9/5/2006pre-calculus r r { [ 4, ) } { (- , 3 ] } { r \ { 2 } } { r \ { 1 } } { r \ { -3, 0 } } r {...

$: 9/5/2006Pre-Calculus R R { [ 4, ) } { (- , 3 ] } { R \ { 2 } } { R \ { 1 } } { R \ { -3, 0 } } R { (- 3, ) } { (- , 4 ] U [ 2, ) } { (- , -1)$
9/5/2006Pre-Calculus

R

R

{ [ 4, ) }

{ (- , 3 ] }

{ R \ { 2 } }

{ R \ { 1 } }

{ R \ { -3, 0 } }

R

{ (- 3, ) }{ (- , 4 ] U [ 2, ) }

{ (- , -1) U [ 0, ) } { [ 0, ) }

R

{ [ -8, ) }

{ [ 0, ) }

{ [ 0, ) }

{ R \ { ½ } }


continuous discontinuousinfinite

discontinuousremovable

continuous discontinuousremovable

discontinuousjump

discontinuous - jump

continuous

discontinuous - infinite

continuous

continuous

discontinuous - infinite


(3x+4)(x<1)+(x-1)(x>1)

jump

(x^3+1)(x0)+

(2)(x=0)

removable

(3+x2)(x<-2)+(2x)(x>-2)

(x<1)+(11-x2)(x>1)

jump


incr: (- , ) decr: (- , 0 ]incr: [ 0, )

decr: (- , 0 ]incr: [ 0, )

decr: [ - 1, 1 ]incr: (- , -1 ], [ 1, )

decr: [ 3, 5 ], incr: [ , 3 ]constant: [ 5, )

decr: [ 3, ), incr: ( 0 ]constant: [ 0, 3)

decr: ( - , )

decr: (- , -8 ]incr: [ 8, )

decr: ( - , 0 ]incr: [ 0, 3 )

constant: [ 3, )

decr: ( 0, )incr: ( - , 0 )

decr: ( 2, )incr: ( - , 2)

constant: [ -2, 2 ]

decr: ( - , 7 )decr: ( 7, )


unbounded bounded belowb = 0

bounded belowb = 1

unbounded bounded aboveB = 0

boundedb= -1, B = 1

bounded belowb = 0 bounded below

b = -1bounded below

b = 0bounded above

B = 0

Right branch:bounded below

b = 5

Left branch: bounded above

B = 5


y-axis

EVEN functions

The graph looks the same to the

left of the y-axis as it does to the right

For all x in the domain of f,

f(-x) = f(x)

x-axis

The graph looks the same above

the x-axis as it does below it

(x, - y) is on the graph whenever

(x, y) is on the graph

origin

ODD functions

The graph looks the same upside

Down as it does right side up

For all x in the domain of f,

f(-x) = - f(x)


f (x)=4x4 −3x2 +1

g(x)=4x5 −2x3

Algebraic Test for even/odd/neither:

Replace all x with –x:


f (x)=4x4 −3x2 +1

f(−x) =4(−x)4 −3(−x)2 +1

=4x4 −3x2 +1

g(x)=4x5 −2x3



f(x)=f(-x) so even!

Now try g(-x):


f (x)=4x4 −3x2 +1

f(−x) =4(−x)4 −3(−x)2 +1

=4x4 −3x2 +1

g(x)=4x5 −2x3

g(−x) =4(−x)5 −2(−x)3

=−4x5 +2x3



f(x)=f(-x) so even!

g(-x)=-g(x) so odd!


f (x)=3x4 −3x3 +1




f (x)=3x4 −3x3 +1

f(−x) =3(−x)4 −3(−x)3 +1

=3x4 + 3x3 +1



This is neither the same nor opposite so it is neither!


Odd Even Even

Odd Even Neither

Even Neither

Even Odd


Students will be able to determine the vertical and

Horizontal asymptotes of functions by inspecting their graphs

Do Now: use your graphing calculator to graph:

f (x)=x

x2 −x−2


verticallyhorizontally


horizontally

vertically

will not actually touch

asymptotes

tan and cot

x = -1 x = 2

y = 0

End behavior

x

lim f(x)

x

lim f(x)

Limit notation

x

lim f(x) 0

x

lim f(x) 0


Students will be able to determine the vertical andHorizontal asymptotes of functions by inspecting

their graphs

g(x)=5x

x−2

Graph the function above and determine the verticalAnd the horizontal asymptotes.


Students will be able to determine the vertical andHorizontal asymptotes of functions by inspecting

their graphs

g(x)=5x

x−2

The end behavior of the function above is related to the Horizontal asymptote .

lim f(x ) =5 x → ∞

lim f(x ) =5 x → −∞

x =2

y=5

Vertical:

Horizontal:


lim f(x ) = x → ∞

We read this as “the limit as x approaches infinity is”Which means “as we look to the far right of the graph,

What y value does it head near”


Vertical: x = - 3 Horizontal: y = 0Vertical: x = 2, -2

Horizontal: y = 0Vertical: x = 3

x

lim f(x) 5

x

lim f(x) 5

x

lim f(x) 3

x

lim f(x) 0

x

lim f(x) 1

x

lim f(x) 1

x

lim f(x) 0

x

lim f(x) 7

x

lim f(x) 0

x

lim f(x)

x

lim f(x) 4

x

lim f(x) 4


Student will be able to use function properties to Analyze a Function.


Yes

{ ( - , -1 ) U (-1, 1) U (1, ) }

Infinite discontinuities

Decreasing: (- , -1), (-1, 0 ]

Unbounded

Left piece: B = 0, Middle piece b = 3, Right piece B = 0

Local min at (0, 3)

Even

Horizontal: y = 0, Vertical: x = -1, 1

Each x-value has only 1 y-value

{ ( - , 0) U [ 3, ) }

Increasing: ([ 0, 1), (1, )

lim f(x) 0 x → ∞

x

lim f(x) 0


What we just did was to “analyze a function”

The sheet provided to you lists the aspects ofA function to include in an analysis.


In-class ExerciseSection 1.3

In-class ExerciseSection 1.3

•Domain

•Range

•Continuity

•Increasing

•Decreasing

•Boundedness

•Extrema

•Symmetry

•Asymptotes

•End Behavior


Yes

{ ( - , ) }

continuous

Decreasing: (- , 0 ]

Bounded below b = 0

Absolute min = (0,0)

Neither even or odd

none

Each x-value has only 1 y-value

{ [ 0, ) }

Increasing: [ 0, )

x

lim f(x)

x

lim f(x)

{ ( - , -3 ] U [ 7, ) }


10 Basic Functions10 Basic Functions

3f(x) x f(x) sinx

f(x) cos x

f(x) x

f(x) x

2f(x) x

1

f(x)x

f(x) x

xf(x) e

f(x) lnx


Do Now:Add:

Subtract:

(x2 + x)+ (3x−7)

(2x3 −4)−(5x3 −9)

What are the domains of : f (x)=3x3 + 7

g(x) =x2 −1


f(x) + g(x)f(x) – g(x)

f(x)g(x)f(x)/g(x), provided g(x) 0

3x3 + x2 + 63x3 – x2 + 83x5 – 3x3 + 7x2 – 7

x2 – (x + 4) = x2 – x – 4

3

2

3x 7

x 1


Restricted Domains:Find the domains of the following functions:

f (x)=x2

g(x) = x+ 3



f (x)=x2 : domain: (−∞,∞)

g(x) = x+ 3 domain: [−3,∞)

1.Find: and include the overlapping domain

2.Find and include the domain f

g(x)

f −g(x)



f (x)=x2 : domain: (−∞,∞)

g(x) = x+ 3 domain: [−3,∞)

f

g(x)=

x2

x+ 3,

f −g(x) =x2 − x−3, Domain:

Domain:

[−3,∞)

(−3,∞)


Do Now:Simplify this “complex” fraction

Student will be able to compose two functionsAnd find their domains

22

x− 2

x2


x2


x2 sinx+,-,x,/

The squaring function the sine functioncomposition

composition

f ○ gf(g(x)

x2


1. Write the first function2. Replace “x” with ( )

3. Place the 2nd function in ( )4. Decide on the domain and simplify

Steps for composition:


x2 sinx+,-,x,/

The squaring function the sine functioncomposition

composition

f ○ gf(g(x)

x2

4x2 – 12x + 9

2x2 – 3

x4

1

5

4x-9


The domains of compositions

We need to “inherit” the domain of the second functionAnd consider the new function as well!

Use the graphing calculator to verify!


x 2

4

1

Domain is all reals except -2


For part b:(see “vars” which is next to clear for the y)


x 2

4

4x

1 2x

1


Domain is all reals exc. 0 and -.5


x 2

4

4x

1 2x

1

1

x1/ x

2(x 2)

x 4


Domain is all reals exc. 0 and -.5

D: all reals except 0

D: all reals except -2 and -4


Decomposition: Students will be able to decomposea composite function.

Do Now: perform the composition:

f (x)= x

g(x) =x2 +1f og(x) =


Given the function, find what f(x) and g(x) could be.

f og(x)= x2 +1Recall the steps for composition:

1. Write the first function2. Replace “x” with ( )

3. Place the 2nd function inside ( )4. Decide on the domain and simplify

Ask yourself: what could be in the ( ) ?


f og(x)= x2 +1

f (x)= x

g(x) =x2 +1


inside function

outside function

x2 + 1 x 2 2f(g(x)) f(x 1) x 1 h(x)x2

x 1 2 2f(g(x)) f(x ) x 1 h(x)


Practice examples:

a) h(x)=(2x+1)3

b) h(x) = x2 −5

c) h(x) =(x−1)2 −3(x−1)+ 4

if

h(x)= f og(x)

find f(x) and g(x)


3

g(x) 2x 1

f(x) x

inside function

outside function

x2 + 1 x 2 2f(g(x)) f(x 1) x 1 h(x)x2

x 1 2 2f(g(x)) f(x ) x 1 h(x)

2g(x) x 5

f(x) x

2

g(x) x 1

f(x) x 3x 4


Hand in this example on an “exit card”

h(x)=4(x−2)7 −5


inverse

functions

horizontal line test

original relation

Graph is a function

(passes vertical line test.

Inverse is also a function (passes

horizontal line test.)

both vertical and horizontal

line test like A one-to-one function

is paired with a unique y

inverse function

is paired with a unique x

f –1 f –1 (b) = a, iff f(a) = b

Graph is a function

(passes vertical line test.

Inverse is not a function (fails horizontal line

test.)


Do Now:State the domain and range of the following function:

f (x)=x

x+2

Student will be able to find the inverse function and stateThe inherited domain and range


DomainRange:

f (x)=x

x+2

x=y

y+2y=xy+2xy−xy=2xy(1−x) =2x

f−1 =2x1−x

,x≠1

x ≠−2y≠1

Get y terms together!Factor y out

Note the “inherited“ domain


D: { ( - , ) }R: { ( - , ) }

D: { [ 0, ) }R: { [ 0, ) }

D: { ( - , - 2) U ( -2, ) }R: { ( - , 1) U (1, ) }

D: { ( - , ) }D: { [ 0, ) }

D: { ( - , 1) U (1, ) }

x 2y 3

1 x 3

f (x)2

x y

1 2f (x) x

y

xy 2

1 2xf (x)

x 1


Functions that are not one to one require restrictionsOn their domains in order to make them 1 – 1 and

Find their inverses!

g(x)=x2 + 3

What is the domain and range?




g(x)=x2 + 3

Domain: restricted to Range:

(−∞,∞)→y≥3

x ≥0




g(x)=x2 + 3,

x=y2 + 3

x−3=y2

g−1 = x−3,

Domain: Range:

y≥3x ≥0

x ≥3y≥0


{ ( - , ) } 1 5

( x 5) (2x 10) x 52 2

f(x) and g(x) are inverses

1 3

( x 5) (2x 10) x 152 2

21( x 5)(2x 10) x 5x 502

1( x 5)2

(2x 10)

1

(2x 10) 5 x 5 5 x2

1

2( x 5) 10) x 10 10 x2

{ ( - , ) }

{ ( - , ) }

{ ( - , - 5) U ( - 5, ) }

{ ( - , ) }

{ ( - , ) }


( 3,6.75)

( 2,2)

( 1,0.25)

(0,0)

(1, .25)

(2, 2)

(3, 6.75)

1x 4

y

1y

x 4

Yes

passes horizontal line test

Yes

(6.75, 3)

(2, 2)

(0.25, 1)

(0,0)

( .25,1)

( 2,2)

( 6.75,3)

D: { ( - , 0 ) U ( 0, ) }

R: { ( - , 4 ) U ( 4, ) } D: { ( - , 4 ) U ( 4, ) }

1 1f (x)

x 4

2g(x) x 2

f(x) x 2 2f(g(x)) f(x 2) (x 2) h(x)


D: { ( - , - 2 ) U ( - 2, 1 ) U ( 1, ) }

3x 2f(g(x))3

1x 2

3g(f(x))

x2

x 1

xf x 1(x)

3gx 2

D: { (- , - 2) U (- 2, 1) U ( 1, ) }

3

xy 2 D: { ( - , 0 ) U ( 0, ) }

3

1 x

3x 3

3x 2D: { ( - , 2/3 ) U ( 2/3, 1 ) U ( 1, ) }

2x 2x

3x 3

3 2xy

x


STUDENTS WILL BE ABLE TO NAME THE TRANSFORMATIONS OF A FUNCTION

THAT TOOK PLACE WHENGIVEN AN EQUATION

DO NOW: worksheet


add or subtract a constant to the entire function

f(x) + c up c units

f(x) – c down c units

add or subtract a constant to x within the function

f(x – c) right c units

f(x + c) left c units

y cos(x) 5 y x 2

2y (x 3) 4


multiply c to the entire function

Stretch if c > 1

Shrink if c < 1

multiply c to x within the function

A reflection combined with a distortion

complete any stretches, shrinks or reflections first

complete any shifts (translations)

f1

cx

⎛

⎝⎜⎜

⎞

⎠⎟⎟ Stretch by C

Shrink by

gc f(x)

f c • x( )

1

c


reflections

negate the entire function y = – f(x)

negate x within the function y = f(-x)

f(x) 2

3x 1

x 2

2

3x 1

x 2

f( x)

2

3( x) 1

( x) 2

2

3x 1

x 2

2

3x 1

x 2

2

3x 1

x 2 2

3x 1

x 2


1. What transformations took place withThe basic absolute value function?

y=5 2x + 32. Write the equation that results by taking the squaring F

function and:

A vertical stretch factor of 4, shift left 6


1. What transformations took place withThe basic absolute value function?

y=5 2x + 3

2. Write the equation that results by taking the squaring function and:

A vertical stretch factor of 4, shift left 6

Vertical stretch by 5, horiz. Shrink by ½, vertical shift up 3

y=4(x+6)2


Answers

Answers

y = 1/x4

y = x, y = x3, y = 1/x, y = ln (x)

y = sqrt(x)y = ln(x)

y = 2sin(0.5x)

Stretch by 8 Shrink ½

Shrink by 1/8 Stretch by 2

9/5/2006pre-calculus r r { [ 4, ) } { (- , 3 ] } { r \ { 2 } } { r \ { 1 } } { r \ { -3, 0 } } r {...

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