VCE Methods Question Thread!

[knightrider]

I'm pretty sure you should still get the same answer as the final answer... Essentially when you're writing down your transition matrix, there's two correct ways in writing it out and you've so happened to have written the other way (look at why they've defined their matrix like that and how you've defined it and it's exactly the same thing but the positions are flipped around).

thanks so much IndefatigableLover

yep i tried it out the answers are the same just in different positions as i have written my transition matrices the other way compared to book.Thanks for clarification i thought i was initially wrong
so the way i wrote my matrix was right and so was the book.
i was just wondering does the answers always end up working fine  if you write your matrix right but the position is just different as you have defined it differently

[IndefatigableLover]

thanks so much IndefatigableLover

yep i tried it out the answers are the same just in different positions as i have written my transition matrices the other way compared to book.Thanks for clarification i thought i was initially wrong
so the way i wrote my matrix was right and so was the book.
i was just wondering does the answers always end up working fine  if you write your matrix right but the position is just different as you have defined it differently

Haha no worries

And yep it should always work so long as you've written them correctly
You just have to remember that your 'final answer' will not be in the same position if you defined your matrix the other way since writing your final answer is crucial as well!

[knightrider]

Haha no worries

And yep it should always work so long as you've written them correctly
You just have to remember that your 'final answer' will not be in the same position if you defined your matrix the other way since writing your final answer is crucial as well!

Thank you very much  IndefatigableLover

[lzxnl]

In Methods, you have 2x2 transition matrices. There is a neat way to work out what the matrix means:

You know, from the steady state matrix, that the vertical column entries must sum to 1. Thus, each column represents a given event occurring. Each row represents an event that may occur. The main diagonal elements represent the chance that an event will occur given it occurred before. I'll make this clearer with an example.

Suppose the chance a girl scores a basket from a free throw is 0.8 if she gets it from the previous throw, and the chance she misses the shot is 0.7 if she missed the previous shot. Let the event 'basket' be B and let the event 'no basket' be b
Now, the vertical columns of the transition matrix have entries that sum to 1. It makes logical sense, therefore, to assign the first vertical column to refer to the probabilities that the girl will score or miss her next free throw given that she has already scored.
Then, the second column should refer to the chance that she'll score or miss her next free throw given that she just missed.
Now, the top left corner should refer to the chance that she scored given that she already scored, which is 0.8. Putting all the other information together, my labelled transition matrix looks like this:

                    B already occurred            b already occurred
B                 0.8                                    0.3         
b                 0.2                                    0.7

See if you can make sense of what I did here. This is essentially what I did in year 9 to ensure I'd remember how the heck to set a transition matrix up (because I was terrible at this back then).

[SE_JM]

Now, the final one - |(1-3x)/2|. I'm going to break this up:

Could you explain why you had put (3x-1)/2 when x is bigger and equal to 1/3?
Does that mean it cannot be (1-3x)/2 when x is equal to 1/3? (since it's only valid for x<1/3

Could you tell me why? How would you know where the equal and greater than/less than signs need to be in other situations?


Also, could I ask this as well:
find dy/dx for y=|(x+2)(x-1)|

[keltingmeith]

Could you explain why you had put (3x-1)/2 when x is bigger and equal to 1/3?
Does that mean it cannot be (1-3x)/2 when x is equal to 1/3? (since it's only valid for x<1/3

Could you tell me why? How would you know where the equal and greater than/less than signs need to be in other situations?

I didn't have to, actually. Let's define the function as two parts - and . Well, at x=1/3:



Since they meet at the same point, it doesn't matter which is ALSO equal to 1/3 - it's only important that one of them is, so that the function remains continuous. This holds for all modulus you will see in methods.

[SE_JM]

I didn't have to, actually. Let's define the function as two parts - and . Well, at x=1/3:



Since they meet at the same point, it doesn't matter which is ALSO equal to 1/3 - it's only important that one of them is, so that the function remains continuous. This holds for all modulus you will see in methods.

thank you

[keltingmeith]

Also, could I ask this as well:
find dy/dx for y=|(x+2)(x-1)|

You need to stop editting your post at the last minute with extra questions... I have a habit of answering one, only to find a new one suddenly appearing.

So, first we split into hybrid form. So, we need to use whatever methods to figure out when the inside is positive/negative. After doing it, you should get that it's positive for x<-2 and x>1, and negative in-between. So, we have:


Remember to ditch the equalities, since y is not differentiable at those points.

[SE_JM]

Sorry

How did you know to test the points -2 and 1? (was it because x-intercepts occurred there?)

 if you got a similar question, like find dy/dx of |4x-x²-5|
would you need to factorise it first before knowing which points to test whether it's positive/negative? or is there a faster/easier way to do it?

Thanks!

[Zues]

how do you graph these sort of graphs?  (1st attachment)
and in the second attachment why is the asymptote 2 with the flip? if there was say a translation of +1 would it be three?

difference between maximal/implied and normal domain?

thanks

[dankfrank420]

For hybrid functions in general, does it matter which expression has the "equal to" sign attached to it? Or is it just important that one domain has the "equal to" sign on it?


         

[Conic]

It depends on the function. For example the function

is basically the same as 

while is not the same as

In the first case, both f(x) and g(x) are equal at x=0, while in the second case f(x) and g(x) are not equal at x=0, so they are different functions.

[Chang Feng]

@dankfrank420:
As long as one of the hybrid functions has the "equal to"- which I suppose you mean by ie, greater than or equal to thingy. It will suffice. -cause technically you can't have to hybrid functions that overlap- by having same domain (I'm pretty sure-someone may want to clarify this).

[Zues]

?

[keltingmeith]

Sorry

How did you know to test the points -2 and 1? (was it because x-intercepts occurred there?)

 if you got a similar question, like find dy/dx of |4x-x²-5|
would you need to factorise it first before knowing which points to test whether it's positive/negative? or is there a faster/easier way to do it?

Thanks!

The reason I knew to test those points was because I graphed the function inside the modulus. By graphing that function, I can see when it was positive and when it was negative, then used the property:

how do you graph these sort of graphs?  (1st attachment)
and in the second attachment why is the asymptote 2 with the flip? if there was say a translation of +1 would it be three?

difference between maximal/implied and normal domain?

thanks

Let's approach these graphs logically - first, graph f(x). Now, to sketch f(|x|), we know that any negative x numbers that go in, they're going to become positive, so you have to treat all negative x-values like positive x-values. So, you should see the function mirror itself across the y-axis. To sketch |f(x)|, any negative outputs (or negative y-values) are going to become positive. So, you should see the negative portions flip over the x-axis. This explains the flip in your second attachment.

There is no such thing as a "normal" domain. You have either the domain defined by the function, or the maximal domain. Let's consider the function f:[1,2]-->R, f(x)=sqrt(x). In this case, the domain of the function is [1, 2], however the MAXIMAL domain of sqrt(x) is x>=0, since those are all the possible numbers you could put into the function. If no domain is given for a function, its domain is implied to be its maximal domain. Sometimes, the specified domain is the maximal domain.

?

... I'm assuming they mean y=blah, not y+blah. In which case, it's B. Graph the two functions sqrt(4+x) and sqrt(4-x), use addition of ordinates, see what comes out. Or, alternatively, graph it on your calculator.

Can someone please help with this just to simplify:

+

I get to the point but how do we know to stop here? Because using the exponential laws we can simplify it further to

+ 1 which is which is more acceptable? Thanks

.

Erm... Want to show me the steps you took to simplify to your final form? Because that's not right... For example, when x=4:

First:
Second:

Also, I highly suggest not stopping tex tags mid-equation. It's easier to follow when it's all as one - for example, your looks like .