Thank you for the video! Can you please make a video on exam technique and where the marks are awarded because that is a part I slightly struggle on as I don't know the amount of working out I should do I.e. If the question is asking to differentiate sinxcosx I can easily do that in one step but it is three marks in the exam so I don't know whether I'll be awarded full marks, hopefully that makes sense.
@@BicenMaths Sorry to revive an old thread, but by saying it's defined for x>0, do you mean that the root they're asking for (between 1.7 and 1.8) is in an interval greater than 0? So from sketching the graph, you'd know that the graph would be continuous at that point (since the region the point's in is past the vertical asymptote)?
@@rayislam2031 OK this is gonna be a project - some of the software I need can't be downloaded on the Mac because it's so old. What is wrong with me! I wish UA-cam had a screen cast built in.
Continuous means there’s no gaps in it - so polynomials will always be continuous, as will simple log and exponential graphs. If a graph has a vertical asymptote, that usually means it’s not always continuous. You can tell this if there’s a fraction and a value of x that would make the denominator 0. Dividing by 0 results in an asymptote. Usually though, if they’re asking you to do numerical methods, you can just say that the graph is continuous otherwise they wouldn’t be asking the question!
Hi Bicen, there is an exam question: Figure 3 shows that the price reaches a local minimum between 9 and 11 hours after trading begins. (b) Using the Newton-Raphson procedure once and taking to =9.9 as a first approximation, find a second approximation of when the price reaches a local minimum. I checked it's answer and it uses the the second derivative to get the right answer. Is the textbook and in your video, nowhere is mentioned that we need to use the second derivative. Can you please clarify this for me. Thanks
It's because it says it reaches a local minimum - so to find a minimum, you need to differentiate. Then you are trying to find where that function now has a solution of 0, hence needing to differentiate it again to use NR method. So first time is to get the gradient function, second time is for NR.
Best way is always to use the homepage, scroll to the section you need, and click on the more playlists bit - it will have them all, and in the correct order too! @@jammydodger1277
i have this question i know a) but with b) how would i work it out The equation In (x/3) - (x ^ 2)/4 + 2 = 0 has two solutions. a) Show that one of the solutions equals 0.425 correct to three significant figures. b) The other solution lies between positive integers and k+1 Find the value of k.
You’d need to use an iterative method to find the other solution- so make an iterative formula by rearranging the original equation so x is the subject, then perform iteration. It should hopefully give you the second solution so you can then say which integers it lies between!
@@C-wd1nl Even if you can't visualise it, I'm trying to say that there are no points where there is a vertical asymptote - we know this as exponentials have no vertical asymptote, and nor does the linear part that goes with it.
@@C-wd1nluse desmos. You can type in graph functions and see what they look like. 1/x is not continuous because there’s no value when x=0 as 1/0 = an undefined (no) number
![Lp. Сердце Вселенной #60 РОЖДЕНИЕ ЛОЛОЛОШКИ [Финал] • Майнкрафт](http://i.ytimg.com/vi/YoR0pAV9FVQ/mqdefault.jpg)
Again massively impressed by the way pf teaching. The rate of students getting good at maths is increasing exponentially by your videos.
Thank you!
Nice pun there
Gonna watch this video with some popcorn
🍿📚
Yup my favourite movie
Great video👍
Lol you got no reply 😅😂
i thought my laptop was frying up i was so confused lol
My laptop was fighting for its life here. This needs a re-record!
Laptop bout to set on fire 😭 i feel your pain
Literally thought it was gonna melt. MacBook hanging on since 2011 👏🏼 Well done to it for surviving 🔥 (Hopefully it wasn’t too distracting)
Trust it’s louder than a PS4
H4M2 I’m working on using our normal mic now so we don’t have that constant fan noise 😤
Btw sir to get a more continuous ink flow try increase/decrease the pen tip sensitivity
Thanks, I’ll play around with it - doesn’t help that this laptop is crushed by the amount I’m trying to make it do!
Thank you for the video! Can you please make a video on exam technique and where the marks are awarded because that is a part I slightly struggle on as I don't know the amount of working out I should do I.e. If the question is asking to differentiate sinxcosx I can easily do that in one step but it is three marks in the exam so I don't know whether I'll be awarded full marks, hopefully that makes sense.
Yes that makes sense! I've got quite a long list of projects I'm currently/about to work on, but I'll add this on!
Hello Sir, for the question at 17:55, why is f(x) continuous? I thought that 1/x wasn't a continuous function.
Hi I think that if you plot the graph of y = ln x - 1/x that graph would be continuous
Yep, this is correct! Note that it’s also only defined for x>0.
@@BicenMaths Thank you very much, Sir and also WaterMelon
@@BicenMaths Sorry to revive an old thread, but by saying it's defined for x>0, do you mean that the root they're asking for (between 1.7 and 1.8) is in an interval greater than 0? So from sketching the graph, you'd know that the graph would be continuous at that point (since the region the point's in is past the vertical asymptote)?
@@nerimaken482 Yes that is correct!
Woah a new set up!
It feels so weird teaching without any students 😭
Bicen Maths why don’t u do a livestream
What’s the benefit of a livestream?
@@BicenMaths live questions and answers like a normal class minus the downside of us catting so much
@@rayislam2031 OK this is gonna be a project - some of the software I need can't be downloaded on the Mac because it's so old. What is wrong with me! I wish UA-cam had a screen cast built in.
How do i know when an equation is continuous and also when there is a vertical asymptote just by looking at it?
Continuous means there’s no gaps in it - so polynomials will always be continuous, as will simple log and exponential graphs. If a graph has a vertical asymptote, that usually means it’s not always continuous. You can tell this if there’s a fraction and a value of x that would make the denominator 0. Dividing by 0 results in an asymptote. Usually though, if they’re asking you to do numerical methods, you can just say that the graph is continuous otherwise they wouldn’t be asking the question!
thank you so much
Hi Bicen,
there is an exam question:
Figure 3 shows that the price reaches a local minimum between 9 and 11 hours after trading begins.
(b) Using the Newton-Raphson procedure once and taking to =9.9 as a first approximation, find a second approximation of when the price reaches a local minimum.
I checked it's answer and it uses the the second derivative to get the right answer. Is the textbook and in your video, nowhere is mentioned that we need to use the second derivative. Can you please clarify this for me.
Thanks
It's because it says it reaches a local minimum - so to find a minimum, you need to differentiate. Then you are trying to find where that function now has a solution of 0, hence needing to differentiate it again to use NR method. So first time is to get the gradient function, second time is for NR.
hey sir, there is no playlist for chapter 10 numerical methods- can you please make one??
There is! It's linked in the description, but I'll paste it here too - ua-cam.com/play/PL0SSkmc4r_BYt4mkC-iGQi1tS5R8Uy0Ca.html
@@BicenMaths oh thanks- couldnt find it when i searched your channel
Best way is always to use the homepage, scroll to the section you need, and click on the more playlists bit - it will have them all, and in the correct order too! @@jammydodger1277
i have this question i know a) but with b) how would i work it out
The equation In (x/3) - (x ^ 2)/4 + 2 = 0 has two solutions.
a) Show that one of the solutions equals 0.425 correct to three significant figures.
b) The other solution lies between positive integers and k+1 Find the value of k.
You’d need to use an iterative method to find the other solution- so make an iterative formula by rearranging the original equation so x is the subject, then perform iteration. It should hopefully give you the second solution so you can then say which integers it lies between!
thankyou
How did u know gx was continuous? 13:21
Because there’s nowhere that might produce a vertical asymptote, like you would get with say 1/x.
how do you visualise the graph? We never learnt the graph in 13:21@@BicenMaths
@@C-wd1nl Even if you can't visualise it, I'm trying to say that there are no points where there is a vertical asymptote - we know this as exponentials have no vertical asymptote, and nor does the linear part that goes with it.
@@C-wd1nluse desmos. You can type in graph functions and see what they look like. 1/x is not continuous because there’s no value when x=0 as 1/0 = an undefined (no) number
this topic has got to be my least favorite so far
(maybe my least favourite to teach, too? It’s not very inspiring…)