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# GCSE further maths question

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1. Find the points of inflection of the curve and prove that they are indeed inflection points.
I tried dy/dx=0 but it didn't work.
2. (Original post by Ano9901whichone)
Find the points of inflection of the curve and prove that they are indeed inflection points.
I tried dy/dx=0 but it didn't work.
should give you the stationary points. should help you determine which ones are points of inflection.
3. (Original post by RDKGames)
should give you the stationary points. should help you determine which ones are points of inflection.
Tried it, it doesn't work.
4. (Original post by L33t)
Hi,

Have you differentiated set 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
5. (Original post by RDKGames)
should give you the stationary points. should help you determine which ones are points of inflection.
So you're saying that all inflection points are stationary points?
6. (Original post by L33t)
have you got that far?
Yes.
7. (Original post by B_9710)
So you're saying that all inflection points are stationary points?
No. The first derivative can distinguish them though.
8. (Original post by RDKGames)
No. The first derivative can distinguish them though.
How do you mean?
9. (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.
10. 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
11. (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.
12. (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
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.
13. (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
Root of a half goes to a quarter...?
14. (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?
15. (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?
16. (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.
17. (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.
18. (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
19. (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.
20. (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,

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