The Student Room Group
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>
Competition - post a photo you've taken of nature >>

GCSE further maths question

Find the points of inflection of the curve y=x4−3x2+x y=x^4-3x^2+x and prove that they are indeed inflection points.
I tried dy/dx=0 but it didn't work.

Scroll to see replies

Original post by Ano9901whichone
Find the points of inflection of the curve y=x4−3x2+x y=x^4-3x^2+x and prove that they are indeed inflection points.
I tried dy/dx=0 but it didn't work.


dydx=0\frac{dy}{dx}=0 should give you the stationary points. d2ydx2=0\frac{d^2y}{dx^2}=0 should help you determine which ones are points of inflection.
(edited 7 years ago)
Original post by RDKGames
dydx=0\frac{dy}{dx}=0 should give you the stationary points. d2ydx2=0\frac{d^2y}{dx^2}=0 should help you determine which ones are points of inflection.


Tried it, it doesn't work.
Original post by L33t
Hi,

Have you differentiated set dydx=0 \frac{dy}{dx}=0 and factorised? Try that. Also in case you don't know to determine the nature of turning point you need to find out if the second differential is less than zero or greater than zero.

Hope that helps :smile:


Read last comment.
Reply 4
Avatar for B_9710
B_9710
18
Original post by RDKGames
dydx=0\frac{dy}{dx}=0 should give you the stationary points. d2ydx2=0\frac{d^2y}{dx^2}=0 should help you determine which ones are points of inflection.


So you're saying that all inflection points are stationary points?
Original post by L33t
dydx=4x3−6x+1=0 \frac{dy}{dx}=4x^3-6x+1=0 have you got that far?


Yes.
Original post by B_9710
So you're saying that all inflection points are stationary points?


No. The first derivative can distinguish them though.
Reply 7
Avatar for B_9710
B_9710
18
Original post by RDKGames
No. The first derivative can distinguish them though.


How do you mean?
Original post by L33t
Have you definitely copied down the question correctly? Perhaps have another read because in order to solve that equation you would need to use the cubic formula which is far beyond the scope of GCSE further

Yes, definitely.
not sure if this is correct, but at points of inflection, the second derivative is = 0. so d2y/dx2=0.
since dy/dx=4x^3-6x+1, d2y/dx2 = 12x^2-6.
so
12x^2 - 6 = 0.
This means that x^2= 1/2 therefore x = root plus or minus 1/2
Someone please correct me if im wrong
:wink:
(edited 7 years ago)
Original post by B_9710
How do you mean?


Stationary points can be good candidates for points of inflection? Just plug their values through the second derivative and see if you get a 0.
Reply 11
Avatar for B_9710
B_9710
18
Original post by Uni12345678
not sure if this is correct, but at points of inflection, the second derivative is = 0. so d2y/dx2=0.
since dy/dx=4x^3-6x+1, d2y/dx2 = 12x^2-6.
so
12x^2 - 6 = 0.
This means that x^2= 1/2 therefore x = 1/4 or -1/4.
Someone please correct me if im wrong
:wink:


Yes you are right but to prove that they are actually proper inflection points you have to check that the second derivative either side of the point in question has opposite signs.
Original post by Uni12345678
not sure if this is correct, but at points of inflection, the second derivative is = 0. so d2y/dx2=0.
since dy/dx=4x^3-6x+1, d2y/dx2 = 12x^2-6.
so
12x^2 - 6 = 0.
This means that x^2= 1/2 therefore x = 1/4 or -1/4.
Someone please correct me if im wrong
:wink:


Root of a half goes to a quarter...?
Reply 13
Avatar for B_9710
B_9710
18
Original post by RDKGames
Stationary points can be good candidates for points of inflection? Just plug their values through the second derivative and see if you get a 0.


But what does a point being a stationary point have to do with anything?
Original post by B_9710
Yes you are right but to prove that they are actually proper inflection points you have to check that the second derivative either side of the point in question has opposite signs.


thanks, im not sure i understand what you mean though, i thought that if the 2nd derivative is = 0 it's already proved to be a point of inflection?
Original post by B_9710
But what does a point being a stationary point have to do with anything?


Knowing where the stationary points lie and their nature allow you to graph somewhat complicated functions.
Reply 16
Avatar for B_9710
B_9710
18
Original post by Uni12345678
thanks, im not sure i understand what you mean though, i thought that if the 2nd derivative is = 0 it's already proved to be a point of inflection?


An inflection point is where a curve changes from concave to convex or the other way round. There are cases where the second derivative is 0 but this condition is not necessarily met.
Original post by RDKGames
Root of a half goes to a quarter...?


oh im so stupid. ive done a similar mistake on another maths question on tsr where i thought id found the right way and then i made a primary school error, sorry, should be root plus or minus 1/2 :wink:
Reply 18
Avatar for B_9710
B_9710
18
Original post by TheRedGoldfish
Knowing where the stationary points lie and their nature allow you to graph somewhat complicated functions.


Have you seen the question in the OP, we don't need to graph it, only find the points o inflection.
Original post by B_9710
Have you seen the question in the OP, we don't need to graph it, only find the points o inflection.


Well there kinda isn't any other way to prove points of inflection without finding the stationary points first because by definition a point of inflection is a stationary point with opposite gradients at either side,

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you