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June 12, 2026, 04:05:46 pm

Author Topic: VCE Methods Question Thread!  (Read 6134645 times)  Share 

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Thorium

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Re: VCE Methods Question Thread!
« Reply #6585 on: November 03, 2014, 02:14:11 pm »
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Question - if they ask to find the inverse of a function, should you always answer it in function notation (including its domain)?
Yep, that is only when you are asked to find the "inverse function". However, you only need to find the inverse equation if you are asked to find the "rule" for the inverse function.
« Last Edit: November 03, 2014, 02:16:59 pm by Thorium »
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speedy

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Re: VCE Methods Question Thread!
« Reply #6586 on: November 03, 2014, 02:16:32 pm »
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To determine graph from its average value, VCAA says:
Spoiler

I'm a bit confused with what the regions should be for other graphs...

So, for example, would these regions have to be equal?
Spoiler

Edit: or is it the full region including the rectangle (beneath)
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Robert123

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Re: VCE Methods Question Thread!
« Reply #6587 on: November 03, 2014, 02:23:46 pm »
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To determine graph from its average value, VCAA says:
Spoiler

I'm a bit confused with what the regions should be for other graphs...

So, for example, would these regions have to be equal?
Spoiler

Edit: or is it the full region including the rectangle (beneath)

If y=2 is the average value of the function, then the red area equals the blue area how we, I don't think the line y=2 would be the average value in your scenario

Edit: actually now that I think about it, that's probably entirely wrong. If you were to find the average value of the function y=2 for the interval (0,6), the average y value would be 2 (obviously) and that would not have an equal area above and below the average value.
So everything I said before is wrong so my current answer is I don't  know, sorry :(
« Last Edit: November 03, 2014, 02:31:32 pm by Robert123 »

speedy

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Re: VCE Methods Question Thread!
« Reply #6588 on: November 03, 2014, 02:33:10 pm »
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If y=2 is the average value of the function, then the red area equals the blue area how we, I don't think the line y=2 would be the average value in your scenario

Edit: actually now that I think about it, that's probably entirely wrong. If you were to find the average value of the function y=2 for the interval (0,6), the average y value would be 2 (obviously) and that would not have an equal area above and below the average value.
So everything I said before is wrong so my current answer is I don't  know, sorry :(

That's not meant to be the correct answer... I'm sorry if I didn't make that clear.

This is a multiple choice question, and one this was one of the wrong answers. I'm just trying to figure out how to check that it's wrong/right! :)
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silverpixeli

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Re: VCE Methods Question Thread!
« Reply #6589 on: November 03, 2014, 02:41:03 pm »
+1
The average value of the function over an interval is the height of a rectangle that could replace that interval and have the same area as the function had.

So here's how we figure out that height, in wordy maths:

Rectangle area = Function area

L * W = integral from a to b of f(x) dx


so let's say width is the 'height' we were talking about earlier, and length is how far along the x axis it goes (the distance from a to b) and we get

(b-a) * average value = integral from a to b of f(x) dx

divide by width (b-a);

average value = integral from a to b of f(x) dx    / (b-a)

(or in formula-speak where f(c) is the average value, thanks paul's calc notes for saving me from writing latex)

which is a formula you can put in your rules book and use to find the value required!


it's not necessarily the value that has half the area above it and half below it! (though it would be an interesting exercise to try to find that value.) it's mroe like the height you could flatten out a function into if it was a pile of stones and you had to smooth out the stones into a flat surface. How high would that surface be above the x axis? that's the average value of the function.
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Re: VCE Methods Question Thread!
« Reply #6590 on: November 03, 2014, 02:51:39 pm »
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Thankyou :)
But I'm still not sure how to go about doing this question: (VCAA 2013 Q15)
Spoiler

How would you do it?

I know it may be intuitive, but yeah...
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cwiddicombe

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Re: VCE Methods Question Thread!
« Reply #6591 on: November 03, 2014, 03:03:23 pm »
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Thankyou :)
But I'm still not sure how to go about doing this question: (VCAA 2013 Q15)
Spoiler

How would you do it?

I know it may be intuitive, but yeah...


I'd draw a line through y=2 and see in which graph the area above y=2 is approximately equal to the area below y=2. The only option that fits is C

silverpixeli

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Re: VCE Methods Question Thread!
« Reply #6592 on: November 03, 2014, 03:07:37 pm »
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Okay I understand, this isn't a formulaic question because we don't have the function definitions to figure out each absolute value, so we have to use intuition and a graphical interpretation of the average value so we can pick an answer.

Here's my thought process;

A. hmm, a trig graph with a 'mean value' of 2, isn't mean the same as average? well yeah it is, but this isn't an unrestricted trig graph so while a single cycle of this graph (and indeed any trig graph with no domain restriction) would have avg = 2, we have 1.5 cycles which shifts the average up a fair bit. Not this one!

B. one way to think of the average value is that 'flattening out stones' thing and it looks like, if we set the surface of these stones to be at y=2, we'll have too many stones to nicely fit in the two little gaps under 2 that this graph has. It's not rigorous, but we can suspect that this one isn't the answer.

C. This one's tempting, we can figure out the area using two triangles and remembering that the part below the axis needs to be subtracted and see if it works;


and by our formula, width * average value = area


it looks like avg = 2, C is the answer


let's look at the last 2 anyway,

D. simple triangle, area = 24 so avg value = 4

E. not so simple, it looks like this one works because the triangle has an area of 8 and a width of 4, but if you notice, this graph extends all the way to x=0 with a height of 0 so its actual width is 6 and avg = area/width = 8/6 not 2
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silverpixeli

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Re: VCE Methods Question Thread!
« Reply #6593 on: November 03, 2014, 03:11:03 pm »
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I'd draw a line through y=2 and see in which graph the area above y=2 is approximately equal to the area below y=2. The only option that fits is C

actually this isn't quite right, it's not the area below the line that the area has to equal, it's kind of the area that ISNT below the line that the area above has to equal. the area that isnt below the line is the same as the area between the function and your new line, instead of the x axis.
« Last Edit: November 03, 2014, 03:13:20 pm by silverpixeli »
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speedy

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Re: VCE Methods Question Thread!
« Reply #6594 on: November 03, 2014, 03:20:42 pm »
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Reus

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Re: VCE Methods Question Thread!
« Reply #6595 on: November 03, 2014, 03:47:39 pm »
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Any predictions for exam 1 guys?
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Phy124

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Re: VCE Methods Question Thread!
« Reply #6596 on: November 03, 2014, 03:50:58 pm »
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Any predictions for exam 1 guys?
I'll hazard a guess and predict a few maths related questions to be on it.
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Reus

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Re: VCE Methods Question Thread!
« Reply #6597 on: November 03, 2014, 03:52:17 pm »
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I'll hazard a guess and predict a few maths related questions to be on it.
I'm still laughing!!!
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vcestudent123

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Re: VCE Methods Question Thread!
« Reply #6598 on: November 03, 2014, 03:56:17 pm »
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Any predictions for exam 1 guys?
I predict questions that will confuse everybody.

speedy

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Re: VCE Methods Question Thread!
« Reply #6599 on: November 03, 2014, 04:00:03 pm »
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Any predictions for exam 1 guys?

Have you seen this thread? Can't remember how much exam 1 was talked about but may be good to look through

Methods Exams 2014: Predictions
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