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YoHoJo
02-08-2010, 12:35 AM
Thanks in advance. Sorry for tying it all out, I hope you still understand.
(I have solved most of these, some are only 1/2 correct, I'd just like to see others' answers please)

1) f(x)= square root (32-6x^2) SOLVED

2) y=-3sin(2X + 3pi/2) ALL SOLVED
a) What is the amplitude?SOLVED
b) What is the period?SOLVED
c) What is the phase shift, and which direction?SOLVED

3) Let f(x)= x/ (x-4) and g(x)= 1/(x^2) ALL SOLVED
a) Find f(g(x)) / f○g(x) (Same thing, different notations)SOLVED
b) What is the domain of a SOLVED
c) Find g(f(x)) / g○f(x) (Same thing, different notations) SOLVED

6) Let f(x)= 3x^2 + 12x + 6 ALL SOLVED
a) Restrict the domain of f so that f is one to oneSOLVED
b) Find f^-1 (Inverse)SOLVED
c) What is the domain of bSOLVED

7) Let f(x)= 4th root (x-2) and g(x)= X^4 + 2. Are f and g inverses of each other? Why/why not? SOLVED

8) Graph the parametric equations SOLVED
x=1-t
y=2 + t^2
(Just explain to me how to do this one please)

9)
a) 2lnx = ln2 + ln(3x - 4)SOLVED
b) 2e^(2x) - 5e^(x) - 12 = 0NOT SOLVED
c) log(base 3) (x-1) + log(base 3) (x+1) = 1NOT SOLVED


I apologize for the horrid typing of mathematical problems, I tried the best I could, I'll clarify anything you cant read. Thanks again!

Cigue
02-08-2010, 01:24 AM
1) What is the question?
2) a) amplitude is 3.
b) period is 1/2.
c) 3pi/2 is the phase shift, since it's added I'd say it goes to the right.

Someone else take over :)

3Garrett3
02-08-2010, 01:27 AM
You're going to have to explain Phase shift.

I go by Vertical or Horizontal translation. It's probably the horizontal translation..

Method
02-08-2010, 01:33 AM
3. f(x)= x/ (x-4) and g(x)= 1/(x^2)

a) f(g(x)) = f(1/(x^2)) = (1 / x^2) / ((1 / x^2) - 4)
1/x^2 / (1 - 4x^2)/x^2
x^2/(x^2(1 - 4x^2))
1/(1 - 4x^2)

b) domain is x is all reals except x = +- sqrt(1/4)

c) g(f(x)) = g(x / (x-4)) = 1 / (x/(x-4))^2
1 / (x^2/(x-4)^2)
(x-4)^2/x^2

9a) 2lnx = ln2 + ln(3x - 4)
ln(x^2) = ln(2(3x - 4))
ln(x^2) = ln(6x - 8)
x^2 = 6x - 8
x^2 - 6x + 8 = 0
(x - 4)(x - 2) = 0
x = 4, 2 (both of these work if you plug them back into the original equation, so they are both ok)

I got bored so I only ended up doing two, sorry. :p

YoHoJo
02-08-2010, 01:41 AM
1) What is the question?
2) a) amplitude is 3.
b) period is 1/2.
c) 3pi/2 is the phase shift, since it's added I'd say it goes to the right.

Someone else take over :)

Amplitude is 3 yeah.



Straight from an old precal book I have:
y= a sin kx y=a cos kx
and
Period=2pi/k
We are working with
y=-3sin(2X + 3pi/2)
............^This '2' would be our K?
So period is 2pi/2 = pi
The period should be pi?



I was told/got that the phase shift is actually 3pi/4!
A apparently you need to take out the 2 attached to the x to get x alone, and then you have the phase shift:

y=-3sin(2X + 3pi/2)
y=-3sin2(X + 3pi/4)
so phase shift is 3pi/4

3Garrett3
02-08-2010, 01:45 AM
Thanks in advance. Sorry for tying it all out, I hope you still understand.
(I have solved most of these, some are only 1/2 correct, I'd just like to see others' answers please)

1) f(x)= square root (32-6x^2)

2) y=-3sin(2X + 3pi/2)
a) What is the amplitude? 3
b) What is the period? pi (the reciprocal times 2pi)
c) What is the phase shift, and which direction? 3pi/2 to the left. (the horizontal translation is the negative of what is added to the x value)

3) Let f(x)= x/ (x-4) and g(x)= 1/(x^2)This one would take me a long time, if you can't figure it out, message me. I've never been good at composite functions. This one is really complex too.
a) Find f(g(x)) / f○g(x) (Same thing, different notations)
b) What is the domain of a
c) Find g(f(x)) / g○f(x) (Same thing, different notations)

6) Let f(x)= 3x^2 + 12x + 6
a) Restrict the domain of f so that f is one to one Don't really get what you mean. I don't know what one to one means..
b) Find f^-1 (Inverse) As far as I got was y^2+4y = 2- x/3 I haven't really done this much lately, but your teacher gives you crazy hard things to do, it's impossible to factor it too. I had x/3 = y^2+4y+2 but it wont factor. IB Math SL just doesn't cut it man.
c) What is the domain of b

7) Let f(x)= 4th root (x-2) and g(x)= X^4 + 2. Are f and g inverses of each other? Why/why not? Yes, they are. I'll explain:
y = x^4 + 2
x = y^4 + 2 (inverse)
y^4 = 2 - x
y = fourth root (2-x)

8) Graph the parametric equations
x=1-t
y=2 + t^2
(Just explain to me how to do this one please)

I'm going to need to look up parametric equations. The name doesn't ring a bell, but IB uses messed up words for stuff. Is this like vectors?

9)
a) 2lnx = ln2 + ln(3x - 4)
b) 2e^(2x) - 5e^(x) - 12 = 0
c) log(base 3) (x-1) + log(base 3) (x+1) = 1


I apologize for the horrid typing of mathematical problems, I tried the best I could, I'll clarify anything you cant read. Thanks again!


I've included my answers in red in the quote. Get back to me on msn

Method
02-08-2010, 01:46 AM
Yes, the period is pi and the shift is 3pi/4 to the left I believe.

3Garrett3
02-08-2010, 01:53 AM
Yes, the period is pi and the shift is 3pi/4 to the left I believe.

Oh. Well I got the period right. I haven't reviewed this since last year. It makes sense to factor out the 2 though.

Santa_Clause
02-08-2010, 01:59 AM
Question 6:

a)

http://latex.codecogs.com/gif.download?\\f(x)&space;=&space;3(x^2&space;+&space;4x&space;+&space;2)&space;=&space;3[(x+2)^2&space;-&space;2]\\&space;$Therefore,&space;for&space;f&space;to&space;be&space;one&space;to&space;one:&space;$&space;x\geq&space;-2&space;$&space;or&space;$x\leq&space;-2


b)

http://latex.codecogs.com/gif.latex?\\x%20=%203y^2%20+%2012y+6,%20y%20\geq%2 0-2\\%20y^2%20+%204y%20+%202%20=%20\frac{x}{3}\\%20( y%20+%202)^2%20=%20\frac{x}{3}%20+%202\\%20y%20+%2 02%20=%20\sqrt{\frac{x}{3}%20+%202}%20$%20(positiv e%20root%20as%20$%20y\geq%20-2$)$\\%20y=%20\sqrt{\frac{x}{3}%20+%202}%20-%202


c)

http://latex.codecogs.com/gif.download?\\x\geq&space;-6

Question 7: g(x) does not even have an inverse function, as it does not pass the horizontal line test.

Question 8:

http://latex.codecogs.com/gif.download?\\x&space;=&space;1&space;-&space;t,&space;t&space;=&space;1&space;-x&space;\\&space;y&space;=&space;2&space;+&space;t^2&space;=&space;2&space;+&space;(1-x&space;)^2\\&space;$Parabola&space;with&space;vertex&space;(1,&space;2)&space;and&space;y-intercept&space;$(0,3)

YoHoJo
02-08-2010, 02:11 AM
Santa, explain how you did 8?
GarretT, 3 looks great, not sure about the rest, Method solved them properly above :).

Brb, thanks EVERYONE! I'll be back in a few hours and we can compare answers more.
Math is FUN!

3Garrett3
02-08-2010, 02:11 AM
Question 7: g(x) does not even have an inverse function, as it does not pass the horizontal line test.

Yes it does?

X^4 + 2 is a parabola.

YoHoJo
02-08-2010, 03:25 AM
Method/Santa_Clause, everything I have matches up with your guys' answers! (And for whatever I didn't solve, you solutions are understandable and make sense!)

Also thanks to cigue (:p) and 3GarretT3!

Thanks a lot everyone! Just a few log/ln ones left.

marpis
02-08-2010, 07:42 AM
log(base 3) (x-1) + log(base 3) (x+1) || Formula: log(x) + log(y) = log(x*y)
= log((x-1)(x+1)) = 1
= log(x^2 - 1) = 1

==> (x^2 - 1)^1 = 3
==> x^2 -1 = 3
==> x^2 = 4

x = +-2
but (-2) is not a solution because log(-2 - 1) = log(-3) non real

ANSWER: x = 2

Markus
02-08-2010, 06:45 PM
log(base 3) (x-1) + log(base 3) (x+1) = 1

log((x+1)(x-1)) = (base 3)log 3

log (x² - 1) = log 3

x²-1 = 3
x² = 4
x = 2 or x = -2

X = -2 is invalid solution because log(-3) doesn't exist.

edit: ninja'd =(, didn't really take a look at any post except the first

Killerdou
02-15-2010, 06:35 PM
2e^(2x) - 5e^(x) - 12 = 0
define y = e^x, thus:
2y^2 -5y -12 = 0 ==> (y -4)*(y + 3/2) = 0
thus y = 4 or y =-3/2
e^x = 4 (e^x = -3/2 is not in R)
x = ln(4) or did I just fail epically?

IwriteCode
02-18-2010, 07:58 AM
http://img21.imageshack.us/img21/8545/mathy.gif (http://img21.imageshack.us/i/mathy.gif/)
lol they been solved already

Santa_Clause
02-27-2010, 10:12 AM
Yes it does?

X^4 + 2 is a parabola.

Complete parabolas do not have inverse functions. They have to be restricted in the domain such that the parabola does not pass its stationary point (ie. only half-parabolas or less have an inverse function).