Maths Methods 3/4 Help Thread 2011

[kefoo]

Hey there, it's great you're helping out people with their maths and i'm hoping you could help me too.
I have a couple of questions;

1. Find the implied domain of sqrt(9-sqrt(x^2-19))
2. Find the coordinates of g(x) if g(x) = 4 + f(3-x) and f(x) = 2x^3 - 8x + 7 and the TP is (2,7) [not sure if i copied it down correctly lol]
3. What would be the domain of each f(x) and g(x) so that fog can exist if f(x) = sqrt(6-3x) and g(x) = x^2 -3

thanks in advance

[thushan]

Q1: Lol this question looks familiar to me...

OK so you need to find when stuff under the root is positive or 0.

So when is 9 - sqrt(x^2-19)>=0 and when is x^2 - 19 >=0

Solve the inequations for x and take the intersection, and you'll have the domain.

2. So we're talking about a transformation of a point. Looking at g(x) we're:

- translating -3 in x-axis
- reflecting in y axis
- translating +4 in y axis

So the point (2,7) would become (-(2-3), 7+4) = (1,11)

3. This is an odd question, because you can define many domains of f(x) so long as they are in the subset (-inf, 2] -> unless the maximum point of the domain was <-3 (eg (-inf, -8) would not work)

So let's assume dom f = (-inf, 2]. We want ran g to be a subset of dom f - i.e we want x^2-3<=2 which gives us -root5<=x<=root5, which is your corresponding dom g.

[kefoo]

Cool, thanks for that
one more question though

Find the inverse of the function with rule f(x) = (x-2)/(x+1)

i swapped the x with the y, but dunno how to solve it for y w/o calc

[vea]

For inverse, interchange x and y.

[Bonifacio]

EDIT: VEA bet me to it
f(x) = (x-2)/(x+1)

f(x) = x+1-3 / x+1

f(x) = 1 -3/x+1

y= 1 - 3/x+1

Inverse

x= 1-3/y+1

x-1 = -3/y+1

1-x = 3/y+1

y + 1 = 3/1-x

y = 3/1-x  -1


Enjoy

[Bonifacio]

Vea I think you have made a mistake:

you have to do this


Inverse

x= 1-3/y+1

x-1 = -3/y+1

1-x = 3/y+1

[Bonifacio]

Could somebody clarify if I or Vea are correct? I am curious!

[vea]

We're both correct... it's just the our answers are in different forms.

x-1 = -3/y+1

1-x = 3/y+1

You're trying to say that you have to do this when finding the inverse but you don't have to? Not sure where you learnt that from but the only thing you HAVE to do when finding an inverse is swap the x and y? Plus, you're just multiplying both sides by -1 :S

[Water]

Both Vea and Bonifacio is correct Lol. Same Equations xD

[Bonifacio]

Yes but your final answer is a relection in the x-axis. Mine is a reflection in the y-axis.

I'm not trying to push forth my own answer, trying to work out why we have different answers.

[Water]

Look more carefully, they're the same, even if you graph it

[Bonifacio]

Ahhh Ok, got it!

Thanks for your help in me helping OP!

[kefoo]

Question::
Solve for x
loge(x) + loge(3x+1) = 1
[e = base]
i tried to make 1 = loge(e) but dunno how they get the actual answer

[xZero]

loge(x)+loge(3x+1)=1
loge(x*(3x+1))=1
loge(3x^2+x)=loge(e)
3x^2+x=e
3x^2+x-e=0
use the quadratic formula and solve for x but only take the positive solution since the x>0 from loge(x)

[kefoo]

thanks xZero, one more question while im still doing this exercise

Find x if;
8e^-x - e^x = 2