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March 29, 2024, 10:09:47 pm

Author Topic: VCE Methods Question Thread!  (Read 4803405 times)  Share 

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a weaponized ikea chair

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Re: VCE Methods Question Thread!
« Reply #19140 on: June 08, 2021, 04:41:27 pm »
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Hello. Please see attached.

1. How many points of inflection are on the blue function shown? Is it 1, 2, or 3? I read somewhere that where f''(x)=0 is a point of inflection (not necessarily a stationary point of inflection).

2. Regardless, what is the formal definition of a point of inflection?

Can anyone shed any light onto this?

1729

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Re: VCE Methods Question Thread!
« Reply #19141 on: June 08, 2021, 04:51:16 pm »
+1
Hello. Please see attached.

1. How many points of inflection are on the blue function shown? Is it 1, 2, or 3? I read somewhere that where f''(x)=0 is a point of inflection (not necessarily a stationary point of inflection).

2. Regardless, what is the formal definition of a point of inflection?

Can anyone shed any light onto this?
Note: I've edited this post as before it didn't make sense and I didn't want to potentially mislead anybody.

Considering you are using second derivative test I'm assuming you have knowledge about concaves.

Point of inflection is defined as a point at which the concavity of the graph changes, the point at which a function transfers from concave up to concave down (or vice versa).

However, please note \(f''(x)=0\) does not always result to point of inflection it can be local max, local, min or inflection. You need to further consider it's domain and if satisfies the above definition of an inflection point. For example, \(f(x)=x^4\) has second derivative of \(f''(x)=12x^2\) and from it's first derivative, we can see it's stationary point is located at \((0,0)\), so we sub that in the second derivative to see it's nature and we get \(f''(x)=12x^2=12(0)^2=0\). Some students may just assume straight away it has a stationary point of inflection at \(x=0\), however considering domains (\(f''(x)>0\) for all \(x\)) you can see that it's second derivative is strictly positive meaning the whole graph is concave up and therefore can never have a point at which changes from concave up to concave down (or vice versa).

Note: Second derivative test is not in the study design, you are not required to know about concavity in methods, you are welcomed to use the textbook method that determines the nature of stationary point by examining the first derivative around that point. Some internal assessments may want you to use the textbook method.

Finding the nature of a stationary point by using first derivative is like so:
If we have a stationary point at \(x=0\) and we have \(f'(x)<0\) for when \(x<c\) and \(f'(x)>0\) for when \(x>c\) that will give a local min. And if we have  \(f'(x)>0\) for when \(x<c\) and \(f'(x)<0\) for when \(x>c\) will give a local max. And if we have \(f'(x)>0\) for when \(x<c\) and \(f'(x)>0\) for when \(x>c\) we have a stationary point of inflection.

Refer to 2010 MCQ 17 if you want a question where you need to use the textbook (first derivative) method.

If it is more convenient for you, you can draw up a increase/decreasing diagram. For a local min the diagram will look like this \_/
For a local max the diagram will look like this /-\ and for an inflection it should look like this \_\ or /-/ (kind of like a linear cubic wave). Where _ represents 0 gradient, / represents positive gradient and \ represents negative gradient. (like a linear function would).
« Last Edit: August 02, 2021, 09:27:55 pm by 1729 »

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Re: VCE Methods Question Thread!
« Reply #19142 on: June 08, 2021, 04:56:14 pm »
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Considering you are using second derivative test I'm assuming you have knowledge about concaves.
Point of inflection is defined as a point at which the concavity of the graph changes.

f"(x)=0 does not always result to point of inflection, it can be local max, local, min or inflection. You need create a table of values to determine the nature of stationary point when f"(x)=0.

For example


Has stationary point at x=0 and second derivative of

The second derivative is zero but it isn't point of inflection. If we set up table of values

x -1   0    1
y 12  0    12     

As we can see there are no changes in concavity and it is a local mininum as it is (concave up or second derivative is more than 0)

Thank you for your response. That clears up what a point of inflection is.

Regarding f(x), would that therefore mean that there is only one POI?

1729

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Re: VCE Methods Question Thread!
« Reply #19143 on: June 08, 2021, 05:01:08 pm »
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Thank you for your response. That clears up what a point of inflection is.

Regarding f(x), would that therefore mean that there is only one POI?
Yes.

a weaponized ikea chair

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Re: VCE Methods Question Thread!
« Reply #19144 on: June 08, 2021, 05:18:02 pm »
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Yes.

Sorry for asking, but could you please explain which one is the POI, and why the other candidates aren't. This is from a practice SAC where the teacher said there were two (supposedly the correct answer), I put three, and others only put one.

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Re: VCE Methods Question Thread!
« Reply #19145 on: June 08, 2021, 05:50:24 pm »
+3
Sorry for asking, but could you please explain which one is the POI, and why the other candidates aren't. This is from a practice SAC where the teacher said there were two (supposedly the correct answer), I put three, and others only put one.
Apologies there is actually three non-stationary points of inflection. The picture below should help, I've put arrows to explain the changes in concavity.

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19146 on: June 08, 2021, 05:50:36 pm »
+4
Should be three points of inflection; the second derivative changes sign three times. \(x^4\) has its second derivative equal to zero at zero but doesn't have a POI because its second derivative is strictly greater than or equal to zero. Here, you can see that the second derivative changes sign three times. It's actually pretty hard to see the changes in concavity but they are there - zoom in if you need to convince yourself again! :)

As for why your teacher said two, I can only think of two reasons: a) the domain you're considering might be restricted or b) your teacher might've made a mistake. Either way, it's safer to reserve judgement until more context is provided - if you can provide us with more context we might be able to resolve this with fewer queries :D

Hope this helps!
« Last Edit: June 08, 2021, 06:16:41 pm by fun_jirachi »
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a weaponized ikea chair

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Re: VCE Methods Question Thread!
« Reply #19147 on: June 08, 2021, 06:04:13 pm »
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Should be three points of inflection; the second derivative changes sign three times. \(x^4\) has its second derivative equal to zero at zero but doesn't have a POI because its second derivative strictly greater than or equal to zero. Here, you can see that the second derivative changes sign three times. It's actually pretty hard to see the changes in concavity but they are there - zoom in if you need to convince yourself again! :)

As for why your teacher said two, I can only think of two reasons: a) the domain you're considering might be restricted or b) your teacher might've made a mistake. Either way, it's safer to reserve judgement until more context is provided - if you can provide us with more context we might be able to resolve this with fewer queries :D

Hope this helps!

Thank you for your responses 1729 and jirachi. So I guess I was right all along!

The only thing was that the domain was between -2 and 2 though I don't think this changes the answer.

@Jirachi, what do you mean by:

"x^4 has its second derivative equal to zero at zero but doesn't have a POI because its second derivative strictly greater than or equal to zero"

Are you saying that x=0 is not a POI? Cause I got (approx rounded) the points x=-1.1, x=0, and x=1.1.

Thanks!
« Last Edit: June 08, 2021, 06:06:04 pm by a weaponized ikea chair »

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19148 on: June 08, 2021, 06:15:06 pm »
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Fair enough, seems like your teacher was probably wrong (or I'm missing something) - though it definitely is more helpful to provide the full question in future (just to avoid unnecessary tangents and/or the XY problem).

In reference to your second question, no, I was referring to the example 1729 provided - should have made it clearer that that was in fact a separate example. I did think it was somewhat clear that that particular example had no relation because your graph clearly has three points of inflection while \(f(x) = x^4\) has none. I am not in any way saying that x = 0 is not a POI (because I did say there were three). Wouldn't contradict myself, after all. I'll be much clearer in future, apologies :)
« Last Edit: June 08, 2021, 06:35:35 pm by fun_jirachi »
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Samueliscool223

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Re: VCE Methods Question Thread!
« Reply #19149 on: June 13, 2021, 07:20:09 pm »
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Why is my CAS giving me the wrong answer? I found d/dx of (850/(x^2)) + ((1700/((12-x)^3))
by hand multiple times, and I got (-2*850(x^-3)) + (1700*2((12-x)^-3)) yet my CAS says both fractions are negative. Please explain, I am confused AF.

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Re: VCE Methods Question Thread!
« Reply #19150 on: June 14, 2021, 01:58:04 pm »
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For Methods Exam 1 2013 Question 10C,
Shouldn't the answer be 55/2*ln(2) -15/2, rather than 25/4*ln(4) -15/2 as stated in the assessors report?  (which is derived from the definite integral above (from the right side of "or")
Would someone be able to help me check?
Thanks!!

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19151 on: June 14, 2021, 02:18:48 pm »
+1
Next time, please link the question in your post, easier for people viewing and answering your question :)

The answer in the solutions is correct - can you show us your working, so we can see where you went wrong? It's difficult to help you out without much context.

The area is just the area of the trapezium minus the definite integral over the same domain. You seem to have gotten only the area of the trapezium wrong.
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Re: VCE Methods Question Thread!
« Reply #19152 on: June 14, 2021, 03:37:45 pm »
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Next time, please link the question in your post, easier for people viewing and answering your question :)

The answer in the solutions is correct - can you show us your working, so we can see where you went wrong? It's difficult to help you out without much context.

The area is just the area of the trapezium minus the definite integral over the same domain. You seem to have gotten only the area of the trapezium wrong.
Apologies
The attached image (which is the alternative way of solving the question and the method I chose) does not equate to the answer provided in the solution?
Unless I have excluded the trapezium?
Regards
« Last Edit: June 14, 2021, 03:40:43 pm by aspiringantelope »

fun_jirachi

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Re: VCE Methods Question Thread!
« Reply #19153 on: June 14, 2021, 04:31:42 pm »
+10
You're calculating the area of a larger trapezium that does not match the one in the question.

The correct integral is \(\int_0^{5\ln 4} \frac{-3x}{10\ln 4} + 2 - 2e^{-\frac{x}{5}}\). If you split up the integral into the integral of the curve and the integral for the trapezium it should become abundantly clear that at \(x = 5\ln 4\) the corresponding y-value is not 1/2 for the function that defines the top of the trapezium, but rather 7/2. This is why you're returning a value much larger than the correct answer - you haven't excluded the trapezium but you've calculated an additional area on top of the trapezium.

Hope this helps :)
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Corey King

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Re: VCE Methods Question Thread!
« Reply #19154 on: June 14, 2021, 05:22:50 pm »
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https://snipboard.io/px8q0n.jpg

https://snipboard.io/6ykVvc.jpg

https://snipboard.io/2rtf8j.jpg


Hey guys,

Does anyone happen to know how to solve these questions?
I've given them my best shot, but clearly don't know how to solve them.

Many thanks,
Corey