SRL Archive - Mathematics

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!

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 :)

You're going to have to explain Phase shift.

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

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

From: Cigue
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=-3sin[B]2(X + 3pi/4)[/B]
so phase shift is [B]3pi/4[/B] 

From: YoHoJo
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

Yes, the period is pi and the shift is 3pi/4 to the left I believe.

From: Method
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.

Question 6:

a)



b)



c)


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

Question 8:

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!

From: Santa_Clause
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.

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.

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

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

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?

lol they been solved already

From: 3Garrett3
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).