# AS Maths - Differenciation - HELP!!

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Right we've started P1 and we're doing this section. Last lesson we did triple differenciation for point of inflextion, and he set homework. Now, I don't understand and we are allowed to get help so what we have to do is differenciate this equation to work out min/max/poi. I'm guessing it has summat to do with triple, but when I do it....it doesnt.

Here's the equation:

y= (1 - x^2) (1 - 4x)

So if you collect and everything don't you get:

y = 1 + 4x.x^2 - x^2 - 4

Which differs. to:

6x - 4 after collecting and such?

I dont understand! HELP! Can you walk me through it simply?! thanks.

NOTES:

^ = To the power of.

Here's the equation:

y= (1 - x^2) (1 - 4x)

So if you collect and everything don't you get:

y = 1 + 4x.x^2 - x^2 - 4

Which differs. to:

6x - 4 after collecting and such?

I dont understand! HELP! Can you walk me through it simply?! thanks.

NOTES:

^ = To the power of.

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#2

i think multiplying out gives you 1-4x-x^2+4x^3

the differential of this is 12x^2-2x-4

which factorises to (3x-2)(4x+2)

where dy/dx=0, x=2/3 or x=-1/2

if you find the gradients of the x points just above and below these values of x using 12x^2-2x-4 you will find that x=2/3 is a min and x=-1/2 is a max

the differential of this is 12x^2-2x-4

which factorises to (3x-2)(4x+2)

where dy/dx=0, x=2/3 or x=-1/2

if you find the gradients of the x points just above and below these values of x using 12x^2-2x-4 you will find that x=2/3 is a min and x=-1/2 is a max

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(Original post by

i think multiplying out gives you 1 - 4x - x^2 + 4x^3

**whitewitch**)i think multiplying out gives you 1 - 4x - x^2 + 4x^3

Oh. Thought yu couldnt multiply 'em like that.

NOTES:

* = multiply

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#4

(Original post by

i think multiplying out gives you 1-4x-x^2+4x^3

the differential of this is 12x^2-2x-4

which factorises to (3x-2)(4x+2)

where dy/dx=0, x=2/3 or x=-1/2

if you find the gradients of the x points just above and below these values of x using 12x^2-2x-4 you will find that x=2/3 is a min and x=-1/2 is a max

**whitewitch**)i think multiplying out gives you 1-4x-x^2+4x^3

the differential of this is 12x^2-2x-4

which factorises to (3x-2)(4x+2)

where dy/dx=0, x=2/3 or x=-1/2

if you find the gradients of the x points just above and below these values of x using 12x^2-2x-4 you will find that x=2/3 is a min and x=-1/2 is a max

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#5

(Original post by

So -x^2 * -4x gives 4x^3 ?

Oh. Thought yu couldnt multiply 'em like that.

**virtuality**)So -x^2 * -4x gives 4x^3 ?

Oh. Thought yu couldnt multiply 'em like that.

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#6

Yeah you can't add coefficients of x to different powers, but you can easily multiply them.

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I'm a bit confused when you say to factorise the answer. We normal do something like this:

12x^2 - 2x - 4 = 0

12x^2 - 2x = 4 (so divide by 4)

3x^2 - 0.5x= 0 (common x)

x (3x - 0.5) = 0

THEREFORE

x = 0 or x =1/6

?!?!?

12x^2 - 2x - 4 = 0

12x^2 - 2x = 4 (so divide by 4)

3x^2 - 0.5x= 0 (common x)

x (3x - 0.5) = 0

THEREFORE

x = 0 or x =1/6

?!?!?

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#8

(Original post by

I'm a bit confused when you say to factorise the answer. We normal do something like this:

12x^2 - 2x - 4 = 0

12x^2 - 2x = 4 (so divide by 4)

3x^2 - 0.5x= 0 (common x)

x (3x - 0.5) = 0

THEREFORE

x = 0 or x =1/6

?!?!?

**virtuality**)I'm a bit confused when you say to factorise the answer. We normal do something like this:

12x^2 - 2x - 4 = 0

12x^2 - 2x = 4 (so divide by 4)

3x^2 - 0.5x= 0 (common x)

x (3x - 0.5) = 0

THEREFORE

x = 0 or x =1/6

?!?!?

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Ah yes! Lol. I still dont know how to get from (3x-2)(4x+2)=0 to the x= bits.

(I'm only 16, I did my GCSE early.)

(I'm only 16, I did my GCSE early.)

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#10

**virtuality**)

I'm a bit confused when you say to factorise the answer. We normal do something like this:

12x^2 - 2x - 4 = 0

12x^2 - 2x = 4 (so divide by 4)

3x^2 - 0.5x= 0 (common x)

x (3x - 0.5) = 0

THEREFORE

x = 0 or x =1/6

?!?!?

which doesnt really help. it is best to use the

12x^2 - 2x - 4 = 0 and factorise this, find 2 brackets which multiply out to give the expression on the LHS.

these brackets are (3x-2)(4x+2), where each of these brackets = 0 there is a turning point.

if you find it hard to factorise quadratic expressions you can always use the quadratic formula, although this takes longer

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#11

(Original post by

Ah yes! Lol. I still dont know how to get from (3x-2)(4x+2)=0 to the x= bits.

(I'm only 16, I did my GCSE early.)

**virtuality**)Ah yes! Lol. I still dont know how to get from (3x-2)(4x+2)=0 to the x= bits.

(I'm only 16, I did my GCSE early.)

if 3x-2=0 then 3x=2 so x=2/3

if 4x+2=0 then 4x=-2 so x=-1/2

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I think I've got it now. It has nothing to do with triple differenciation! thats where I got confused, cos he taught us that. What a dipstick!

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#13

**virtuality**)

Ah yes! Lol. I still dont know how to get from (3x-2)(4x+2)=0 to the x= bits.

(I'm only 16, I did my GCSE early.)

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Yes, I did get it. Because I did it as I posted, I have that "I must be wrong" feeling.

Right d^2y/dx^2 = 24x - 2

When x=2/3 .....that gives 14 which is +ve therefore MIN

When x=-1/2....thats -14 which is -ve therefore MAX.

Done.

Right d^2y/dx^2 = 24x - 2

When x=2/3 .....that gives 14 which is +ve therefore MIN

When x=-1/2....thats -14 which is -ve therefore MAX.

Done.

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