Notes from 2.4: Operations on Functions/ Composition of Functions

*Sorry these notes weren't up sooner, but I posted them before, and then they somehow didn't show up on the blog. Sorry guys*

1. Composition of a Function:
-(fog)(x) or “f circle g of x” aka f(g(x))
--f(x)=x2–5
--g(x)=3x-4
(fog)(x)=f(g(x))
=f(3x-4)
=(3x-4)2-5
=9x2-24x+16-5
(fog)(x)=9x2-24x+11
(gof)(x)=g(f(x))
=g(x2–5)
=3(x2–5)-4
=3x2-15-4
(gof)(x)=3x2-19
2. Inverses:
- use PEMDAS in reverse (SADMEP)
-inverses are always functions
-1/2 of inverse (quadratics) will show because the other half does not pass the vertical line test. (Seen in graphing on calculator)
--f(x)=2x-1
f-1(x)=
g(x)=(x-3)2
g-1(x)=
or:
g-1(x)= +3
3. Graphing Inverses:
-inverse will be a reflection of equation over the y=x line
-points switch from (x,y) to (y,x)

-graph vs. y=x

4. Algebra of Inverses:
--f(x)=
f-1(x) » y=
y=
x= (x switches places with y)
x2=y-3
y=x2+3 » f-1(x)= x2+3
D , D-1:
5. Algebra Cont. Checking for Inverses:
--f(x)=x3+1
--g(x)=
--(fog)(x)= (gof)(x)=x
f(g(x)): g(f(x)):
f( ) g(x3+1)
( 3+1 ) -1
x-1+1=x ) =x YES, they are inverses!
Homework (due Sept. 28) - 180:15-35 odd, 197: 1-15 odd, 33-41 odd, 71-80
Quiz (Sept. 26) – Quadratics, Solving with Calculator, Absolute Value, Inequalities, Application Problem

Notes 9/28/2007

September 21, 2007 Class Notes

September 21, 2007
*We went over homework*
*NOTE: Timeliness has been added to the blog rubric and a link to WIKI has been added to the page to access handouts*

-Absolute Value Inequalities
Equations: x=# x=(+#) or (-#)
Case 1
x># x>(+#) or x<(-#) EX: x>2
x>2 or x<-2 Case 2
x<# x<(+#) or x>(-#)
EX: x<2>-2

*And vs Or*
Less than--->and
-less thand
Greater than--->or
-Greator than

Application
3x-2 ≤ 1
3x-2 ≤ 1 3x-2 ≥1
x ≤ 1 and x ≥ 1/3

-Interval Notation
-Third way to depict inequality answers
-Easier way/less time consuming
[ ] ----≤ ≥

( ) ----< >
EX: 3x-2≤1 [1/3, 1]
EX: x >2 (∞, -2)∪(2, ∞)

·means union to connect two sets together

-Inequalities by Calculator
EX: (x+3)/(x-2)> 0 *Enter left side in Y1
*Start with standard window
* Answers above the x axis
* ( -∞,1) ∪(2,∞)

*Homework quizzes returned*Use homework quizzes to show your areas of weakness and to gauge where you are in understanding*

-Functions –“last thing before we jump into trig waters”-Mr. M
*Operations and functions
-arithmetic EX: f(x)=x2-4
f(x)+g(x) g(x)=x+3
f(x)-g(x) (f+g)(x)=f(x)+g(x)
f(x) (f+g)(x)=x^2-4+x+3
-composition = x^2+x-1
(fog)(x)
*cont’d on Tuesday, September 25th*
*Homework: Pg 168 #29-59/odd, 63, 65, Pg 179 #1-9/odd*

*Alright hope everyones weekend went well, see ya'll tomorrow!*

-Lauren

Well I'm finally posting my notes from two weeks ago

So here they are. Just double-click the pictures to get a bigger picture of the notes

Homework
Pg 155 #21-27 odd, 53-69 odd, 73, 76, 77

9/18/07 How to pick a method

How to pick a method:
-What characteristics do we use to solve an equation?
-When is it more efficient to us one method over another?


An important side note: For the quiz coming up tomorrow make sure you know how to use each of these methods, and be able to tell why you used the method to solve the problem. Absolute value:
X= 5 for this equation x= 5 or -5
X-3= 5 we are trying to find a way to make this 5, -5
-So what you would need to do is eliminate the absolute value signs, and it will look like this:
X-3= 5 or X-3= -5 so X would be 8 or -2
-You want to get the absolute value alone like you would get a(2) alone
X2+X-6= 4 or X2+X-6= -4 X2+X-10= 0 or X2+X-2= 4
So X=-2,1 or (X+2) (X-1)
If you plugged this into the quadratic it would look like this: -1+/-√1-4(1)(-10)/2

which would equal: -1+/-√41/2 (You could also check your answer graphically.)


Radicals:
(√X-4) 2 =(6) 2 (√X+8) 2 =(X+2) 2
X=40 X-4= 36 X+8= (X+2) 2
X+8= X2 +4x+4
0= X2 +3x+4
(X+4) (X-1)

We have to make sure they both work . 1 would be the actual solution because when you plug -4 back into the original equation it equals 2 and -2
X≠4 X=1

More difficult example:
√X +√2x =4 -make sure you isolate one of the radicals
(√2x) 2 = (4-√x) 2
2x= (4-√x) 2
2x= 16- 8√x +x -divide the whole equation by negative 8
(-1/8 (x-16)) 2 =(√x) 2
1/64 (x2 -32x +256)= X
1/64 x2 -32/64x+ 256/x=X
X2 -32x+256 = 64x
X2 -96x+256= 0 -Use the quadratic formula to finish


Last week we also began talking about inequalities:
Inequalities: divide by (-) or multiply by (-) switch the order
- compound inequality two ways to solve

You can use either method for compound inequalities!
-3<>

x> -7 x≤ 5
-7<>

Domain


Key things to look at when finding domain with a function involving square roots.
Ø Square roots cannot be negative


Ex. Pg 132 #28
K(x)= x^2+1 set the function as an equality

X^2+1>0 anything times its self is positive so the domain of this function is all real numbers.


Ex. Pg 132 # 27
H(x)= 2-x set the function up as an equality

2-x >= 0 subtract –2 from both sides
-2 -2
-x >= -2 divide by –1 and since you are dividing by a negative the inequality sing flips
-1 -1
x <= 2

Ex. Pg 130 #25

F(x)= ` X `
X^2 * 2X factor the denominator

F(x)= ` X `
X (X-2) set the denominator equal to 0

X (X-2) =0
X=0
X-2=0
+2 +2
X=2
So X cannot equal 0 or 2 because then the denominator is 0

Key things to remember are no negatives under a radical and the denominator cannot equal 0.





Piecewise

Important thing to remember is that you find the value that the two functions share.

F(x) { radical over 1-x if x<1
X^2-1 if x>= 1 plug in 1 into both functions

Radical over 1-1=0
1^2-1=0


Homework
9/12/07
pg 155 #1-10 even, 33-42even 43-52even

You may not have figured this out yet, and I'm sorry for not telling you yet, but if you double click on the scanned notes, they ARE readable. I probably should've told you that.

Brad.