VCE Methods Question Thread!

[Hancock]

Could someone explain how to do this question:
A spinner has three equally likely sectors numbered from 1 to 3.
Find Pr(X=2 | X≠1)

I'm not completely sure because I haven't done probability in a while.

Pr(X=2 | X≠1) = Pr(X=2 N X≠1)/Pr(X≠1) = (1/3) / (2/3) = 1/2

Could also think about it like this. Find the Pr(X=2) given X is not equal to 1. Therefore, we only have two options, 2 or 3. So it's a 0.5 chance of X=2.

[soccerboi]

What did the question define the random variable X as?

X was the number spun

I'm not completely sure because I haven't done probability in a while.

Pr(X=2 | X≠1) = Pr(X=2 N X≠1)/Pr(X≠1) = (1/3) / (2/3) = 1/2

Could also think about it like this. Find the Pr(X=2) given X is not equal to 1. Therefore, we only have two options, 2 or 3. So it's a 0.5 chance of X=2.

Yes thats correct, but  i dont understand why Pr(X=2 N X≠1) is equal to 1/3??

[TrueTears]

Oh okay, note that assuming it is spun once.

[dinosaur93]

hey guys, just wondering, is U-substitution of indefinite integrals part of the methods course?

[brightsky]

no. don't use it explicitly in any of methods sac/exam, unless of course your teacher gives you permission to do otherwise in a sac.

[soccerboi]

Hi guys, im having trouble understanding this:

if the original graph is p(x) and we are asked to sketch the graph of p(x+0.5), why is this new graph moved 0.5 units to the left? i.e why do we subtract 0.5 from the x coordinates of the original graph p(x)?

I know we let the stuff inside the bracket equal 0 and solve but why do we do this? I'm confused because inside the bracket its x+0.5, so i keep thinking that we just add 0.5 to the x coordinates of the original graph p(x).

Can someone explain this to me, Thanks

[TrueTears]

Hi guys, im having trouble understanding this:

if the original graph is p(x) and we are asked to sketch the graph of p(x+0.5), why is this new graph moved 0.5 units to the left? i.e why do we subtract 0.5 from the x coordinates of the original graph p(x)?

I know we let the stuff inside the bracket equal 0 and solve but why do we do this? I'm confused because inside the bracket its x+0.5, so i keep thinking that we just add 0.5 to the x coordinates of the original graph p(x).

Can someone explain this to me, Thanks

Let y = p(x) be the original graph now let y'=p(x'+0.5) be the new (transformed) graph.

Clearly thus y=y' and x = x'+05 -> y'=y and x' = x-0.5

Thus (x,y) ->(x-0.5, y)

the result follows

[Lasercookie]

I'll attempt at a simple example and crappy analogy.





Originally we have coordinates:


But now we have:


Here's an attempt at an analogy, where we're travelling along the curve for lols trying to get to a certain coordinate. So let's say we're starting at x=-2, moving along in a right-wards direction (increasing x-value) and we're trying to get to a point where the y-value is 9.

So for the original p(x), as we travel along from x=-2 the y-values are x^2. So we're starting at x=-2 and we trace along left to right until we reach our destination, which in this case is the coordinate (3,9)

For the p(x+1), for the x-value we're on, we now have a value of 1 added onto it, kind of like a speed boost I guess. So this means we're going to get to a y-value of 9 quicker right? Since we're tracing along from left to right, and we're reaching the destination quicker, this means that the value would be more to the left than it was for p(x).

If we had p(x-1), we're going to take longer to reach that coordinate with a y-value of 9. Continuing the analogy, it'd be like if something was stuck on your shoe and you had to travel along slower.

I hope that made sense (not too sure if it did).

[kamil9876]

Suppose the point is on the curve this means that . You ask why is the point on the curve ?, well let's check to see if it is... by definition this means we have to check that but this is clearly true since this equation turns out to be , which we know is satisfied since we have assumed to be a point on the original curve.

Same idea works if you do dilations, i.e there is a "compensation" as laseredd mentioned.

Same idea works if you play around with values in such a way, the graph of is just the graph of shifted DOWN by 0.5.

[#1procrastinator]

How do you write as a composition of functions?

[Jenny_2108]

How do you write as a composition of functions?

Let
     

write in composite function: y= g(f(x))

[#1procrastinator]

^ thanks - is that as far as we can go?

[Jenny_2108]

I think so because the question asks to write as a composite of function and we did it

[#1procrastinator]

^ ah ok, thanks. That actually wasn't the original question, it's actually a derivative problem that I tried to use the chain rule on until I realised I was doing the completely wrong thing. Got me curious about it though

[soccerboi]

The function f has the rule f(x)= (x-2)²+1. Which of the following sets is a possible domain for f if the inverse function f ^-1 exists?
A [1,∞)
B (-∞,0]
C [0,5]
D (-2,∞)
E [0,∞)

Can someone explain to me why the answer is option B?