AS Maths - Differenciation - HELP!!

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virtuality
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#1
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#1
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.
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whitewitch
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#2
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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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virtuality
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#3
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(Original post by whitewitch)
i think multiplying out gives you 1 - 4x - x^2 + 4x^3
So -x^2 * -4x gives 4x^3 ?
Oh. Thought yu couldnt multiply 'em like that.

NOTES:
* = multiply
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lgs98jonee
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#4
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(Original post by 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
if u differentiate a second time, with values of x, when dy/dx=0, the max point(s) is when d2y/dx2 is negative and min point is when d2y/dx2 is positive
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m:)ckel
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#5
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(Original post by virtuality)
So -x^2 * -4x gives 4x^3 ?
Oh. Thought yu couldnt multiply 'em like that.
you can
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john !!
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#6
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Yeah you can't add coefficients of x to different powers, but you can easily multiply them.
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virtuality
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#7
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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

?!?!?
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lgs98jonee
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#8
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(Original post by 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

?!?!?
4 divided by 4 is NOT zero..it is 1 (see line 4)
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virtuality
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#9
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Ah yes! Lol. I still dont know how to get from (3x-2)(4x+2)=0 to the x= bits. :confused:

(I'm only 16, I did my GCSE early.)
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whitewitch
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#10
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(Original post by 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

?!?!?
dividing by 4 on line 2 gives 3x^2 - 0.5x= 1 (4/4 =1)
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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whitewitch
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#11
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(Original post by virtuality)
Ah yes! Lol. I still dont know how to get from (3x-2)(4x+2)=0 to the x= bits. :confused:

(I'm only 16, I did my GCSE early.)
if two numbers multiplied by each other are 0 it follows that one of them must be equal to 0. if you treat each of the brackets as one number then the expression in each bracket must be equal to 0.
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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virtuality
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#12
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#12
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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lgs98jonee
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#13
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(Original post by virtuality)
Ah yes! Lol. I still dont know how to get from (3x-2)(4x+2)=0 to the x= bits. :confused:

(I'm only 16, I did my GCSE early.)
well it is gcse not alevel that one of 4x+2 or 3x-2 is 0 so x=-1/2 or 2/3
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virtuality
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#14
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#14
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.
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