VCE Methods Question Thread!

[lzxnl]

As it is in the study design it is entirely possible that VCAA may one day chuck it in Exam 1 to throw off people who may have just ‘skipped’ over a part of the study design.

You leave out anything in the course at your own peril.

[Lear]

You leave out anything in the course at your own peril.

Agreed

[Sine]

Does the ‘sum and product of functions’ come up often in methods? Like graphing (f+g)(x) and (fg)(x). I assume it’s important because it’s in the textbook... but my teachers have told us to skip the topic so I’m unsure whether it’s actually worth going over.

pretty sure the last time it came up was like 1997-1999 where we had it was graphing via addition of ordinates and although it's not too common - VCAA has left question styles  in various subjects for 10+ years only for them to come up when no one expects them.  It's quite intuitive if you learn the rules so you will have plenty of time to properly learn them.

[AR1472]

Ok thanks everyone, I guess I'll go over the topic but not into as much detail as what I've been asked to learn.

[WhompingWillow]

Need some help with composition functions

Let f:[0,∞)→R,f(x)= x+1
 Let h:R→R,h(x)=x^2 +3

State the range of f (h(x)).

[Aaron]

Need some help with composition functions

Let f:[0,∞)→R,f(x)= x+1
 Let h:R→R,h(x)=x^2 +3

State the range of f (h(x)).

Have you got any working out so far? If not, can you explain your understanding so we can work with something. Just show us what you currently know so we can help get you to the end step. Doing the Q for you won't help much.

[WhompingWillow]

Have you got any working out so far? If not, can you explain your understanding so we can work with something. Doing the Q for you won't help much.

I haven’t wrote anything down because I don’t know how to solve it. First time seeing this.

[RuiAce]

\begin{align*}f(h(x)) &= f(x^2+3)\\ &= (x^2+3)+1 \\ &= x^2+4 \end{align*}

Note: It may be worth mentioning that the domain of \(f\circ h\) did not need to be restricted here. But I don't know if that's even covered in methods which is why I'm not commenting on it too much yet.

[WhompingWillow]

\begin{align*}f(h(x)) &= f(x^2+3)\\ &= (x^2+3)+3 \\ &= x^2+4 \end{align*}

Note: It may be worth mentioning that the domain of \(f\circ h\) did not need to be restricted here. But I don't know if that's even covered in methods which is why I'm not commenting on it too much yet.

Thanks

[WhompingWillow]

Oops just realised that the answer in the book is [2,infinity)

[RuiAce]

Oops just realised that the answer in the book is [2,infinity)

That most certainly cannot be the case.


That would be the case, however, if \(f(x) = x-1\) instead.

[WhompingWillow]

That most certainly cannot be the case.
(Image removed from quote.)

That would be the case, however, if \(f(x) = x-1\) instead.

my mistake it’s Let f:[0,∞)→R,f(x)=square root x+1
And Let h:R→R,h(x)=x2 +3.

[RuiAce]

my mistake it’s Let f:[0,∞)→R,f(x)=square root x+1
And Let h:R→R,h(x)=x2 +3.

This is a more generic method. It can also be adapted to the previous problem.

[WhompingWillow]

This is a more generic method. It can also be adapted to the previous problem.

Thanks a lot

[aspiringantelope]

That most certainly cannot be the case.
(Image removed from quote.)

That would be the case, however, if \(f(x) = x-1\) instead.

Omg @RuiAce, what application is that? Or what did you use to graph them?