points of inflexion and horizontal p.o.i (1 Viewer)

izzy_xo

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just wondering if anyone could help explain the difference between the two?
:)

cheers.

so say for y=(x-1)^3
and you find the first deriv and make it equal zero and x=1
and there is no change in concavity, how do you know if its a
point of inflexion or horizontal point of inflexion.
 
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annabackwards

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For all points of inflection, y'' = 0

For horizontal points of inflection, y''= 0 AND y' = 0 :)

** remember to show a change in concavity when proving a point is a point of inflection ;)
 

gibbo153

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a point of inflexion is just a point where concavity changes on a curve. the 'line' that can be drawn through the point is not horizontal

a horizontal point of inflexion = point where concavity changes and the 'line' that can be drawn at the point IS horizontal.

kinda hard to explain without drawing each but does that answer your question?
 

izzy_xo

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a point of inflexion is just a point where concavity changes on a curve. the 'line' that can be drawn through the point is not horizontal

a horizontal point of inflexion = point where concavity changes and the 'line' that can be drawn at the point IS horizontal.

kinda hard to explain without drawing each but does that answer your question?

yep thanks:)
 

gibbo153

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hmm potentially, but its not covered in the 2-unit course so might not be best option for OP if they only do 2-unit
 

Cazic

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Eh? You're looking for mins and maxs of the function y', so the second derivative test applied to any stationary points of the function y' (ie, when y''(x) = 0) would surely be the best option? Just because we're working with the third derivative of some function (ie, y) doesn't mean we should rule out the second derivative test .. surely?
 

Cazic

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how bout i just make up my own test while were on the topic, the 20th derviative test, you noob
The 20th derivative test: If the 20th derivative of a function y is the first non-zero derivative when evaluated at a stationary point of the first derivative of y, then that point is not a point of inflection of y.
Example: y = x^20 does not have a point of inflection at (0,0) since the 20th derivative y^(20)(x) = 20x19x...x2x1 is the first non-zero derivative at the point (0,0) (ie, the only stationary point of y').
(Note: not recommended for tests :p)
 

Uncle

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what the fuck is with the 3rd order and higher derivatives.
absolutely pointless and unnecessary, ignore those who suggest 3rd or higher order.

2nd order is sufficient, even engineering mathematics doesnt use 3rd or higher order derivatives in finding local max or min of functions with 2 independent variables i.e. f(x, y)
 

Cazic

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what the fuck is with the 3rd order and higher derivatives.
absolutely pointless and unnecessary, ignore those who suggest 3rd or higher order.

2nd order is sufficient, even engineering mathematics doesnt use 3rd or higher order derivatives in finding local max or min of functions with 2 independent variables i.e. f(x, y)
What do functions of two variables and their mins and maxs have to do with this thread? And can you explain to me how the second derivative of y = x4 tells me that (0,0) is a minimum of that function? Or isn't the second derivative as sufficient as you suggest?

At the end of the day you're going to have to use some other method to determine that (0,0) is a minimum of y = x4.

And now getting back to the actual topic of this thread, points of inflection, when looking for them you're just looking for mins and maxs of the first derivative. In doing so, applying the second derivative test to the any stationary points of the first derivative is exactly the same method as applying the second derivative test to the stationary points of any other function. The fact that you're working with the third derivative of some function in this case is no reason to ditch the method. After all, every function is the third derivative or some function ...
 
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Uncle

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What do functions of two variables and their mins and maxs have to do with this thread? And can you explain to me how the second derivative of y = x4 tells me that (0,0) is a minimum of that function? Or isn't the second derivative as sufficient as you suggest?

At the end of the day you're going to have to use some other method to determine that (0,0) is a minimum of y = x4.

And now getting back to the actual topic of this thread, points of inflection, when looking for them you're just looking for mins and maxs of the first derivative. In doing so, applying the second derivative test to the any stationary points of the first derivative is exactly the same method as applying the second derivative test to the stationary points of any other function. The fact that you're working with the third derivative of some function in this case is no reason to ditch the method. After all, every function is the third derivative or some function ...
The HSC should not ask you to find max and min to polynomials of a very high order and if they do, they must supply you the roots.
However for us there is something called numerical methods where we can numerically find roots of almost any function, more than current analytical methods, even 39th order and above polynomials to help us find the max and min of a function.

The HSC only gives you a speck of dust out of the galaxy of tools in mathematics in finding what you want.
My point is, regardless if the function contains one or two independent variables, the second derivative test is sufficient.
 

Cazic

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The HSC should not ask you to find max and min to polynomials of a very high order and if they do, they must supply you the roots.
However for us there is something called numerical methods where we can numerically find roots of almost any function, more than current analytical methods, even 39th order and above polynomials to help us find the max and min of a function.

The HSC only gives you a speck of dust out of the galaxy of tools in mathematics in finding what you want.
My point is, regardless if the function contains one or two independent variables, the second derivative test is sufficient.
I disagree with pretty much everything you wrote, but I think it's too far off topic again. What you said in your last message though:
what the fuck is with the 3rd order and higher derivatives.
absolutely pointless and unnecessary, ignore those who suggest 3rd or higher order.

2nd order is sufficient, even engineering mathematics doesnt use 3rd or higher order derivatives in finding local max or min of functions with 2 independent variables i.e. f(x, y)
is just wrong. For the third (heh) time, the third derivative of a function is needed to apply the second derivative test to the stationary points of the derivative of your function, which you would definitely consider using when looking for points of inflection. Why? Since you need to find the stationary points of the derivative anyway, you've got half the information you need before you even consider whether you'll use the second derivative test again - unless you have a really shitty function it's pretty much the obvious way to try and find your points of inflection.
 

tehrobzorz

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although it seems the 3rd derivative sounds relatively good...

can't imagine using the chain rule more than once....let alone quotient rule =X
 

Uncle

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I disagree with pretty much everything you wrote, but I think it's too far off topic again. What you said in your last message though:

is just wrong. For the third (heh) time, the third derivative of a function is needed to apply the second derivative test to the stationary points of the derivative of your function, which you would definitely consider using when looking for points of inflection. Why? Since you need to find the stationary points of the derivative anyway, you've got half the information you need before you even consider whether you'll use the second derivative test again - unless you have a really shitty function it's pretty much the obvious way to try and find your points of inflection.
never used the third derivative test, never bothered to, never will.
 

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