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March 29, 2024, 09:40:23 am

Author Topic: VCE Methods Question Thread!  (Read 4802839 times)  Share 

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Perryman

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Re: VCE Methods Question Thread!
« Reply #14520 on: February 16, 2017, 09:01:31 pm »
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ok thanks heaps...
bit more simpler that way than the way i tried!

thanks again!!

deStudent

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Re: VCE Methods Question Thread!
« Reply #14521 on: February 18, 2017, 07:24:52 pm »
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http://m.imgur.com/a/esPHj
For this question, is this interpretation entirely correct? Why couldn't we directly use V = 1/3*pi*r^3, where r = 6 to get us the solution?

I'm not entirely sure what the image attached is doing?

http://m.imgur.com/ySbsXak

For part c) what is the point of the "g(x)=f(x)"? It didn't seem to effect the final answer (-infinite, 0] but what was its purpose, if it had one?

Thanks

Shadowxo

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Re: VCE Methods Question Thread!
« Reply #14522 on: February 18, 2017, 07:55:11 pm »
+3
http://m.imgur.com/a/esPHj
For this question, is this interpretation entirely correct? Why couldn't we directly use V = 1/3*pi*r^3, where r = 6 to get us the solution?

I'm not entirely sure what the image attached is doing?

http://m.imgur.com/ySbsXak

For part c) what is the point of the "g(x)=f(x)"? It didn't seem to effect the final answer (-infinite, 0] but what was its purpose, if it had one?

Thanks

Hi,
For 6, the formula for a cylinder is V = πr2h. In this case, the radius of the sphere is 6, but the radius of the cylinder is not 6. The radius can change depending on how wide and tall the cylinder is and it fits inside the sphere. The solution is using a2+b2=c2 to find the radius in terms of h, as the diameter of the cylinder, height of the cylinder and the diameter of the sphere (as the edges of the cylinder touch the sphere) form a right angled triangle which can be used to find r in terms of h. Then sub in these values to find V in terms of h.

For c), they're saying g(x) = f(x) because g(x) is just f(x) with a different domain that you have to find. It's just saying it's a different graph as it has a different domain, instead of saying "f(x) is restricted to the domain S"

Hope this helps :)
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Gogo14

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Re: VCE Methods Question Thread!
« Reply #14523 on: February 20, 2017, 04:02:26 pm »
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Really confused with this question about the order in whic you do ht etransformations.
Question 17.
I did translations of 3 units in negative y, 5 units in negative x
and then reflections in X and Y axis
and dilation of 1/2 from y .

However, the textbook had a really different answer and order of sequences which I am very confused about,.
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Sine

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Re: VCE Methods Question Thread!
« Reply #14524 on: February 20, 2017, 04:13:27 pm »
+2
Really confused with this question about the order in whic you do ht etransformations.
Question 17.
I did translations of 3 units in negative y, 5 units in negative x
and then reflections in X and Y axis
and dilation of 1/2 from y .

However, the textbook had a really different answer and order of sequences which I am very confused about,.
nah you are fine it's all correct


Translate 3 units in negative direction of the y axis


Translate 5 units in the negative direction of the x axis



Reflect in the y axis


Reflect in the x axis


Dilate by scalar factor 1/2 from the y axis


« Last Edit: February 20, 2017, 04:15:16 pm by Sine »

Gogo14

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Re: VCE Methods Question Thread!
« Reply #14525 on: February 20, 2017, 04:48:37 pm »
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Can someone do a proof of why f(f^-1(x))=x? I attempted it but here is what i got.
Let (x,y) be a coordinate on f(x) and hence (y,x) be a coordinate on f^-1(x)
Thus f(f-1(y))= f(x)=y. Hence the composite function lies on (y,y) for y elemtn of Real.hence y=x

Is my proof correct?
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Shadowxo

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Re: VCE Methods Question Thread!
« Reply #14526 on: February 22, 2017, 12:48:03 pm »
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Can someone do a proof of why f(f^-1(x))=x? I attempted it but here is what i got.
Let (x,y) be a coordinate on f(x) and hence (y,x) be a coordinate on f^-1(x)
Thus f(f-1(y))= f(x)=y. Hence the composite function lies on (y,y) for y elemtn of Real.hence y=x

Is my proof correct?

Hi, sorry for the late response. I'll give it a go :)
Your proof looks to be good, I might give it a little more detail though
Let (x,y) be a coordinate on f(x) and hence (y,x) be a coordinate on f^-1(x)
Therefore f-1(y)=x
Therefore f(f-1(y))=f(x)=y
Therefore the point (y,y) lies on the composite function, therefore the composite function lies on the line y=x therefore f(f-1(x))=x
Sorry if the notation isn't perfect, but your proof should be good :)
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Monkeymafia

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Re: VCE Methods Question Thread!
« Reply #14527 on: February 22, 2017, 06:25:56 pm »
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Hey ,

How would we graph x = -y(y-2)^2 ?

Thanks!
« Last Edit: February 22, 2017, 06:39:02 pm by Monkeymafia »

zhen

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Re: VCE Methods Question Thread!
« Reply #14528 on: February 22, 2017, 06:45:32 pm »
+1
Hey ,

How would we graph x = -y(y-2)^2 ?

Thanks!
I would graph it like I would graph a cubic, but switch the axes. So having y-intercepts on (0,0) and (0,2) and drawing it like a cubic on its side.

MightyBeh

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Re: VCE Methods Question Thread!
« Reply #14529 on: February 22, 2017, 06:45:52 pm »
+1
Hey guys,

How would we graph x = -y(y-2)^2 ?

Thanks!
If you don't need it to be a function, turn your page sideways and draw it the same way you would \(y = -x(x-2)^{2}\) ::). Make sure you mark which is the positive end of your axes or you might end up with a flipped version. Otherwise restrict the domain and find the inverse the conventional way. Not really sure what you'd restrict the domain to in this case though because there's a lot of cross over.

Otherwise it's standard cubic drawing; find the axis intercepts/turning points and then you're done, pretty much.
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Shadowxo

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Re: VCE Methods Question Thread!
« Reply #14530 on: February 22, 2017, 07:09:16 pm »
+1
Hey ,

How would we graph x = -y(y-2)^2 ?

Thanks!

As the others have said, graph it like you would normally. You know it's a negative cubic, with y intercepts at 0 and 2, bouncing off the 2. Since it's a negative cubic, as y gets larger x becomes more negative.
Alternatively, draw a graph of y=-x(x-2)^2 and flip it along the line y=x (the inverse). I wouldn't flip the page sideways as it wouldn't result in the right graph
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Sine

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Re: VCE Methods Question Thread!
« Reply #14531 on: February 22, 2017, 07:24:43 pm »
+1
Hey ,

How would we graph x = -y(y-2)^2 ?

Thanks!
if you are ever clueless i graphing something just sub numbers in and plot each point. Extremely slow method but still gets the job done. Once you have enought points you can join them up (although you still need to keep the graph neat with no kinks).

Pretty much a last ditch tactic.

Rowan1999

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Re: VCE Methods Question Thread!
« Reply #14532 on: February 22, 2017, 07:46:22 pm »
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Hi, just having trouble in finding the inverse of the following function:



For a problem like this, I would normally apply long division or the likes to put it into a more workable form, but I can't seem to figure this one out.

Thanks!

zhen

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Re: VCE Methods Question Thread!
« Reply #14533 on: February 22, 2017, 07:55:56 pm »
+1
Hi, just having trouble in finding the inverse of the following function:



For a problem like this, I would normally apply long division or the likes to put it into a more workable form, but I can't seem to figure this one out.

Thanks!
This is how I did long division to put it into a more workable form. From here just swap x and y and be careful of what you put as the domain and range of the inverse function.

Syndicate

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Re: VCE Methods Question Thread!
« Reply #14534 on: February 22, 2017, 08:08:49 pm »
+1
Hi, just having trouble in finding the inverse of the following function:

f : R\{2/3}->R ; f(x) =(2x + 3)/(3x − 2)

For a problem like this, I would normally apply long division or the likes to put it into a more workable form, but I can't seem to figure this one out.

Thanks!

Put this function in a \( \frac{A}{x-B} +b \) form.


This function is already 1:1, therefore you do not need to restrict the domain.



Domain and range are swapped.

EDIT: realised zhen has already put up an answer. I have just used a different method, which would yield the same solution.
« Last Edit: February 22, 2017, 08:13:18 pm by Syndicate »
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