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Calculus?
k
A Pre-calculus
Review
A. Functions
The lifeblood ofprecalculus is functions. A function is a set of points .(x, Y) such that for every x, there is one
and only nney. In short, iÿ a function, the x-values cannot repeat while the y-values can. In AB Calculus, atl of
your ÿaphs wilt come from functions.
The notataon for functions is rather ) = or f(x) - . In the f(x) notation, we are stating a rule to findy
given a value ofx.
e) f(x+h)-f(x)
1. If f(x)=x2-5x+8, find a) f(-6) b) f(3)
h
:) f(oÿ+h)-f(x)
b
a) 1(-4) = (-6)2 - 5 (-6)+ s
36+30+8
74
3
32
h
3
9 15
4 2
h
.... ÿ- 8
xÿ- + 2xh+h'--5x-5h+8-x'- +5x-8
1!
5
4
h2 + 2xh- 5h
= h(h+2x-5) _h+2x_5
h
h
Functions do not always use the variable x. In calculus, other variables are used liberally.
b) A(2s)
o) A(r + l)- a(r)
i
"J
2
A(r + l)- A(r)= g(r + l)-- :cr
[Z(2s)= ZC(2S)2 =47ÿSz]
:ÿ(>+t)
One concept that comes up in AP calculus is composition of functions. The format of a composition of
functions is: plug a value into one ftmction, determine an answer, and plug that answer into a second fimction.
3. Iff(x)=x2-x+l andg(x)=2x-1, a) find f(g(-1)) b) find g(f(-l)) c) showthatf(g(x))¢g(f(x))
9 - 1)+ 1
f(g(x)) = f(2x- 1) = (2x- 1)- -(-x
g(-l) = 2(-1)- ! =-3
:(21)=>!+1=3
f(-3)=9+3+1=13
g(ÿ) = 2(3)-t = s
=4x2-4x+1-2x+l+l=4x2-6x+3
g( f (x))= g(xz - x + l)= 2(x2-x+l)-I
=2x2-2x+1
Final!y, expect to use piecewise functions. A piecewise function gives different roles, based on the value ofx.
..
fx2-3, x>0
4. Iff(x)=12x+l, x<O'finda) f(5)
I.r(s)=25-3==l
t
i
b) f(2)-f(-l)
t:(ÿ)-:(-,) =,-(-ÿ)= ÿI
c) f(f(1))
t,r(1)=-2, :(-2)=-3t
A. Function Assignment
'
i: i:
• if f(x)= 4x--xz, find
3. f(x+h)-f(x-h)
. :(4)-:(-4)
2h
" If V(r) = 4rcr3, find
5 V(r+l)-V(r-O
iii::
• If ,f(x) and g(x) are given in the graph below, find"
7. (i-g)(3)
8. f(g(3))
• If f(x)=x2 -5x+3 and g(x)=l-2x, find
9. f(g(ÿ))
f:7
I-fÿ- 2, x>_2
• If f(x) = ÿxÿ - 1,
0 -< x < 2, find
!
I--X,
10. f(O)-f(2)
x<O
z :(:(3))
B. Domain and Range
/ First, since questions in calculus usually ask about behavior of functions in intervals, understand that intervals
can be written with a deseriptinn in terms of<, _<, >, > or by using interval notation.
Description Interval
Description
Interval
notation
[Description
! Interval
notation notation
x>a
[a,ÿ)
a<x<b
(a,b) - open interval a<x<_b
x < a
(--ÿ,a)
a <_x <_ b
(a,bJ
[a,b]-closedinterval Allrealnumbers (--%ÿ)
If a solution is in one interval or the other, interval notation will use the connector ÿ. So x _< 2 or x > 6 would
be va-itten (-ÿ,2] vJ (6,ÿ,) in interval notation. Solutions in intervals arc usually written in the easiest way to
define it. For instance, saying that x < 0 or x > 0 or (-ÿ)-<o--ÿ,ÿ is best expressed as x ÿ 0.
The domain of a function is the set of allowable x-values. The domain of a function/" is (---ÿ,ÿo) except for
values of x which create a zero in the denominator, an even root of a negative number or a logarithm of a non-
positive number. The domain of ap('ÿ) where a is a positive constant and p(x) is a polynomial is @%ÿ,,).
• Find the domain of the following functions using interval notation:
l, f(x)=x2-4x+4
2x
2. y= 6
3, y=xz_2x_3
x-6
[x;ÿ-l,x ÿ3l
4. y=,jTÿ
x2 +4x+6
5. y= 3 x-ÿ
6.
y=ÿ
The range of a function is the set of allowabley-vatues. Finding the range of fuuctions algebraically isn't as
easy (it really is a calculus problem), but visually, it is the [lowest possible y-value, highest possible y-value].
Finding the range of some functions are fairly simple to find if you realize that the range of y = x2 is [0,ÿ] as
any positive number squared is positive. Also the range of y = ÿx is also positive as the domain is [0,ÿ] and
the square roo* ofuny positive number is positive. The range of y = a" where a is a positive constant is(0,ÿ)
as constants to powers must be positive.
• Find the range of the following fimctions using interval notation:
7. y=I-xz
1
8. y=x-7
, Find the domain and range of the following functions using interval notation.
10. ,ÿ ÿ!!\\:iCÿÿ
/'i '7 \, I
Domain: (0,4)
11, Range: '[0,4)
9
B. Domain and Range Assigÿament
.... : • Find the domain of the following functions using interval notation:
t. f(x)=3
X3 --X2 ÷X
2. y=X3--Xz+X
3. y=
x
x-4
4. y- xZ_16
1
5. f(x)= 4xz_4x_3
7. f(t) :
6. y= 2ÿx-9
9. y = 5xÿx-z
10. y:log(x-lO)
tt. y:
12. y=-logx
: ii!ÿ#?
Find the range of the following functions.
13. NmX4+X2--I
14. y=lO0ÿ'
15. y= xÿ+l+l
Find the domain and range of the fonowing functions using interval notation.
16.
17.
18.
:-:
' ÿ k'.'i ti/) ;
r
f'"-7"'ÿ
i
lO
C. Graphs of Common Functions
There are certain graphs that occur all the time in calculus and students should know the general shape of them,
where they hit the x-axis (zeros) andy-aXJs @-intercept), as wel! as the domain and range. There are no
assignment problems for this section other than students memorizing the shape ofalt of these functions, in
section 5, we will talk about transforming these graphs.
y =2g
y.i nlcrÿv'0ÿ a
I
1
/
Function: y = x
Function: y = a
Function: y = x"ÿ
Domain: (-ÿ,ÿ)
Domain: (-ÿ,ÿ)
Range: [a,a]
Range: [0,ÿ)
"
2=4'7-
Function: y = x3
Range:
t
y=Itlo;
x-lalmr cÿrx: 0
yq'nt erÿpl; {I
x-iÿ,ÿo=pi: i
x4 mÿ'ÿ'ÿx: noÿ
qmÿcÿa
),.intÿrc*pt: 13
t
Function: y = -jTx
Fmÿction: y = [xl
Domain: [0,ÿ)
Domain: (--ÿ ÿo)
Domain: x ÿ 0
Domain: (0,ÿ,)
R ge:
Range:
Range: y € 0
Range: (--ÿ,ÿ)
yÿea
/
X
Function: y = tnx
y= COSX
y ÿsinx
.ÿqraÿpt; norÿ
aqrJter¢¢ÿ: r0alÿe
4 ÿtemÿ: t
,ÿ
Function: y = -
x-tÿlef,.ÿt...., 2 ' 2 ' 2, 2 ....
1
Function: y = ex
Function: y = e-*
Function: y = sin x
Function: y = cosx
Domain: (-ÿ,ÿo)
Dome:
Domain: (--ÿ,ÿ)
Domain: (--ÿ,ÿ)
Range:
Range: [-1,1]
Range: [-1,1]
Range: (0,ÿ,)
i
!1
D, Even and Odd Functions
Functions that are even have the characteristic that for alt a, f(-a) = f(a). What this says is that plugging in
a positive number a into the function or a negative number -a into the fimction makes no difference ... you will
get the same result. Even functions are symmetric to the y-axis.
Functions that are odd have the characteristic that for all a, f(-a) = -f(a). What this says is that plugging in
a negative number-a into the fimcfion wilt give you the same result as plugging in the positive number and
taking the negative of that. So, odd functions are symmetric to the origin, if a graph is symmeWic to the x-axis,
it is not a function because it fails the vertical-line test.
1. Of the common functions in section 3, which are even, which are odd, and which are neither?
Even: y=a, y= x', y=Ixt, y=cosx
Odd: y= x, y= x3, y=-, y=sin x
Neither: y -- ÿxx, y = !ax, y = eÿ, y = e-ÿ
2. Show that the following functions are even:
a) f(x)=x4-xZ+l
b) f(x)= 1
C) f(x) = Xva
:(-x)7(-x:
f(-x) = @x)4 -(-x)z ÷1
=x4-x2 +!= f(x)
3. Show that the following functions are odd:
a) f(x)=x -x
b)
e) f(x) = e"ÿ - e-'ÿ
(-x: +x
=x-x3 =-f(x)
I:(-x)
4. Determine if f(x) = x3 - x'- + x- ! is even, odd, or neither. Justify your answer.
]f(-x)=-xa-x2-x-i# f(x)sof isnoteven,
-f(x)=-x3+x2-x-lc f(-x)sof isnotodd.I
Graphs may not be functions and yet have x-axis or y-axis or both. Equations for these graphs are usually
expressed in "implicit form" where it is not expressed as "y =" or "f(x) =". If the equation does not change
after making the following replacements, the graph has these symmetries:
x-axis: y with -y
y-axis:xwith-x
origin: both x with- x and y with-y
5. Determine the symmetry for xÿ + .ÿ,+ y2 = 0.
x- axis: x'- + x(-y) + (-y)= = 0 ÿ x2 - xy + y-' : 0 so not symmetric to x - axis I
y-axis: (-x)2 +(-x)(y)+f = 0 ÿx"-xy+y= =0 so not syinmetrie to y-axist
%.
origin: (-x)'-+(-x)(-y)+y2=Oÿx2+xy+y==Ososymmetrietoofigin
1
13
D. Even and odd functions - Assignment
..... • Show work to determine if the following functions are even, odd, or neither.
I. f(x)=7
2, /(x)=2x2-4x
3. f(x) = -3x3 - 2x
8. s(±)--tsxt
9 f(x):lsxl-ax
4_ i(x): 4727
7. s(ÿ):8ÿ-ÿ
8x
Show work t6 determine if the gIaphs of these equations are symmetric to the x-axis, y-axis or the origin.
10. 4x=l
13. x=ly[
!!. yZ=2x4+6
!4. N:Iyl
12.3xZ=49
15. N:S-+2y+!
!4
E. Transformation of Graphs
:<
A curve in the form y = f(x), which is one of the basic common functions from section C can be transformed
in a variety of ways. The shape of the resulting cmwe stays the same but zeros and y-intercepts might change
and the graph could be reversed. The table below describes transformations to a general function y = f(x) with
the parabolic function f(x) : xÿ as an example.
Notation
How f(x) changes
S(x)+a
Moves graph up a units
Example with f(x) = xa
i ,.
:(x)-
Moves graph down a units
:(x+a)
Moves graph a units left
:(x- a)
Moves graph a units right
a > 1: Vertical Stretch
all(x)
i
0 < a < I: Vertica! shrink
]
J
F
:(=)
a > 1: Horizontal compress
(same effect as vertical stretch)
:(=)
0 < a < !: Horizonta! elongated
(same effect as vertical shrink)
-:(x)
Reflection across x-axis
:(-x)
. Reflection acrossy-axis
i
!5
E. Transformation of Graphs Assignment
" Sketch the following equations:
I. y =--x2
2. y = 2xz
3. y--O--2)
,t,
.... 7!
!,
F
'ÿ ,i .L iÿ= I : a
f
"
!:
f,
i"
f:
4ÿ y =2-ÿxx
5. y=ÿ!x+l+l
6. y=ÿx
T'
F
7
f,
......
i
......
7_ y=lx+ll-3
8. y=-21x- 1I+4
9. Y=-2-1
P
• k: - ,
I'
I
r,
tl. y=-2*+2
10. y=2ÿ-2
12, y = 2-zÿ
r
t'
tÿ
, .... }ÿ
1
13. y=-x-2
f,
!i
, ,
-2
14. y=-x+l
15. y=
3
16
F. Special Factorization
: :;!:: While
factoring skills were more important iÿ the days when A topics were specifical!y tested, students still
must know how to factor, The special forms that occur most regularly are:
Common factor: x3 + x2 + X = x(x2 + x + i)
Difference of squares: x' - ya = (x+ y)(x-y) orx'" - yÿ" = (x" +y")(x" -yÿ)
Perfect squares: xz +2ay+yz =(x+y)z
Perfect squares: xÿ - 2xy + y'- = (x - y)ÿ
Sum of cubes: x3 + S3 = (x + y)(x2 - xy + y2)_ Trinomial unfactorable
Difference of cubes: x3 - y3 = (x- y)(x2 + xy + 3,2)- Trinomial unfactorable
Grouping: xy+xb+ay+ab=x(y+b)+a(y+b)=(x+a)(y+b)
The term "factoring" usuatly means that coefficients are rational numbers, For instance, xz - 2 could
tenlmioal,y ÿe ÿtorÿ ÿs (x +.ÿ)(x- ÿ) b.t since ÿ is not ÿtional. we sÿ. ÿat x'ÿ-ÿ is oot ÿaotoÿUle
It is important to know that x2 +y2 is tmfactorable.
• Completely factor the folloÿdng expressions.
1. 4az + 2a
/
2. x2 +16x+64
3.4xa - 64
[4(ÿ+4)(x'4)1
4. 5x4-5y4
5. 16x2-Sx+l
l<xÿ+i>+l>-ill
7. 2X2 -- ZIOx ÷ 200
9. 8x3 + 273,'3
1(2x+ 3y)(4xÿ- 6xy * 9y2)1
10. x4 + 11x2-80
11, x4-10x2 +9
(x+l)(x-l)(x+3)(x-3)l
13. x3-xÿ-+3x-3
14. x3 + 5x2-4x- 20
12. 36xZ- 64
14(3x+ 4)(3x-4)t
15. 9-(x2+2xy+y2)
x2(x- 1)+ 3(x- 1)
xZ (x + 5)-4(x + 5)
9-(x+y)ÿ
(x-l)(xZ +3)
(x + 5)(x- 2)(x + 2)
(3+x+y)(3-x-y)
x..
t7
F. Special Factorization - Assignment
- Completely factor the following expressions
1. x3-25x
2. 30x-9x2-25
3. 3x2-5x2+2x
6. 9a4-aÿb2
4. 3xZ-3
5. t6x4-24x2y+9y2
7. 4x4 +7x2- 36
N, 250X3 -128
10. xS+17x3 +16x
11. 144+32X2- x4
t2_ 16x4" _ySÿ
13. x3-_ÿ+x2y-y3
14. xÿ-9x4-81x2+729
15. x2-8xy+1637-25
16. Xÿ+xa+x2+I
t7. x6 -I
18. x6+l
rÿL
18
G. Linear Functions
Probably the most important concept from preeatcutus that is required for differential calculus is that of linear
functions. The formulas you need to know backwards and forwards are:
Slope: Given two points (xl,yÿ) and (x,,yÿ), the slope of the llne passing through the points can be written as:
/TZ
rise hy Y2-Yÿ
run Ax xz - xÿ
Slope intercept form: the equation of a line with slope rn and y-intercept b is given by y :mx + b.
Point-sinpe form: the equation of a line passing through the points(x, yl) and slope m is given by
y-yÿ =re(x-x1). White you raight have preferred the simplicity of the y=ÿro:+b form in your algebra
course, the y - yÿ = m(x - xÿ) form is far more useful in calculus.
Intercept form: the equation of a line with x-intercept a andy-intercept b is given by x_+_y = 1.
a b
General form: Ax + By + C = 0 where A, B and C are integers. While your algebra teacher might have required
,'our changing the equation y- 1 = 2(x - 5) to general form 2x - y- 9 = 0, you will find that on the AP
calculus test, it is sufficient to leave equations for a lines in point-slope form and it is recommended not to
waste time changing it unless you are specifically totd to do so.
Parallel lines Two distinct lines are paralM if they have the same slope: mÿ = rrÿ.
Normal lines: Two lines are normal (perpendicular) if their slopes are negative reciprocals: ma .mz = -I.
Horizontal lines have slope zero. Vertical fines have no slope (slope is undefined).
1. Find the equation of the line in slope-intercept form, with the given slope, passing through the given point.
a. m=-4, (1,2)
b. m=2,(-5,1)
Iy-2:-4(x-!)ÿy---4x+6[
c. m=0,(-I,3)
y-1:2(x-5) ÿ y =-ÿ-7i
2. Find the equation of the line in slope-intercept form, passing through the following points.
b. (0,-3) and (-5,3)
a. (4,5)
m 3+3
m= 5+! =1
4+2
t
y-5=x-4=ÿy=x+l
-6
-- --T2-6- o -- %-
y+3:@xÿ y:@x-
e.
,-
and 1,
m (½+1)(41 2+4 6
{
l
[y-l:6(x-I)ÿ y=6x-l@j
3. Write equations of the line through the given point a) parallel and b) normal to the given line.
a. (4,7), 4x-2y:l
a) y-7:2(x-4)
b) Y-7:@(x-4)I
a) y-l=@(x-2) b) y-l=5(x-2)'
19
G. Linear Functions - Assignment
1. Find the equation of the line in slope-intercept form, with the given slope, passing through the given point.
=.
m =
-7, (-3,-7)
b. m = ÿt, (2,-8)
2 f
1)
o.
2. Find the equation of the line in slope-intercept form, passing through the following points.
a. (-3,6) and (-1,2)
b. (-7,1) and (3,-4)
e. (-2,2) alld (1,1)
k 3)
3. Write equations of the line tlgough the given point a) parallel and b) normal to the given line.
a. (5,-3), x+yy4
b. (ÿ5,2), 5x+2y=7
c. (-3,-4), y=-2
4. Find an equation of the Iine containing (4,-2) and parallel to the line containing (-i, 4) and (2,3). Put your
answer in general form.
5. Find k if the Iines 3x- 53' = 9 and 2x + ky = 11 are a) parallel and b) perpendicular.
!
20
H. Solving Quadratic Equations
:ÿi - Sotving quadratics in the form of ax2 + bx + c = 0 usually show up on the AP exam in the form of expressions
that can easily be factored_ But occasionally, you will be required to use the quadratic formula. When you have
a quadratic equation, factor it, set each factor equal to zero and solve• If the quadratic equation doesn't factor or
if factoring is too time-consuming, use the quadratic formula:
• -b +- .fbZ - 4ac
x=
2a
. The diseriminant b2 - 4ac will tell you how many solutions the quadratic has:
> 0, 2 real solutions (if a perfect square, the solutions are rational)
b2-4ac j=0, 1 real solution
1< 0. 0 real solutions (ÿÿon,ÿ hut A P ÿÿ
1. Solve for x.
a. xa+3x+2=0
b. x2-t0x+25 =0
C. X2-64= 0
(x+2)(x+t)=O
x = -2, x = -i
d. 2x" +9x = 18
,-
= sÿ(x- 5)2 =0
x=8,x=-8
e. 12xz +23x =-iO
j
f. 48x-64xÿ = 9
(2x-3)(x+6) =0
3)-' = o
x
5x
2
/
J(4 +5)(3x+2t--olf
g. X2 +5x=2
h. 8x- 3xz = 2
i. 6xZ+5x+3=O
x=-5-+ÿ.
o real solutions
j. x3-3x" +3x-9=0
k. _- -3
1. x'ÿ-7x2-8=O
3 2 x
6x(X 5 -31
(x-3)(xZ-3)=O
2x"-15x+18 = 0
(ax-3)(x-6) =0
x = 3, x =_+x/3
3
x= ,x=6
2. If y = 5xz -3x + k, for what values of k will the quadratic-have two real solutions?
(-3)2 -4(5)k >0 ÿ 9-20k>O=k < 91
20ÿ
21
H. Solving Quadratic Equations Assignment
f
1. Solve forx.
a. xa +7x-18=0
b. x2+x+l=O
c. 2xZ-72=0
& 12xz -5x =2
e. 20x2- 56x+t5 =0
f. 81xZ+72x+16=O
g. xZ +lOx=7
h, 3x-4x2 =-5
i. 7xÿ-7x+2= 0
k. xs -5x2 + 5x- 25 = 0
1. 2x4-15xÿ +18x2 =0
j.X+
4
1 17
--
x 4
2. If y = xÿ + kx - k, for what vatues ofk will the quadratic have two real solutions?
2x--I
3. Find ÿe domain of y = 6x2_5x_6.
22
L Asymptotes
Rational functions in the form of y= ÿ possibly have vertical asymptotes, lines that the graph of the cltrve
approach but never cross. To fmd the vertical asymptotes, factor out any common factors of numerator and
denominator, reduce if possible, and then set the denominator equal to zero and solve.
Horizontal asymptotes are lines that the graph of the fianction approaches when x gets very large or very small.
While you learn how to find these in calculus, a rule of thumb is that if the highest power ofx is in the
denominator, the horizontal asymptote is the tiney = 0. If the highest power ofx is both in numerator and
denominator, the horizontal asymptote will be the line y - highest degree doefficient in numerator . If the
highest degree coefficient in denominator
highest power ofx is in the numerator, there is no horizontal asymptote, but a slant asymptote which is not used
in calculus.
1) Find any vertical and horizontal asymptotes for the graph of y =
--X2
--X2
Y=x2-x-6- (x-3)(x+2)
Verticalasymptotes: x-3=0ÿx=3 and x+2=0ÿx=-2
Horizontal asymptotes: Since the highest power ofx is 2 in both numerator and
denominator, there is a horizontal asymptote at 3' = -1.
i
i.:iiiÿ :!
This is confirmed by the graph to the right. Note that the curve actually crosses its horizontal asymptote on the
left side of the graph.
.............
3x+ 3
2) Find any vertical and horizorÿtal
asymptotes for
the
graph of v ,=,
•
"
xZ-2x-3"
3x+3
- 3(x+!)
3
. ÿ;4:'
i 2' " .
., '¢, ' , : i :
ÿ--4ÿ-....' .:.. :
Y:x2-2x-3 (x-3)(x+l) x-3--@ÿ'ÿ i
" 'F \i
VerticaI asymptotes: x- 3 = 0 ÿ x = 3. Note that since the (x + 1) cancels, there
is no vertical asymptote at x = 1, but a hole (sometimes called a removable discontinuity) in the graph.
Horizontal asymptotes: Since there the highest power ofx is in the denominator, there is a horizontal asymptote
aty = 0 (the x-axis). This is confirmed by the graph to the right.
2xz - 4x
3) Find any vertical and horizontal asymptotes for the graph of y = x2 + 4
I"
2x2-4x 2x(x-2)
Y- xz+4
m
iH.•
xÿ+4
Vertical asymptotes: None. The denominator doesn't factor and setting it equal
to zero has no solutions.
Horizontal asymptotes: Since the highest power ofx is 2 in both numerator and
denominator, there is a horizontal asymptote at y = 2. This is confirmed by the graph to the right.
23
L Asymptotes - Assignment
• Find any vertical and horizontal asymptotes and if present, the Iocation of holes, for the graph of
x-1
1. y=--
x+5
2xz + 6x
4. y=
xZ + 5x +6
4+3x-x2
7. y=3x2
xÿ
10. y=
x2 + 4
13, y =
1
8
2. Y=-v
x"
x
5. y=x: 2ÿ--ÿ
5x+1
8. Y=xz_x_l
x3 + 4x
i1. y=
x3 - 2xz + 4x - 8
2x+16
3. y=-x+8
xz - 5
6. Y=2xZ 12
1-x-5x2
9. y= x2+x+l
10x + 20
12. y=
x3 - 2x2 - 4x + 8
X
(hint: express with a common denominator)
x x+2
/::":-Z
24
J. Negative and Fractional Exponents
In cMcuins, you will be required to perform algebraic manipulations with negative exponents as well as
fractional exponents. You should know the definition of a negative exponent: x-" = ÿ,x € O. Note that
x
negative powers do not make expressions negative; they create fractions. Typically expressions in multiplechoice answers are written with positive exponents and students are required to eliminate negative exponents.
Fractional exponents create roots. The definition of xv2 = xÿxx and xÿ/b = ÿ = (bxff)ÿ.
As a reminder:when we mnhiply, we add exponents: (x'ÿ)(xb)=xÿ+b.
Xa
When we divide, we subtract exponents: ÿ = x"-b,x ÿ 0
When we raise powers, we muIÿiply exponants: (x°)°=xÿ
In your algebra course, leaving an answer with a radical in the denominator was probably not allowed_ You had
to rationalize the denominator: ÿ changed toÿx)tÿxx)- x '
In calculus, you will find that it is not
necessary to rationalize and it is recommended that you not take the time to do so.
• Simplify and write with positive exponents. Note: # 12 involves complex fractions, covered in section K.
::: i::
(-3)-ÿ : 1 :£
= ÿ - 25xs
I(-st t l
4. (36xi°)w
5. (27xÿ)-2/3
;
(x4)-ÿ (-3)ÿ- xÿ 9[
6. (I6x-2)3/4
-I/2
8. (4x -12x+9)
:.
1
1
1
- 22-3
(xÿ,2_x)ÿ ÿ-2x3r- +xÿ
tl. (x+4)vÿ
lo. @8x)-ÿ(8)
-16
_
-16
rc'ÿ-ÿ:, ÿ 3(32)xsf3
(x-4)-1/2
_
t
6xsÿ
(x+4)ÿ (x-4)ÿ = (xÿ -16)//ÿ
12. (x-! + y-I)-1
L+LIC ) y+.
x y)
25
J. Negative and Fractional Exponents - Assignment
• Simplify and write with positive exponents.
1.-122x-s
2. (-12xS)-2
3. (4x-ÿ)-ÿ
6.(ÿ3<)-2
7. (121xS)I/a
8. (8xZ)4/ÿ
9. (-32x-S)-3/5
10. (x + y)-2
11. (x3 +3x2 +3x+l)-:ÿ
tÿ 4ÿ"-x)-=
13. I (!6x2)-3!" (32x)
!4. (xZ - 1)-V2
15. (x-2÷2-T
26
K2 Eliminating Compiex Fractions
::ÿ!ii)ii:ÿ-
Calculus frequently uses complex fractions, which are fractions within fractions. Answers are never left with
complex fractions and they must be eliminated. There are two methods to eliminate complex fractions:
When the probIem is in the form of ÿ, we can "flip the denominator" and write it as a d ad
7
b c bc
However, this does not work when the numerator and denominator are not single fractions. The best way is to
etiminate the complex fractions in all eases is to find the LCD (lowest common denominator) of at1 the fractions
in the complex fraction. Multiply all terms by this LCD and you are left with a fraction that is magically no
l+x-ÿ
I+Iÿ
longer complex. Important: Note that ÿ- can be written as -Yx but 7- must be written as 1X
•EIiminate the complex fractions.
2
2
-
3
t.ÿ
g
3-+5
I+2.
3
5
3, 4 3
l+g
2-g
1
/1+2)ÿ6) 6+4 10
29
22
/
H--
t
x---
1+!X-ÿ
,
2
l+!_x-ÿ
3
5.
2xS/ÿ
2x
6. 5
x2 q- ÿx2
5
7
1
l+3ÿflk6x2 6x+21
7. --
xÿ+ ! ][,4xÿ) 4x'+l
)
(x_i),ÿ x(x-*)-ÿ
x-3 q-x
x-=+l
8. t(2x + 5)-ÿ3
9.
-2
2
x-1
3
"(x-l)v'- 2(xXl)v2)U2(x_l)w-
6_
-3
a- 4(i iV'
FT_.-E'ÿ_..
>.(x-1)-x = x-____L_2
.2(x-t)3/2 2(x-l)3/2
>
27
K. Eliminating Complex Fractions - Assignment
• Eliminate the complex fractions.
5
I. ÿ
-2
--
3
1
. mx
x--i
X+--
2
4---2.
2+__7+__3
3.
9
4
3+--
2 5
3
4
5----
3
5. --
l+x-ÿ
,
I_x-z
X
x-2+x-ÿ +1
.
X-2 -- X
8. l(3x- 4)-ÿ/4
-3
2x(2x_l)
.
(2x-l)
4
!
;
28
L. Inverses
- 115
No topic in math confuses students more than inverses. Ira function is a role that maps x to y, an inverse is a
rule that brings y back to the original x. If a point (x,y) is a point on a ftmctionfi then the point (y,x) is on the
inverse function f-I . Students mistakenly beIieve that since x-ÿ = 1, then f_ÿ = 1 This is decidedly incorrect.
x
f
If a function is given in equation form, to find the inverse, replace all occurrences ofx with y and all
occurrences ofy with x. If possible, then solve fory. Using the "horizontal-line test" on the original functionf
wilt quickly determine whether or not f-ÿ is also a function_ By definition, f (f-ÿ (x)) = x. The domain of f-ÿ
is the range of f and the range of f-a is the domainoff
4x+5
1. Find the inverse to y = -- and show graphically that its inverse is a function.
Inverse: x- 4y+5 =,xy_x=4y+5ÿy=X+5]
y-I
x-4]
Note that the function is drawn in bold and the inverse as dashed. The
fimction and its inverse is symmetrical to the line y = x. The inverse is a
function for two reasons: a) it passes the verticaI line text or b) the function
passes the horizontal line test.
2. Find the inverse to the following ftmcfions and show graphically that its inverse is a function.
b. y=x-'+!
a. y =4x-3
Inverse: x = 4y- 3
x+3
= -- (function)
4
c. y:x'ÿ +4x+4
Inverse: x = y'- +4y+4
inverse: X = Yÿ-- + 1
1
= ±4ÿ-I (not a function)/
[
x:(y+ 2)z ÿ +ÿx : y+2
3' = -2-+ ÿxx (not a function)
3. Find the inverse to the following functions and show that f(f-i (x)) = x
a. f(x)=7x+4
b. f(x)= 1
c. f(x)= x3-1
x-i
1
Inverse: x = 1 ÿ ÿ Ix = 1
y-1
Inverse: x = 7y+4
y= f-I(x)=X+l
Xÿ
y= f-l(x)=X-4
7
Inverse: x: y3 --I
y= f-I(x)= 3 xÿ
3
3
:mÿx
x+t--x
X
4. Without finding the im, erse, find the domain and range of the inverse to f(x) = -,Ix- 2 + 3_
[Function: Domain: [-2,ÿ), Range: [3,ÿ)
Inverse: Domain: [3,,,@ Range: [-2,ÿ)[
29
L. Inverses - Assignment
/....
1. Find the inverse to the following fanctions and show graphically that its inverse is a fanction.
a. 2x-6y=l
b, y=ax+b
c. y=9-x2
9
d.y= lÿ-xÿ
e.y=-
2x+1
£Y=3-ÿ2--'ÿ
7
2. Find the inverse to the following functions and show that f(f-i (x)) = x
1
4
a, f(x)=ÿ-x-7
b. f(.::)= x2 -4
x2
o. f(x)= x2 +1
4;7-7
3. Without finding the inverse, fred the domain and range of the inverse to f(x) = xZ
7ÿ: -,
>.
3O
M. Adding Fractions and Sob4ng Fractional Equations
•. z
There are two major probtem types with fractions: Adding/subtracting fractions and solving fractional
equations. Algebra has taught you that in order to add fractions, yon need to find an LCD and multiply each
fraction by one ... in such a way that you obtain the LCD in each fraction. However, when you sulve fractional
equations (equations that involve fractions), you still fred the LCD but you multiply every term by the LCD.
When you do that, all the fractions disappear, leaving you with an equation that is hopefully solvable. Answers
should be checked in the original equation.
X X
!. a, Combine: ---
3 4
b. Solve: x-x=12
3 4'
[LOD:,2 3\4) 4\3j
4ÿ- 3x = !44 -ÿ. x - ldd
4x-3x_ x
12
144 144
=48-36=12
3 4
6
12
x=144:
6
2. a. Combine x+-
b. Solve: x+-=5
x
X
x2 +6= 5x ÿ xZ-5x+6=O
(x-2)(x-3)=Oÿ x=2, x=3
x=2:2+-6=2+3=5 x=3:3+6=3+2=5
3. a. Combine:
12 4
x+2 x
LCD: x(x + 2)
-;-/)ÿ 7j- Z,7;5ÿ
(12)(x1
Solve
3
=1
x+2 x
12 (x)(x+2) 4(x)(x+2)=l(x)(x+2)
x+2
x
o
2
12x-4x-8= x- + 2xÿ x -6x+8=0
12x-4x-8
x(x+2)
(x-2)(x-4) =0 =ÿ x= 2,4
8x-8
=2:L2_4=3_2=1 x=4:12_4=2_1=1
x(x+2)
4. m
b.
2
12 4
4 2
X
3
2x-6
xZ--6x+9
b, Solve
x
6
3
4
x-2
2x-6 xZ-6x+9 3x-9
x
3 , =ÿ16(x-3)a
2(x-3) (x--3)-'- ("-)j
LCD: 2(x - 3)2
p-q 3 ,,11
2(x-3)[.x-3j (x-3)- ÿ,2)
;x(x- 3)-18= a(x- 3)(x- 2)
3X2 -9x-I8 = 2xÿ" -10x +12
;a +x- 30: 0 ÿ (x+6)(;ÿ-5)=0 = ÿ:=-6,5
x'- -3x-6'
Fÿ-7"\
, 2(x- 3f
3 -8
. 5 3 3
X=--6 -6
2------=-X=3:------=--
-18 81 -27
4 4 6
3t
M. Adding Fractions and Solving Fractional Equations - Assignment
2
21
b. Solve:
1. a, Combine:---
3 x 6
3x
!
1
2. a. Combine:--+-xÿ3
5
3. a. Combine:
x-1
t
i
b. Solve:--+
ÿ;--ÿ3 x+3
5
2x 3x+t5
2x - 1
4. a. Combine:
5
....
3x
2x+l
b. Solve:
5
5
!0
x2-9
5
2x 3(x+5) x
2x - 1
b, Solve:
x-I
3x
2x+l
xÿ + 11
2xZ-x-1
r7-;
\
32
N. Solving Absolute Value Equations
Absolute value equations crop up in calcnlus, especially in BC calculus. The definition of the absolute value
•
.
ÿ
:
" ÿx
if x -> 0
fi.mctlon Is a p,ecewise function: f(x) : [xI =-[
. So, to solve an absolute value equation, split the
x ifx<0
absolute value equation rote two equations, one with a positive parentheses and the other with a negative
parentheses and solve each equation. It is possible that this procedure can lead to incorrect solutions so solutions
must be checked.
• Solve the following equations.
t. Ix-11:3
2. 13x+21:9
-x+l=3
X=4
x = -2
I x-t=3 -(x-t)=3
[
i
?7 :
3x=7
3x=-lI
-11
X=__
3
3
4. tx+sl+5=o
x+5+5=o -(x+{)+5=0
-(2x-1)-x = 5
x=6
-3x-2=9
X=-7
3.12x-ll-x= 5
2x-l-x=5
3x+2=9
-3x = 4
x = -10
-x-5+5=0
x=O
-4
Both answers are invalid. It is impossible
3
to add 5 to an absolute value and get 0.
6. Ix-lO!= xÿ'-lOx
5. Ixÿ-xl= 2
-(x- 10) = x2 - 10x
x- 10 = x2 - 10x
xZ-x-2=O
(x-2)(x+l)=0
x = 2,x = -1
-x+!0 = x'- -10x
-x2+x=2
x2-11x+10 =0
x2 -9x-lO=0
0=x2+x+2
(x<)(x-10) =0'
No real solution
x=l,x=10
(x-10)(x+l)=0
x=t0,x=-I
Both solutions cheek
Of the three solutions, only x = -1 and x = 10 are valid.
7. [xl+12x- 21= 8
x+2x-2=8 -x+2x-2 =8 x-(2x-2):8 -x-(2x-2)=8
3x = 10
x = 10
-x = 6
10
Xÿ3
f- ......
% •
x = -6
-3x =6
x = -2
10
3
Of the four solutions, only x = -- and x = -2 are valid
33
N. Solving Absolute Value Equations - Assignment
zÿ
• Solve the following equations.
!. 4Ix + gt = 20
2.11-7x1:13
3.18+2xl+2x=40
4. 14x-5t+ 5x+2 = 0
5. Ix2-2x-11=7
6. }12-x]= xz -lax
7. [xl+14x-4l+x=14
34
O. Solving Inequalities
" You may think that solving inequalities are just a matter of replacing the equal sign with an inequality sign. ha
!: reality, they can be more difficult and are fraught with dangers. And in calculus, inequalities show up more
frequently than solving equations. Solving inequalities are a simple matter if they are based on linear equations.
They are solved exactly like linear equations, remembering that if you multiply or divide both sides by a
negative number, the direction of the inequality sign must be reversed.
However, if the inequality is inure complex than a linear function, it is advised to bring all terms to one side.
Pretend for a moment it is an equation and solve. Then create a number line which determines whether the
transformed inequality is positive or negative in the intervals created on the number line and choose the correct
intervals according to the inequality, paying attention to whether the zeroes are included or not.
If the inequality involves an absolute value, create two equations, replacing the absolute value with a positive
F
solution on your number line. Then determine which intervals satisfy the original inequaIity.
If the inequality involves a rational function, set both numerator and denominator equal to zero, which will give
you the values you need for your number line. Determine whether the inequality is positive or negative in the
intervals created on the number line and choose the correct intervals according to the inequality, paying
attention to whether the endpoints are included or not.
• Solve the following inequalities.
2. 1-3x >x-5
1. 2x-8 <_6x+2
d
-10 _< 4x
3. -5_<6x-I<11
2
2-3x>2x-10
-4x _< 10
-6-<6x_<t2]
--5<x
or
x>-5
2
,
12>5xÿx<ÿ
2
t2x-ll<__x+4
-1-<x_<2 J
5. xZ-3x>18
12x-ll-x-4_< 0
2x-l-x-4=O -2x+1-x-4 =0
Xa-3x-18>Oÿ(x+3)(x-6)>O
x=5
For (x+3)(x-6)=0, x=-3,x=6
x=-l
+++++++0
0+++++++
-1
5
SO-l_<x_<5 or [-1,5]
6ÿ
0+++++++
-3
6
Sox<-3orx>6 or
2x-7 <1
7. Find the domain of
x-5
2x-7
x-5
++++++0
1=0 2x-7 x-5<0 x-2<0
x-5 x-5
x-5
++++++0
2
So 2_<x<5 or [2,5)
ÿ+++++++
5
2(4+x)(4-x) >-0
0+++++++0
-4
4.
o -4 _< x _< -4 or [-4,4]
35
O. SoLving Inequalities - Assignment
• Solve the following inequalities.
t. 5(x-3)_< 8(x+5)
2. 4-ÿ >-(2x+I)
5. (x + 2)-" < 25
6. x3 < 43c2
8. Find the domain of
36
P. Exponential Functions and Logarithms
Calculus spends a great deal of time on exponential functions in the form of bÿ. Don't expect that when you
start working with them in calculus, your teacher wili review them. So learn them now! Students must know
that the definition of a logarithm is based on exponential equations. If y = b xthen x = togÿ y. So when you are
trying to fred the value of logz 32, state that Iogz 32 = x and 2.7 = 32 and therefore x = 5.
If the base of a log statement is not specified, it is defined to be 10. When we asked for log I00, we are solving
the equation: 10ÿ = !00 and x = 2. The fimetion y = tog x has domain (0,ÿ) and range (-ÿ,,ÿ). In calculus,
we prinÿrity use logs with base e, which are called natural logs (In). So finding in 5 is the same as solving the
equation eÿ = 5. Students should know that the vaIue ore = 2.71828...
There are three roles that students must keep in mind that will simplify problems involving logs and natural
logs. These rules work with logs of any base including natural logs.
i. toga+iogb=log(a.b)
1. Find a. log48
iii. Iogab = bloga
e. i0ÿ°g*
b,
log4 8 = x
]
l
x : el!2
0z =4 SO I0'°g4 =4
q
d. fog2 + tog50
i!og4
x and log are inverseÿ
0 to a: power
e. logÿ 192-tog43
Iog(2- 50) = log 100
log464 = 3j
2. Solve a. iOgg(XZ-X+3]=l
J 2
b. log36 x + tog36(x-1) =t
e. Inx-ln(x-t)= I
logÿ x(x- 1) = 1
d.
2-x+3=9v2
x (x - 1) = 36v" -- 6
(x-1)=O
x'--x-6=O
=0,x=t
(x-3)(x+2) =0
x
--= e :::ÿ X = ÿ--e
x--I
e
Only x = 3 is in the domai
d. Y=20
f. 2ÿ = 3ÿ-ÿ
e. e-ZX = 5
[log(5*)= 10ÿ20 ]
! log5: log2O
lne-2" = in5
|
e--1
=@
l°g(2X) = l°g(3ÿ-')
1
xlog2 = (x - 1)log3
-2x=in5ÿx
.
Ix- l°g20 orx: In20I
5-ÿ [ log5
InSI
log3
xlog2 = xlog3-1og3 ÿ x = l°g3-1°g2l
37
P. Exponential Functions and Logarithms - Assignment
I
t. Find a. log2ÿ
d. 5I°ÿ40
g-togz 2+ ]og2 3-ÿ
2, Solve a. Iogs(3x-8)=2
,L logÿ (ÿ- 1) + logÿ(ÿ + 3): s
g. 3ÿ-2 =18
b. logs4
e. ÿln12
I
c, in--
U
£ Ioglz 2- toglz 9- log12 8
h. [ogÿ ÿ-ÿ-[ogÿ i2
3
b. log9 (x2 -x + 3)= ÿ"
e. logs(x+3)-lOgsX=2
h. e3"+j =10
c. log(x-3)+log5 = 2
f. lnxÿ inx2
1
2
i. 8ÿ=5''-1
rÿZ .....
38
Q. Right Angle Trigonometry
ii: Trigonometry
aa.integral
of AP calculus.
must know
the basic
ftmction
definitions in terms
of opposite, is
adjacent
mad part
hypotenuse
as well asStudents
the definitions
if the angle
is in trig
standard
position,
Given a right triangle with one of the angles named 0, and the sides of the triangle relative to 0 named opposite.
(y), adjacent (x), and hypotenuse (r) we define the 6 trig funetions to be:
sin0 = opposite _ y
csc0- hyptotenuse _ r
hypotenuse r
I"/ÿ !
¢/
opposite
hyptotenuse
1"
sec0 ....
cos0 =- adjacent _ x
i
hypotenuse r
opposite
_ _=
-- = -
y
adjacent
x
_ adjacent - x
cÿMÿ
y
adjacent x
opposite y
The Pythagorean theorem ties these variables together: x- + y = r-. Stadents
.,*-J
should recognize right triangles with integer sides: 3-4-5, 5-12-t3, 8-I5-17,
i Y
7-24-25. Also an3' multiples of these sides axe also sides of a right triangle.
Since r is the largest side of a right triangle, it can be shown that th6 range of
sin0 and cos0 is [-1,1],the range of csc0 and sec0 is (--ÿ,-l]u[1,ÿ)
and the range of tan0 and cotO is (--ÿ,ÿ).
Also vital to master is the signs of the trig functions in the four quadrants. A good way to remember this is
i A - S - T - C where A!! trig fimctions axe positive in the 1 ÿ quadrant, Sin is positive in the 2ÿd quadrant, Tan is
-ÿrd
th
:::: positive in the J quaÿant and --Cos is positive in the 4 quadrant.
--
-
1. Let be a point on the terminal side of 0. Find the 6 trig fimctions of 0. (Answers need not be rationalized).
a) P(-8,6)
b. P(1,3)
q
csc 0 =
5
secO=-5
tan0 = ---ÿ
4
5_
3
cos0=-!
=-4K y =-,g, = 4
x=l, y=3, r =.!io
x=-8,y = 6, r=IO
sin 0 = =
c_ P(--Jÿ,-,ff)
cot0 = _4
3
2. K cos0 =2, 0 in quadrant W,
sin0 = ÿ10
cs°0= iÿ3
cos0=ÿl0
secO=.,/ÿ
tan0=3
cot0=-1
3
3. If seo0 =-j3
sin0 = -'f64
cso0 = -ÿ6-6
.cos0 = _ 1ÿ04
[tan0 =
sec0 = --ÿ
cot0
L
4_ Is 3cosO+4=2possible?
fred sin 0 and tan0
f'md sin0 and tanO
0 is in quadrant I or IV
x=2, r=3, y=-4ÿ
3cosO = -2
=1, y= _+./5, ,-=./5
3
ms0 = _2 which is possible.
tanO =ÿ_ÿ5
2
sin 0 = ±ÿ-23-2y tan0 = ±ÿJ
3
39
Q. Right Angle Trigonometry - Assignment
1. Let be a point on the terminal side of 0. Find the 6 trig functions of 0. (Answers need not be rationalized).
a) P(t5,8)
b. P(-2,3)
c. P(-2ÿ/5.-ÿ/5)
19
If tan 0 = 5" 0 in quadrant m,
.
fend sinO and cosO
3. IfcseO = , 0 in quadrant 1I,
find cosO and tanO
4.
cotO=
3
find sinO and cosO
5. Find the quadrants where the following is true: Explain your reasoning.
a. sinO>0 and cosO<0
b. cscO<0 andcotO>0
c. all functions are negative
6. Which of the following is possible? Explain your reasoning.
a. 5sinO=-2
b. 3sina+4cosfl=8
c. 8tan0+ 22 = 85
rÿ
€ ...............
4O
R. Special Angles
Students must be abte to find trig functions of quadrant magles (0, 90°, I80°,270°) and special angles, those
based on the 30°-60°-90° and 45°-45°-90° triangles.
First, for most calculus problems, angles are given and found in radians. Students must know how to convert
degrees to radians and vice-versa. The relationship is 2a- radians = 360° or ¢r radians = 180°. Angles are
assumed to be in radians so when an angle of -ÿ is given, it is in radians. However, a student should be able to
180°
o
picture this angle as -- = 60 . It may be easier to think of angles in degrees than radians, but realize that
3
unl.ess specified, angle measurement must be written in radians. For instance, sin-' (!1 = 1c.
1.2) 6
¢r 3zcÿ
90°,1800,270° or u,ÿ-,zc, ÿ-j can quickly be found. Choose a point
The trig functions of quadrant angles
along the angle and reatize that r is the distance from the origin to that point and always positive. Then use the
definitions of the Irig functions.
i
i ii
i llllll
0
pomt
0
(I,0)
y
1'
ii
0
r
sin0
1
cos0
0
tanO
!
0
, ,
.ÿ
: :
:
-- or 90°
2
( )0,1
zÿ or 180°
(-t,0)
Hill
.......
3ÿor 270°
--
2
does not
i,,ll
0
-1
1
0
1
1
ii, I
0
0
-1
exist
-1
0
-
1
i
does
0 not
1
exist
(0,-1)
ii i ii
csc0
1
I
exist
i
!
does not
exist
exist
does not
1
exist
does not
Does0not
-1
i ill i
i
sec0 I i cot0
0
III
does not
exist
iii
does not
-1
0
exist
If you picture the graphs of y = sin x and y = cos x as shown to
the right, you need not memorize the table. You must know
these graphs backwards and forwards.
• Without looking at the table, find the value of
a. 5cos180° - 4sin270°
5(-i)-4(-1)
-5+4=-1
8(13-6(o3 ( 81 4/
k.
41
Because over half of the AP exam does not use a calculator, you must be able to determine trig functions of
special angles. You must know the relationship of sides in both 30° - 60° - 90° \-ÿ, ÿ,ÿ)
and 45°-45°-90° ÿ,ÿ ÿ) triangles.
43
In a 30°-60°-90° triangle,
tn a 45°-45°-90°
the ratio of sides is t - -]3- 2.
the ratio of sides is 1 - 1- ÿ/2.
\6 3 2]
0
sin 0
30° or
Special angles are any multiple of 30°
cos 0
ÿ
,ÿ,
triangle,
tan 0
ÿ-
or 45°
_ To find trig fimetions of any of these angles, draw
them and find the reference angle (the angle created with the x-axis). Although most problems in calculus will
use radians, you migaht think easier using degrees. This will create one of the triangles above and trig fimetious
can be foand, remembering to include the sign based on the quadrant of the angle. Finally, if an angle is outside
the range of 0° to 360°(0 to 2ÿ:), you can always add or subtract 360°(2ÿ) to fred Ng functions of that angle.
These angles are called co-terminal angles. It should be pointed out that 390° ÿ 30° but sin390° = sin30°.
• Find the exact value of the following
a. 4sin120°-Scos570°
Subtract 360° from 570°
4sinl20°-8cos210°
120° is in quadrant II with reference angle 60°.
(
b. 2cos3r- 5tan
(2cos180°- 5tan3t5°)a
180° is a quadrant angle
315° is in quadrant Ill with reference angle 45ÿ
210° is in quadrant t1I with reference angle 30°.
•-2(-1)- 5(-1)]2 = 9
42
I
R. Special Angles - Assignment
• Evaluate each of the following without looking at a chart,
1. sin2120° + cosÿ 120°
2. 2 tan2300° + 3 sin2150° - cosz 180°
3. cotz 135° - sin 2t 0° + 5 cos2225
4. cot (-30°) + tan (600°)- csc(-450°)
cos 2ÿ tan 3zÿ /2
5. T- T/
6./.s,nT- tan T)(si= T + tÿT)
('. llz
5x)(
llz
5zÿ
• Determine whether each of the following statements are true or false.
• ÿ
. ÿ
. f.
5ÿr
5ÿ
cos -- + l cos --
ÿ:'ÿ
7_ +sinT: s,n( ÷T)
8.
3
_=3
-- sec 5zÿ
-- - 1
tan2 5;,'z-
3
3
oosÿ4ÿ +sin4ÿ
9. 2(ÿ+ sinÿ)(l+cosÿ) > 0
10.
3
3>0
COS2 4ÿ
3
r ......
44
S. Trigonometric Identities
"-' Trig identities are equalities involving trig fimctions that are tree for all values of the oecam4mg angles. While
you are not asked these identities specifically in calculus, knowing them can make some problems easier. The
following chart gives the major trig identities that you should know. To prove trig identities, you usually start
with the more involved expression and use algebraic roles and the fimdamentat trig identities. A good
technique is to change all trig fi.metions to sines and cosines.
Fundamental Trig Identifieg
1
1
1
sinx
cosx
cscx=--,
secx=--,
eotx=--,
tanx=--,
cotx=-sinx
cosx
sm-x+cos-x= ,
tanx
l+tan-x=see x,
COSX
sinx
l+cot-x=csc-x
Sum Identifies
sin(A + B)= sin AcosB +cosAsinB
cos(A + B)=cosAcosB-sinAsinB
Doubte Angle Identifies
sin(2x) = 2sinxcosx
cos(2x)=cos-x-sm-x=l-2sm x=2cos x-!
• Verify the following identities.
I. (tanÿ x + l)(cos2-1) = -tanZ x
2, secx-cosx = sin xtanx
_____cosx(C°sx1
cosx
\cosxj
1 - cosz x sin Zx
COS x
os - x / -
!sec x)
COS X
tatlÿ x
cot2x
3.
l+sinx
1-sinx
.
=
l+cscx
cosa ÿ Sin2 X -sin2 x sin-' x
1+ÿ1
cosx
sinx
cos'- x
sinÿx+sinx
sinx j
l-sinÿx (I +sinx)(1-sinx)
sinx(l+sinx) sin41+sin )
t-sinx
sinx
cosx
t---=2secx
I + sinx
rl+sinx)(l+sinxÿ+( cosx ÿ(cosx)
cosx jkl+sinxj \t+sinxJ\cosxJ
1 + 2 sinx + sin2+ cosz x
cosx(1 + sin x)
t+2sinx+l
2+2sinx
cosx(1 +sinx) cosx(l+ sinx)
2(I +sinx) = 2secx
:cosx(! +sinx)
5, cos42x--sin42x = COS 4x
6. sin(3u-x) = sinx
!(cosZ 2x + sin2 2x)(eosa 2x--sin= 2x)
sin3Jicosx- cos 3ÿr sinx
,,:i.ÿ ,
IEcos2(2x)]
0(cosx) - (-l)sinx = sin x
cos4x
45
S. Trig Identities - Assignment
• Verify the following identities.
2. seca x+3 = tan2x+4
[ - SeCX
3.
- -secx
1 -- cosx
4.
1
l+tÿnY
7. csc2xCSCX
=-2cosx
1ÿtÿntr
sin3 x+ cos3 x
5. cosx-cosy + sinx-siny =0
sinx+siny cosx+cosy
1
.+--=1
.
sinx+cosx
t-sinxcosx
eos3x
8. = t -4sinz x
cosx
46
T. Solving Trig Equations and Inequalities
#- Trig equations are equations using trig functions_ Typically they have many (or infinite) number of solutions so
usually they are solved within a specific domain. Without calculators, answers are either quadrmat angles or
special angles, and again, they must be expressed in radians.
For trig inequalities, set both numerator and denominator equal to zero and solve. Make a sign chart with all
these values included and examine the sign of the expression in the intervals. Basic knowledge of the sine and
cosine curve is invaluable from section R is invaluable.
• Solve for x on [0,2K)
2, tanx+sin2x=2--cos2x
1. XCOSX = 3COSX
Do not divide by cosx as you wilI lose solutions
cos x (x- 3) = 0
cosx=0
zc 37c
t21nx +siliX x +cosX x = 2
tanx+l =2
tanx=l
x-3=0
7c 5Jr
x=3
x=ÿ,7
2'2
Two answers as tangent is positive
You must work in radians.
in quadrants I and III.
Saying x = 90° makes no sense.
4. 3cosx = 2sin2 x
3. 3tan2x-I =0
3cosx-2(1-cos2x)
3tanZx- !
2cosZ x + 3cosx- 2 =O
(2cosx- 1)(cosx + 2) = 0
2Ln2x=!
3
2cosx = 1
nx=+ ÿ-1 =+"]ÿ
-ÿ13 - 3
1
COSX = --
cosx = -2
No solution
2
5Ic 7a" 11To
x:6'6"6' 6
X--
7ÿ 5;'/:
3'3
7. Solve forx on [0,2ÿ) : 2cosx+lsin:____ÿ > 0
-1
2ÿr 4Jr
2cosx=-lÿcosx=--ÿ x=
2
sin- x= 0 ==> x = 0,E
2ÿ
++++++@ .......
3'3
0
27f
-5-
0+++++++
4ÿ
-3
2ÿ
4at
Answer: f0,-ÿ- ) u (-ÿ-,2ÿ")
47
T. Solving Trig Equations and Inequalities - Assignment
.
" Solve for x on [0,2ÿr)
1. sinZx =siux
2. 3 tan3 X = taP.x
3. siIlz x = 3C0S2 X
4. cosx+sinxtanx=2
5. sinx = COSX
6, 2cos2x+sinx-1 =0
:72 i
7, Solve forx on [0,21r): x-iv
COS2---ÿ < 0
48
U. Graphical Solutions to Equations and Enequalities
• You have a shiny new calculator• So when are we going to use it? So far, no mention has been made of it. Yet,
the calculator is a tool that is required in the AP calculus exam. For about 25% of the exam, a calculator is
permitted. So it is vital you know how to use it.
There are several settings on the calculator you should make. First, so you don't get into
rounding difficulties, it is suggested that yon set your calculator to three decimal places.
That is a standard in AP caIcutns so it is best to get into'the habit. To do so, press ÿ
and on the 2nd line, take it offFLOAT and change it to 3. And second, set your calculator to
radian mode from the MODE screen. There may be times you might want to work in
degrees but it is best to work in radians.
N
2 .ÿzet-9
3: Nin imÿ
5: intepseot
"Idÿ_/dx
You must know how to ÿaph functions. The best way to graph a function is to input the
see the graph's total behavior by pressing ÿ 0. To evaluate a fimetion at a specific value of x, the easiest
way to do so is to t3ress these keys: @ [ÿ [1 :Function Iÿ ÿ [] and input your x-value.
Other than basic calculations, and taking trig fimctions of angles, there are three calculator features you need to
know: evaluating functions at values ofx and finding zeros of fimctious, which we know is finding where the
function crosses the x-axis. The other is finding the point of intersection of two graphs. Both of these features
are found on the TI-84+ calculator in the CALC menu ÿ ÿ. They are l:value, 2: zero, and 5: intersect_
Solving equations using the calculator is accomplished bÿsettinÿÿion equal to zero, graphing the
fimction, and using the ZERO feature. To use it, press ÿ ÿ EÿERQl. You will be prompted to type in a
number to the left of the approximate zero, to the right of the approximate zero, and a guess (for which you can
press ÿ. Yon wili then see the zero (the Sohltion) on the screen.
.- Solve these equations graphically.
2. 2cos2x-x= 0 on [0,2rc) and find 2cos(2e)-e.
t. 2x2-9x+3=0
I" ........ " [
xyÿ
%y7ÿ
I14 If j
This equation can be solved with the quadratic formula.
I I¥ÿ(e)
"1.393
If this were the inequality 2cos2x-x > 0,
the answer would be [0,0.626).
x 9_+
= ,NiaN=4 -2
3. Find the x-coordinate of the intersection of y = xÿ and y = 2x - 3
I You can use the intersection feature.I
...... " ..... jJ .... Jl
::::;ÿz,,:-s
.............. i .,y,=
....
I Or set them equal to each other: x; = 2x - 3 or xs - 2x + 3 = 01
tÿoo tlÿtÿ tloÿ
Z]
IIÿ',•ÿ'%'p,ÿ'b-"•'ÿ" l
49
U, Graphical Solutions to Equations and Inequalities - Assignment
. " Sotve these equations or inequalities graphically
1. 3xÿ-x-5=O
2. x3-5xz +4x-l=O
x2 - 4x - 4
5. xÿ-9xZ-3x-15<O
6.
7. xsinxÿ >0 olÿ [0,3]
-' x>°x- OIl
[-1,1}
8ÿ cos
xÿ+1
>Oon[O,8]
5O