VCE Methods Question Thread!

[brightsky]

Mathematics is a beautiful language.

I agree. In fact, it is so beautiful that we can use it to prove God exists:

[keltingmeith]

I agree. In fact, it is so beautiful that we can use it to prove God exists:

(Image removed from quote.)

Having a bit of trouble following that... I think you showed that there must exist some God-like entity with probability 1? Time to brush up on my symbols~

[AirLandBus]

Can someone explain how to find the domain and range of composite functions?

[keltingmeith]

Often when discussing composite functions, whilst the notation fog(x) is more convenient to write, the notation f(g(x)) is easier to understand.

Using the second notation, we can clearly see that in our "machine examples", we have the following case:


Domain of g         Range of g                 Domain of f          Range of f
    input----------->transition------------->transition--------->output

From this, we can clearly see two things:

1. If the numbers coming out of g aren't defined in f's domain, then the composition cannot exist. I.e, the
2. The domain of our composition must be the domain of g.

The range is slightly harder - if the range of g is exactly equal to the domain of f, then the range of f must not have changed, and so the range of the composition is simply the range of f. If the range of g, however, is not exactly equal to f (i.e., is only a subset of f's domain), then you must draw f using the range of g as its domain to find the new range of the composition.

[knightrider]

How come you can simplify the expression to =   Why does this actually work?



Whereas   cannot be simplified to =  Why cant the be cancelled out here whereas it can be in the first expression above ?

[Phy124]

In your working out you have incorrectly cancelled out a 2! when factorisation was required. You need to do as follows:

[SE_JM]

Hello, this is a question that is frustrating me. It's pretty easy, but i don't understand what the question is about

This is the question:
Let p(x)= x⁵-3x⁴+2x³-2x²+3x+1

Given that P(x) can be written in the form of (x²-1)Q(x)+ax+b, where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the remainder when p(x) is divided by x²-1

The reason why i don't understand this question is because it tells you this information:

P(x) can be written in the form of (x²-1)Q(x)+ax+b, where Q(x) is a polynomial and a and b are constants, hence or otherwise,

What am i expected to do with this information? Am i meant to incorporate this information somehow in the process of finding the correct answer?

If the question was Let p(x)= x⁵-3x⁴+2x³-2x²+3x+1
find remainder when p(x) is divided by x²-1, i would just use the remainder theorem and be done with it..

Really long explanation over a super simple problem... oh well, i hope someone can clarify this for me

[SammyBoy]

Sorry this is a really simple question , how do I factorise x^3+27?

[knightrider]

In your working out you have incorrectly cancelled out a 2! when factorisation was required. You need to do as follows:

Thanks Phy124
but the example i did above in my previous post was fine right as i was multiplying.

[knightrider]

Sorry this is a really simple question , how do I factorise x^3+27?

Using the rule

let a=x and b=3

[SammyBoy]

Thanks nightrider!

[keltingmeith]

Given that P(x) can be written in the form of (x²-1)Q(x)+ax+b, where Q(x) is a polynomial and a and b are constants, hence or otherwise, find the remainder when p(x) is divided by x²-1

Check out the bit I've highlighted. If a question EVER says just "hence", you MUST use the previously indicated information to answer the question. HOWEVER, this one says "hence or otherwise" - this means there is a way to use the previous information, and they'd love you if you could, but you will not be penalised if you use any other appropriate method.

I do, however, want to highlight that the remainder theorem is only for linear factors. What I would suggest for this question is to do the actual division, then highlight that the remainder term must be the ax+b - once you've found out what the a and b are, you've found the remainder term.

[SE_JM]

Oh. I thought I could use remainder theorem for any polynomial Thank you telling me!

[keltingmeith]

Oh. I thought I could use remainder theorem for any polynomial Thank you telling me!

You can use it on any polynomial, however the remainder theorem only works when you divide by a linear function. Essentially, you can find the remainder of , r(x) by subbing in P(-b/a). This doesn't hold if the term before Q(x) is not linear.

[knightrider]

When using permutations

how do you know when to multiply things together in situations or add them together?