thank you for making this video, I *did* find it useful. I'm essentially teaching myself further maths a few months before the gcse, and this resource has honestly been invaluable in this process. so, thank you :)
I loved this video I was able to understand concepts perfectly and solve too. I have exam in 1 week this video is amazing. Thank you for such nice content.
hi sir, for question 6, do we HAVE to prove that it's stationary before we prove it's a minimum? or would we get all the marks by just proving it's a minimum? thank you so much
You need to do both Showing that d^2y/dx^2 > 0 does not guarantee that it is a minimum, it just means the curve is curving upwards (we call this convex but you don't need that for this course). For example if y = x^2 dy/dx = 2x d^2y/dx^2 = 2 This means that no matter what x values you look at d^2y/dx^2 is always positive. It doesn't mean that all points are minimums though. So to have a minimum you must have both dy/dx = 0 d^2y/dx^2 > 0 For the x values you are looking at.
@@1stClassMaths I'm pretty sure you are required to be able to know that when the second differential is 0 then it is a point of inflection and are also required to be able to prove this by working it that the gradient on either side of the graph is either positive and positive or negative and negative. I have seen past paper questions where this is required and my teacher taught us this. Sorry if this comes across as rude.
@@itsfrankie9570 Hi. I am afraid you are incorrect. Section 4.7 of the teaching guidance that available to teachers states "Points of inflection will not be tested." There are no past paper questions on points of inflection. You can find a link to this information here: static.wixstatic.com/media/9f3fb0_130401e1809c4366b321c387078708eb~mv2.png/v1/fill/w_600,h_530,al_c,q_85,usm_0.66_1.00_0.01,enc_auto/9f3fb0_130401e1809c4366b321c387078708eb~mv2.png Additionally it is worth noting that you have been taught/understood incorrectly. If the second derivative is 0 at a point then this does not mean it is a point of inflection. It means that it is a possible point of inflection. A simple counter example would be the function f(x) = x^4 Here f'(x) = 4x^3 and f''(x) = 12x^2 At the point when x = 0 f''(x) = 0 but it is not a point of inflection, it is a minimum. Unfortunately a point of inflection is sufficient to claim that the second derivative is 0 but it does not work the other way around. Further to this either side of a point of inflection the gradient will actually not change sign as you describe. If the gradient changes sign either side of a stationary point then this is a minimum or maximum value. I believe you actually mean the second derivative either side of the point, which would change to indicate that the function goes from concave to convex or vice versa. Finally, it might also be worth noting for future study that points of inflection can occur at points where the first derivative is not zero e.g. f(x) = sin(x) has one when x = 180°/(pi rad)
@@1stClassMaths Okay I am sorry. The fact that the sign of the gradient does not change is what I was "attempting" to say when I stated either negative and negative or positive and positive. So the same on either side/does not change. And yes my teacher did mention that d2y/dx2 = 0 did not always mean point of inflection. I must have forgot to mention that. So just to clarify further. Points of inflection was therefore only required knowledge on the previous spec? As in 2019 paper 2 question 27 they asked you to prove that a point was a point of inflection using dy/dx as well as in 2016 they asked a question on points of inflection, although this knowledge may not have carried through to the current spec as you exemplified.
Hi, I'm really confused on question 3- if second part of the answer for d2y/dx2 is 8x-3, how is that equivalent to 8/x3 if you get completely different answers by putting x into the two equations that mean the same thing?
8x^(-3) is the same as 8/x^3 This is because x^-3 can be written as 1/x^3 You get the same values when you substitute a number in for x. If you are getting different ones you may be typing it wrong into the calculator
Hi sir! I was just wondering if you were going to make a summary video about all the topic in further maths as we have our first exam tomorrow?
thank you for making this video, I *did* find it useful.
I'm essentially teaching myself further maths a few months before the gcse, and this resource has honestly been invaluable in this process. so, thank you :)
Thats amazing. Well done for doing this alone, that is exactly what I made these videos for :)
how did you end up doing
I loved this video I was able to understand concepts perfectly and solve too. I have exam in 1 week this video is amazing. Thank you for such nice content.
how do we know when we are supposed to differentiate once or twice?
Once to find the gradient and a second time to find the nature of a point.
@@1stClassMaths thank you
@@1stClassMaths and for minimum value
hi sir, for question 6, do we HAVE to prove that it's stationary before we prove it's a minimum? or would we get all the marks by just proving it's a minimum? thank you so much
You need to do both
Showing that d^2y/dx^2 > 0 does not guarantee that it is a minimum, it just means the curve is curving upwards (we call this convex but you don't need that for this course).
For example if y = x^2
dy/dx = 2x
d^2y/dx^2 = 2
This means that no matter what x values you look at d^2y/dx^2 is always positive. It doesn't mean that all points are minimums though.
So to have a minimum you must have both
dy/dx = 0
d^2y/dx^2 > 0
For the x values you are looking at.
@@1stClassMaths ohhh okay thank you so much sir! You have no idea how much of a help you are!
Good luck!
are we also expected to know about points of inflection where d^2y/dx^2 =0 for the further maths gcse
No
@@1stClassMaths I'm pretty sure you are required to be able to know that when the second differential is 0 then it is a point of inflection and are also required to be able to prove this by working it that the gradient on either side of the graph is either positive and positive or negative and negative. I have seen past paper questions where this is required and my teacher taught us this. Sorry if this comes across as rude.
@@itsfrankie9570 Hi. I am afraid you are incorrect.
Section 4.7 of the teaching guidance that available to teachers states "Points of inflection will not be tested." There are no past paper questions on points of inflection.
You can find a link to this information here: static.wixstatic.com/media/9f3fb0_130401e1809c4366b321c387078708eb~mv2.png/v1/fill/w_600,h_530,al_c,q_85,usm_0.66_1.00_0.01,enc_auto/9f3fb0_130401e1809c4366b321c387078708eb~mv2.png
Additionally it is worth noting that you have been taught/understood incorrectly. If the second derivative is 0 at a point then this does not mean it is a point of inflection. It means that it is a possible point of inflection. A simple counter example would be the function f(x) = x^4
Here f'(x) = 4x^3 and f''(x) = 12x^2
At the point when x = 0 f''(x) = 0 but it is not a point of inflection, it is a minimum. Unfortunately a point of inflection is sufficient to claim that the second derivative is 0 but it does not work the other way around.
Further to this either side of a point of inflection the gradient will actually not change sign as you describe. If the gradient changes sign either side of a stationary point then this is a minimum or maximum value. I believe you actually mean the second derivative either side of the point, which would change to indicate that the function goes from concave to convex or vice versa.
Finally, it might also be worth noting for future study that points of inflection can occur at points where the first derivative is not zero e.g. f(x) = sin(x) has one when x = 180°/(pi rad)
@@1stClassMaths Okay I am sorry. The fact that the sign of the gradient does not change is what I was "attempting" to say when I stated either negative and negative or positive and positive. So the same on either side/does not change. And yes my teacher did mention that d2y/dx2 = 0 did not always mean point of inflection. I must have forgot to mention that. So just to clarify further. Points of inflection was therefore only required knowledge on the previous spec? As in 2019 paper 2 question 27 they asked you to prove that a point was a point of inflection using dy/dx as well as in 2016 they asked a question on points of inflection, although this knowledge may not have carried through to the current spec as you exemplified.
@@itsfrankie9570 yes that is correct. It is not on the new specification. You will need it for A level though. 🎉
Hi, I'm really confused on question 3- if second part of the answer for d2y/dx2 is 8x-3, how is that equivalent to 8/x3 if you get completely different answers by putting x into the two equations that mean the same thing?
8x^(-3) is the same as 8/x^3
This is because x^-3 can be written as 1/x^3
You get the same values when you substitute a number in for x. If you are getting different ones you may be typing it wrong into the calculator
Could you please make a level maths content, as your videos are so good
Thank you for this! I plan to in the future.
I start my a levels this year, do you reckon maths a level content would be available by then?
Hi. It is very hard to say. I am working on content in my free time but if the channel grows I can dedicate more time to it.