VCE Methods Question Thread!

[dc302]

In that case,

2sin(2x-30)=1

sin(2x-30) = 1/2

So for what angles does the sin of said angle = 1/2 ? That would be 30 and 150, as you should remember/be able to look up.

So 2x-30 = arcsin(1/2)

2x-30 = 30, 150

2x = 60, 180

x=30, 90

[Incommensura]

Find the values of x between o and 360^degrees for which 2sin(2x-30)^degrees=1

Okay, well -
dividing by two, sin(2x-30) = 1/2
sin(a) = 1/2 for a = 30, 150, 390, 510, 750, 870
let a = 2x-30
2x-30 = 30, 150, 390, 510 ...
2x = 60, 180, 420, 540, 780 ...
dividing by two
x = 30, 90, 210, 270, 390
but the domain is 0 to 360, so x = 30, 90, 210, 270

[Deceitful Wings]

how can y=6x³+6x²+3x be written in the form y=a(x-b)³+c ?

[dc302]

Find the values of x between o and 360^degrees for which 2sin(2x-30)^degrees=1

Okay, well -
dividing by two, sin(2x-30) = 1/2
sin(a) = 1/2 for a = 30, 150, 390, 510, 750, 870
let a = 2x-30
2x-30 = 30, 150, 390, 510 ...
2x = 60, 180, 420, 540, 780 ...
dividing by two
x = 30, 90, 210, 270, 390
but the domain is 0 to 360, so x = 30, 90, 210, 270

Yeah this is better than mine, forgot he wanted it between 0 and 360.

[brightsky]

how can y=6x³+6x²+3x be written in the form y=a(x-b)³+c ?

not all cubics have SPOIs.

[Deceitful Wings]

thanks but how do you prove it? can i see the working out? i can't seem to understand

[dc302]

thanks but how do you prove it? can i see the working out? i can't seem to understand

To show that a cubic doesn't have an SPOI, first find where the POI is, and then show that it isn't stationary.


For example:

f(x)=6x^3+6x^2+3x
f'(x)=18x^2+12x+3
f''(x)=36x+12

To find the POI, we set f''(x) = 0, so 36x+12=0, x= -1/3

Now show that this is not 'stationary'.

ie. find f'(-1/3)

f'(-1/3) = 18(-1/3)^2 + 12(-1/3) + 3
= 2 - 4 + 3
= 1
=/= 0

Alternatively, you could just show the function has no stationary points at all, but of course this won't work with all cubics.

[Incommensura]

To show that a cubic doesn't have an SPOI, first find where the POI is, and then show that it isn't stationary.


For example:

f(x)=6x^3+6x^2+3x
f'(x)=18x^2+12x+3
f''(x)=36x+12

To find the POI, we set f''(x) = 0, so 36x+12=0, x= -1/3

Now show that this is not 'stationary'.

ie. find f'(-1/3)

f'(-1/3) = 18(-1/3)^2 + 12(-1/3) + 3
= 2 - 4 + 3
= 1
=/= 0

Alternatively, you could just show the function has no stationary points at all, but of course this won't work with all cubics.

Unless my memory hugely fails me this isn't within the scope of Methods. Alternative - just take f'(x), solve f'(x) = 0 and if there is more than one solution then the function has multiple stationary points, in which case it doesn't have a stationary point of inflection.

I'm not sure if this works the other way though - if it has only one stationary point, that may not necessarily mean it's a SPOI... Somebody who knows more than me can answer that one

[b^3]

f'(x)=18x²+12x+3
For there to be a stationary point, f'(x)=0
b²-4ac=12²=4*18*3
=144-216
=-72

As b²-4ac<0 there are no solutions to 18x²+12x+3=0, so there are no stationary points/stationary points of inflexion.

EDIT:

I'm not sure if this works the other way though - if it has only one stationary point, that may not necessarily mean it's a SPOI... Somebody who knows more than me can answer that one

As for the stationary point of inflection for one stationary point, just show that f'(x) on either side are different signs. I.e. plug a value in and gradient table.

[Planck's constant]

how can y=6x³+6x²+3x be written in the form y=a(x-b)³+c ?

By the methods available to you in MM 3/4, the best you can do is to state that it cant be done because the cubic expansion (x+b)³ requires that the coefficients of x³, x² and x must be in the ratio 1:3:3 (after you take 6 out as a common factor)
In other words, you cant 'complete the cube' without some nasty non-constant bits left over

[kamil9876]

It's not necessarily a 1:3:3 ratio unless b=-1 (and don't worry about taking out the common factor of 6, ratios are preserved if u multiply everything by a constant anyway).


So the ratio is and you've got yourself So you would want , which is impossible.

[Planck's constant]

It's not necessarily a 1:3:3 ratio unless b=-1 (and don't worry about taking out the common factor of 6, ratios are preserved if u multiply everything by a constant anyway).


So the ratio is and you've got yourself So you would want , which is impossible.

I was waiting for you to tighten up my answer, mate

[dc302]

Oh yeah, this is methods. My bad LOL

[Deceitful Wings]

thanks for the help everyone, i am having trouble trying to change y=6x^3+6x^2+2x  into turning point form. I don't know how to do it? it seems like you can't use completing the square to do it, so how is it possible?

[dc302]

One way is to equate coefficients:

set a(x-b)^3+c to be equivalent to 6x^3+6x^2+2x

Expand the first to get ax^3 - 3ax^2b + 3axb^2 - ab^3 + c

From this you will get a=6, b=-1/3, c=-2/9.

Alternatively, you can find the stationary point by differentiating, which turns out to be (-1/3, -2/9). So we know that y=a(x+1/3)^3 - 2/9, and then it becomes obvious that a=6.


edit: also if I used spesh again, someone please tell me :/