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March 28, 2024, 07:19:08 pm

Author Topic: VCE Methods Question Thread!  (Read 4802162 times)  Share 

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Syndicate

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Re: VCE Methods Question Thread!
« Reply #14640 on: March 26, 2017, 06:58:09 pm »
+2
Could someone please guide me with these questions. I get really close to the answer, but not quite there.

Thanks!  :)

b) Using the quotient rule:


c) Using the chain rule:




d) Using the chain rule and the product rule.

« Last Edit: March 26, 2017, 07:04:55 pm by Syndicate »
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geminii

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Re: VCE Methods Question Thread!
« Reply #14641 on: March 28, 2017, 08:20:40 pm »
0
What is the difference between a point of inflection and a stationary point of inflection? I'm doing the following question:

Let f: R->R, f(x) = x(x-2)^2
It is true to say that the graph of y = f(x) has
A. A stationary point of inflection
B. A local minimum located at a point where x<0
C. A local maximum located at a point where x<0
D. A point of inflection
E. Three stationary points.

I have narrowed it down to A and D after graphing it - however I know it doesn't have a point of inflection, so I believe the answer is A. The thing is I don't know what a stationary point of inflection is?

Thanks in advance :)
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Quantum44

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Re: VCE Methods Question Thread!
« Reply #14642 on: March 28, 2017, 08:41:01 pm »
+3
What is the difference between a point of inflection and a stationary point of inflection? I'm doing the following question:

Let f: R->R, f(x) = x(x-2)^2
It is true to say that the graph of y = f(x) has
A. A stationary point of inflection
B. A local minimum located at a point where x<0
C. A local maximum located at a point where x<0
D. A point of inflection
E. Three stationary points.

I have narrowed it down to A and D after graphing it - however I know it doesn't have a point of inflection, so I believe the answer is A. The thing is I don't know what a stationary point of inflection is?

Thanks in advance :)

A point of inflection is when the graph changes concavity, however this can occur on a rising, falling on flat plane. The point of inflection is only stationary if the concavity change occurs on a flat plane. This can be tested using First and second derivatives. If f''(x) = 0 then there is a point of inflection, if f'(x) also = 0 then there is a stationary point of inflection.
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plsbegentle

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Re: VCE Methods Question Thread!
« Reply #14643 on: March 28, 2017, 09:30:41 pm »
0
A point of inflection is when the graph changes concavity, however this can occur on a rising, falling on flat plane. The point of inflection is only stationary if the concavity change occurs on a flat plane. This can be tested using First and second derivatives. If f''(x) = 0 then there is a point of inflection, if f'(x) also = 0 then there is a stationary point of inflection.
Do we need to know this for methods, always thought point of inflection and stationary point of inflection are technically the same thing. Going by what you said, the answer is D?
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Quantum44

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Re: VCE Methods Question Thread!
« Reply #14644 on: March 28, 2017, 09:53:11 pm »
0
Do we need to know this for methods, always thought point of inflection and stationary point of inflection are technically the same thing. Going by what you said, the answer is D?

I'm not sure that it's in the methods course since I learnt it in spesh
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Syndicate

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Re: VCE Methods Question Thread!
« Reply #14645 on: March 28, 2017, 09:53:22 pm »
0
A point of inflection is when the graph changes concavity, however this can occur on a rising, falling on flat plane. The point of inflection is only stationary if the concavity change occurs on a flat plane. This can be tested using First and second derivatives. If f''(x) = 0 then there is a point of inflection, if f'(x) also = 0 then there is a stationary point of inflection.

I would have thought this graph had a point of inflection rather than a stationary one. f'(4/3) =/= 0 (4/3 is the x coordinate for the point of inflection).

Do we need to know this for methods, always thought point of inflection and stationary point of inflection are technically the same thing. Going by what you said, the answer is D?

This is in the methods book (Cambridge). So I would assume that you will need to know it for methods.
« Last Edit: March 28, 2017, 09:57:24 pm by Syndicate »
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Guideme

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Re: VCE Methods Question Thread!
« Reply #14646 on: March 29, 2017, 12:07:08 am »
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Need help with this question pls
:0 :)

lzxnl

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Re: VCE Methods Question Thread!
« Reply #14647 on: March 29, 2017, 08:12:20 pm »
+1
A point of inflection is when the graph changes concavity, however this can occur on a rising, falling on flat plane. The point of inflection is only stationary if the concavity change occurs on a flat plane. This can be tested using First and second derivatives. If f''(x) = 0 then there is a point of inflection, if f'(x) also = 0 then there is a stationary point of inflection.

Not necessarily true. The graph y = x^4 has zero first and second derivatives at the origin, but it clearly has a global minimum there. Strictly speaking, points of inflection occur when the first derivative goes from increasing to decreasing or vice-versa. If the first derivative is differentiable, which it generally will be in VCE maths, a point of inflection occurs then when the second derivative changes sign. If the second derivative is discontinuous, you can actually have a point of inflection without the second derivative ever being zero. If the second derivative is continuous, like you normally have in VCE, then the point of inflection is solved for by setting the second derivative to zero. BUUUUUUUT strictly speaking, you MUST check that the second derivative does actually change sign there; it can be zero without changing sign, like the second derivative of x^4, which is 12x^2 and obviously doesn't change sign at the origin.
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noregret

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Re: VCE Methods Question Thread!
« Reply #14648 on: March 29, 2017, 08:31:18 pm »
0
How come log base 9(1/2)^3=1/2?

lzxnl

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Re: VCE Methods Question Thread!
« Reply #14649 on: March 29, 2017, 08:55:18 pm »
+1
How come log base 9(1/2)^3=1/2?

It definitely isn't.
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Guideme

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Re: VCE Methods Question Thread!
« Reply #14650 on: April 02, 2017, 06:05:07 pm »
0
Is it possible to do question 2 g without a calculator
« Last Edit: April 02, 2017, 06:10:03 pm by Guideme »
:0 :)

zhen

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Re: VCE Methods Question Thread!
« Reply #14651 on: April 02, 2017, 07:48:48 pm »
+4
Is it possible to do question 2 g without a calculator(Image removed from quote.)
This is how I would do it. Someone else should check the answer, since I might be wrong.
« Last Edit: April 02, 2017, 07:51:49 pm by zhen »

cookiedream

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Re: VCE Methods Question Thread!
« Reply #14652 on: April 02, 2017, 09:26:43 pm »
+3
This is how I would do it. Someone else should check the answer, since I might be wrong.

Yep! That's exactly how I'd do it :)
If you wanted to go a step further, you can use the change of base rule like I did below.
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gameboy99

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Re: VCE Methods Question Thread!
« Reply #14653 on: April 03, 2017, 01:23:13 pm »
0
Hi, Can someone help with this exponential transformations question.
Here's my working out. http://imgur.com/a/ie2tp
And here's the question itself. http://imgur.com/a/C8Cvq
If somebody can look at my working for 3b) and tell me where I went wrong it would be great  :D
Thanks
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Syndicate

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Re: VCE Methods Question Thread!
« Reply #14654 on: April 03, 2017, 02:13:44 pm »
0
Hi, Can someone help with this exponential transformations question.
Here's my working out. http://imgur.com/a/ie2tp
And here's the question itself. http://imgur.com/a/C8Cvq
If somebody can look at my working for 3b) and tell me where I went wrong it would be great  :D
Thanks

It should be (x,y) -> (x,2y) ->(x+3, 2y-4) -> (x+3, -(2y-4)) = (x', y')

x' = x + 3
y' = -(2y - 4)

therefore x = x' - 3  and y = (-y' + 4)/2

Eqn. of the transformed function =>

I think the answer from the book is incorrect.
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