horizontal pt of inflexion (1 Viewer)

[wrong_turn]

my question is based on the below question. do you know how when one is trying to prove that the curve will always be concaving up, and there could be diffrent ways of getting to that answer? well here is one way, that i want someone to confirm, or disconfirm.

1. for the stationary point on the curve y= x^3 . shw that it is an inflexion.


a horizontal inflexion always concaves up.

y= x^3
y'= 3x^2
y"= 6x
y"' = 6

therefore it is always positive.

and since it is always positive, can't i then say that there is a horizontal pt of inflexion since it is a cube root and it is always positive?

this is just a trivial query. so can you? thanks

 

[|Axis_]

I don't really understand your argument, but it's easy to disprove your "always" statement with this counter-example:

y = -x^3
y' = -3x^2
y" = -6x
y"' = -6

yet y = -x^3 has a horizontal point of inflexion.

but i dunno if ive understood ur statement either!

 

[Trev]

You say it is a horizontal point of inflection where y'=0 and y''=0.
I do not know much about third derivative, however I am curious at why you have learnt it. It is not in the syllabus (or is it?) especially the mathematics course.....

 

[shafqat]

if y''>0, then it is concave up
if y''<0, then it is concave down
if y=o, more work is needed:
do a table of concavities to determine nature of that point

 

[Trev]

if y''>0, then it is concave up
if y''<0, then it is concave down
if y=o, more work is needed:
do a table of concavities to determine nature of that point

"if y=o, more work is needed:" this is incorrect, yes?
I think you mean y'=0 or y''=0.

 

[shafqat]

yep sorry
i meant y''

 

[Slidey]

my question is based on the below question. do you know how when one is trying to prove that the curve will always be concaving up, and there could be diffrent ways of getting to that answer? well here is one way, that i want someone to confirm, or disconfirm.

1. for the stationary point on the curve y= x^3 . shw that it is an inflexion.


a horizontal inflexion always concaves up.

y= x^3
y'= 3x^2
y"= 6x
y"' = 6

therefore it is always positive.

and since it is always positive, can't i then say that there is a horizontal pt of inflexion since it is a cube root and it is always positive?

this is just a trivial query. so can you? thanks

I think you've touched on something important here!

You can in fact use the third derivative to test for an inflexion point!

See here for details: http://www.boredofstudies.org/community/showthread.php?t=63548

 

[Trev]

I think you've touched on something important here!

You can in fact use the third derivative to test for an inflexion point!

See here for details: http://www.boredofstudies.org/community/showthread.php?t=63548

Yes, but not even extension 2 has taught us what is behind third derivative, why would they teach it to a mathematics student? Is it quicker, rather than testing either side?

 

[shafqat]

its not much quicker
and 2u students should not use it

 

[wrong_turn]

i am a 3u student asking for a piece of trivia about 2u. think before you type. even if it isn't in the syllabas, it doesn't mean you can't use it to your advantage in some shape or the other.

 

[MuffinMan]

a horizontal point of inflexion is a stationary point as well

testing for point of inflexion

f"(x) = 0 at that point
f"(x) changes signs at that point

or

f"(x) = 0
f'''(x) does not equal to zero

 

[Kutay]

Never knew there was a third derivative, thats amazing so what excatly does it find?

 

[shafqat]

f"(x) = 0
f'''(x) does not equal to zero

not necessarily HSC_sUcKsSsS
wat abt x^5 at x = 0?
f'''(0) = =, but it does have a horizontal pt of inflexion
more work is needed to show the horizontal pt of inflexion

 

[buchanan]

What the syllabus doesn't tell you is that the second and third derivative tests can be generalised:

Successive derivative test:

If f⁽ⁿ⁾(x₀)=0 for n=2,3,4,...,2k and if f^(2k+1)(x₀) exists and &ne;0 then (x₀, f(x₀)) is an inflection point.

If f⁽ⁿ⁾(x₀)=0 for n=1,2,3,4,...,2k-1 and f^(2k)(x₀)>0 then (x₀ f(x₀)) is a local minimum.

If f⁽ⁿ⁾(x₀)=0 for n=1,2,3,4,...,2k-1 and f^(2k)(x₀)<0 then (x₀ f(x₀)) is a local maximum.

So for f(x)=x⁵, f(0)=f'(0)=f''(0)=f'''(0)=f''''(0)=0, but f'''''(0)=120&ne;0, so (0,0) is a horizontal point of inflection.

 

[shafqat]

thats why i reckon 2u students should stick to table of concavities

 

[wrong_turn]

what did i fucken say you moron?! i am not a bloody 2u student!!

however, if you were refering to the actual use of finding the pt of horzontal inflexion the normal way, then pardon me.

 

[Nerd Queen]

hey I am a 2u student and this is how i would work out that question ( it is probably wrong) but i hope it helps

y=x^3

y' = 3x^2
y' = 0, for stationary pnt
3x^2 = 0
x^2 = 0
hence x=0

when x=0, y= 0^3
= 0
therefore (0,0)

y''= 6x
y(0)'' = 6(0) =0

since y''=0 there is a possible point of inflexion

check concavity:

x -1 0 +1
y'' -6 0 +6

since concavity changes at (0,0)

ie. f''(x)<0, 0, f''(x)>0

there is a point of inflexion at (0,0)

EDIT: Sorry i think i read the question wrong, i dont get the whole always positive thing. maybe you are supossed to find first derrivative only

ie

y=x^3
y' = 3x^2
y' = 0, for stationary pnt
3x^2 = 0
x^2 = 0
hence x=0

when x=0, y= 0^3
= 0
therefore (0,0)

TURNING POINT at (0,0)

check LHS and RHS

x -1 0 1
y' 3 0 3

since the curve is increasing (positive) on both sides it is a horizontal pnt of inflexion

maybe that is it?

i dunno, sorry if it got you confused (and made me look really dumb)

 

[shafqat]

what did i fucken say you moron?! i am not a bloody 2u student!!

however, if you were refering to the actual use of finding the pt of horzontal inflexion the normal way, then pardon me.

i know ur a 3u student
i was just referring to the normal method

 

[wrong_turn]

i know the normal method, but originally i was asking to have a short-cut confirmed or disconfirmed. whoever, said it was not part of the syllabas was kind of right. i asked my teacher, and she said nto to do it, because it won't be accepted if one rights up the reasoning wrong. so i'll stick to the old method.

 

[Trev]

what did i fucken say you moron?! i am not a bloody 2u student!!

Considering shafqat got 99.95, I assume he is far from being a moron!