0 : y I 
100 I : z I, where x, y, z = the olution el. 
Example : Solve x - 2y + Z = 7 
3x + Y - z = 2 
2x + 3y + 2z = 7. u ing an augmented matrix. 
The augmented matri i 	 I I -2 I: 7 I 
13 I -I : 21 
12 3 2: 71 
Mult iply row I by -3 and add 10 rO\ 2 	 I I -2 I: 7 I 
107-4:-19 1 
12 3 2: 7 1 
MUltiply row I by -2 and add to I' w 3 	 I I -2 I: 7 I 
10 7 -4 :-19 1 
10 7 0: -7 1 
Mu ltiply row 2 by - I and add trow 3 	 II -2 I: 7 1 
I 0 7 -4 : -1 9 I 
100 4 : 12 1 
Add row 3 to I' W 2 	 I 1 -2 I : 7 I 
107 0 :-7 1 
10 0 ~ : 12 1 
Mu lt iply row 2 by 117 	 -2 1 I: 7 I 
01 0: - I I 
004 : 12 1 
Multiply row .... by 1/4 I -2 I : 7 1 
o I 0: -I I 
o 0 I: 3 1 
Multiply row 2 by 2 and add to row I I 0 I : 5 I 
010: -I i 
00 1 : 3 1 
Multiply row by -I and add to r w I 1 00 : 2 1 
o I 0 : -I I 
00 1 : 3 1 
The solution set = 2, -I, 3. 
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