Integration
Scroll to see replies
Original post by just george
Surely that would be a point of inflection (or something like that) if it takes the value of 0.
If you think about what f''(x) actually is, its the gradient of the gradient.
The gradient of the gradient at any maximum is clearly going to be negative
If you think about what f''(x) actually is, its the gradient of the gradient.
The gradient of the gradient at any maximum is clearly going to be negative
Not necessarily, in the case of stationary points of inflection you need to investigate the behaviour of the neighbourhood.
For example:
If y= x^4
dy/dx = 4x^3
dy/dx=0 -> x=0, SP at (0,0) = A
d^2y/dx^2 = 12x^2
At x=0, SD =0
So it would imply you have a SP point of inflection.
But if you further investigate the neighbourhood:
when x=-1, y= 1 = A1
when x=1, y=1 = A2
What this means is that:
A1> A and A2>A
So therefore, we would have a minimum.
Original post by f1mad
Not necessarily, in the case of stationary points of inflection you need to investigate the behaviour of the neighbourhood.
We can also check whether the point is the point of inflection by finding the 3rd derivative, see my previous post.
Now in the case of
y= -x^64
dy/dx = -64x^63
dy/dx=0 -> SP at x=0 = A
d^2y/dx^2 = -4032x^62
At x=0, SD= 0. Which again would imply a SP of inflection.
However again with careful consideration:
When x=1, y= -1 = A1
When x= -1, y = -1 = A2
Doing some analysis: A1< A, A2 < A.
Hence you have a maximum point.
y= -x^64
dy/dx = -64x^63
dy/dx=0 -> SP at x=0 = A
d^2y/dx^2 = -4032x^62
At x=0, SD= 0. Which again would imply a SP of inflection.
However again with careful consideration:
When x=1, y= -1 = A1
When x= -1, y = -1 = A2
Doing some analysis: A1< A, A2 < A.
Hence you have a maximum point.
Original post by raheem94
We can also check whether the point is the point of inflection by finding the 3rd derivative, see my previous post.
That's not really a thorough analysis though is it?
Original post by f1mad
That's not really a thorough analysis though is it?
Its a correct method.
If f(x)=0, f''(x)=0 and f'''(x) does not =0, this means the point is the point of inflection.
There is no need to do the other method of checking the sign of derivative of both sides to find a point of inflection.
Original post by raheem94
Its a correct method.
If f(x)=0, f''(x)=0 and f'''(x) does not =0, this means the point is the point of inflection.
There is no need to do the other method of checking the sign of derivative of both sides to find a point of inflection.
If f(x)=0, f''(x)=0 and f'''(x) does not =0, this means the point is the point of inflection.
There is no need to do the other method of checking the sign of derivative of both sides to find a point of inflection.
That still not an analysis, that's an assertion (or a rule it doesn't necessarily show why).
Yes there is a need to do so, how else do you identify a max or min point? In some cases (i.e here), you don't need to use the first derivative to work out whether either side is + or - if you know the general curvature of the curve, nearby the stationary point.
(edited 12 years ago)
Original post by f1mad
That still not an analysis, that's an assertion (or a rule it doesn't necessarily show why).
Yes there is a need to do so, how else do you identify a max or min point? In some cases (i.e here), you don't need to use the first derivative to work out whether either side is + or - if you know the general curvature of the curve, nearby the stationary point.
Yes there is a need to do so, how else do you identify a max or min point? In some cases (i.e here), you don't need to use the first derivative to work out whether either side is + or - if you know the general curvature of the curve, nearby the stationary point.
Your way is correct for identifying max or min points. But i was just stating a way which can be used to check if a point is a point of inflection or not.
And yes, if we know the general curvature, then we don't need to work out the 1st derivative.
I'll chat to you guys tomorrow, just several questions left.
Original post by raheem94
But i was just stating a way which can be used to check if a point is a point of inflection or not.
There's nothing wrong with your method, except it doesn't say why.
Original post by f1mad
Now in the case of
y= -x^64
dy/dx = -64x^63
dy/dx=0 -> SP at x=0 = A
d^2y/dx^2 = -4032x^62
At x=0, SD= 0. Which again would imply a SP of inflection.
However again with careful consideration:
When x=1, y= -1 = A1
When x= -1, y = -1 = A2
Doing some analysis: A1< A, A2 < A.
Hence you have a maximum point.
y= -x^64
dy/dx = -64x^63
dy/dx=0 -> SP at x=0 = A
d^2y/dx^2 = -4032x^62
At x=0, SD= 0. Which again would imply a SP of inflection.
However again with careful consideration:
When x=1, y= -1 = A1
When x= -1, y = -1 = A2
Doing some analysis: A1< A, A2 < A.
Hence you have a maximum point.
I see.. Sorry
i did do that but somehow miss interpreted the analysis :L oops.
Odd i was never taught this :L
Original post by raheem94
This answer is wrong.
I got my grad as 5 when x is 1, and the points are (1,9)...
Which equation do I use for 42?
Original post by f1mad
There's nothing wrong with your method, except it doesn't say why.
Maybe you can help me with 43....
Original post by King-Panther
Maybe you can help me with 43....
Have you drawn the graph?
It's fairly obvious after that (integration).
Original post by f1mad
Have you drawn the graph?
It's fairly obvious after that (integration).
It's fairly obvious after that (integration).
The graph crosses through the x axis at -2?
I got 3 when subbed in 3 and got 4/3 when subbed in -2... why did I need to differentiate?
(edited 12 years ago)
Original post by King-Panther
The graph crosses through the x axis at -2?
No, if y= x^2-2
Then when y=0 -> x^2 = 2. x= +-sqrt2.
Original post by f1mad
No, if y= x^2-2
Then when y=0 -> x^2 = 2. x= +-sqrt2.
Then when y=0 -> x^2 = 2. x= +-sqrt2.
o.k, why did i need to do that, i dont understand what the question wanted and what would i do next
Original post by King-Panther
o.k, why did i need to do that, i dont understand what the question wanted and what would i do next
It wants you to find the area bounded by the curve, the x axis and the lines x=3 and x= -2.
So what would integrate here by looking at the graph?
Original post by f1mad
Not necessarily, in the case of stationary points of inflection you need to investigate the behaviour of the neighbourhood.
For example:
If y= x^4
dy/dx = 4x^3
dy/dx=0 -> x=0, SP at (0,0) = A
d^2y/dx^2 = 12x^2
At x=0, SD =0
So it would imply you have a SP point of inflection.
For example:
If y= x^4
dy/dx = 4x^3
dy/dx=0 -> x=0, SP at (0,0) = A
d^2y/dx^2 = 12x^2
At x=0, SD =0
So it would imply you have a SP point of inflection.
No , it implies there MIGHT be a point of inflection
Original post by steve2005
No , it implies there MIGHT be a point of inflection
Did you not read the rest of the post?
Original post by f1mad
Did you not read the rest of the post?
Yes, but you should not give a statement which is not true. Your statement should have been worded differently.
You said it implies, it only suggests the possibility.
Uploaded with ImageShack.us
(edited 12 years ago)
Quick Reply
Related discussions
- Mechanics help
- essay question
- Help integration
- Help with differential equations please.
- Help maths urgent ‼️
- integration
- Help urgent maths
- Applied Calculus Help
- Help I don’t understand integration
- A level maths mechanics question (1)
- Does anyone know a sheet with all the differentiated and integrated functions?
- A Level maths mechanics help
- Maths: Integration
- Postgraduate Master's Loan - 2
- Can I get student finance for my 4th year on my MSci course?
- A-level Maths, Integration, Finding area between curves & Lines.
- Further maths integration help
- Parametric integration
- Integrated masters
- Integration question
Latest
Trending
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths
71
Trending
Last reply 2 days ago
Did Cambridge maths students find maths and further maths a level very easy?Last reply 2 weeks ago
Edexcel A Level Mathematics Paper 2 unofficial mark scheme correct me if wrongMaths
71
