Maths Methods 3/4 Help Thread 2011

[Rairiko]

thanks guys knew how to do the positive side but not the negative

got another quick question how do i find  fg when f(x) = root(x-2) and g(x)=root(4-x)

[xZero]

if ur talking about f(g(x)) then you sub x=root(4-x) into f(x)

Edit: on second thought, the range of g(x) is not a subset of f(x) so f(g(x)) doesn't exist

[Rairiko]

if ur talking about f(g(x)) then you sub x=root(4-x) into f(x)

oh sorry i meant (fg)(x)

[thushan]

Multiply the two functions together, and the domain is the intersection of the domains of the original two functions.

[Rairiko]

Multiply the two functions together, and the domain is the intersection of the domains of the original two functions.

ok i tried doing (root(x+2))(root(4-x)) what do i do next?

[xZero]

if a root times another root, u can just times everything inside the roots and put it under a single root (confusing ey)
so root( (x+2) (4-x) )

[Rairiko]

if a root times another root, u can just times everything inside the roots and put it under a single root (confusing ey)
so root( (x+2) (4-x) )

didn't know you could do that thanks again

[m@tty]

if a root times another root, u can just times everything inside the roots and put it under a single root (confusing ey)
so root( (x+2) (4-x) )

This is usually fine, but things can go wrong when you're working with complex numbers (ie. negatives inside the root)

(if you multiply)

But in reality;



So you have to be careful.

Not that this is really relevant to methods students

[xZero]

if a root times another root, u can just times everything inside the roots and put it under a single root (confusing ey)
so root( (x+2) (4-x) )

This is usually fine, but things can go wrong when you're working with complex numbers (ie. negatives inside the root)

(if you multiply)

But in reality;



So you have to be careful.

Not that this is really relevant to methods students

opps ahhaah didnt think of that

[thushan]

Don't need to worry about it for this question, as the domain specifies that x + 2 and 4 - x (the terms under the root) are always >0 anyway.

[pi]

Need help with sketching:


I'm no sure how to approach this problem.

Thanks (in advance)

[brightsky]

Break them down.

For the first one |x| < 4, if x > 0, then x < 4 if x < 0 then x > 4.
For the second one |y| < 2, if y > 0, then y < 2 if y < 0 then y > 2.

Sketch each accordingly. (I'm not sure if this helps but I can't see a way to get a graph up. :/)

[pi]

dw, I got it now

[Water]

Break them down.

For the first one |x| < 4, if x > 0, then x < 4 if x < 0 then x > 4.
For the second one |y| < 2, if y > 0, then y < 2 if y < 0 then y > 2.

Sketch each accordingly. (I'm not sure if this helps but I can't see a way to get a graph up. :/)

hey Rohitpi, howd you do the 2nd equation for the sketching. I"m not sure if what I"m thinking is correct lol.

Cause what I've got is shading from [0,2) for all Ys, I'm probably wrong.

[pi]

What I did was sketch the lines x=4 and x=-4, the required region is between those two lines. Then I sketched y=2 and y=-2 and the required region is between those two lines. The total required region is the overlap of those sections, which looks like a broken-lined shaded rectangle (where shaded = RR) with vertices at (-4, 2), (4, 2), (-4, -2) and (-4, 2).

(I think the question is badly worded though)