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C3 functions

http://www.examsolutions.net/a-level-maths-papers/Edexcel/Core-Maths/Core-Maths-C3/2008-January/paper.php#Q8

how would i go around doing part d of this?

if i split 8x3−11−2x3\dfrac{8x^3 -1}{1-2x^3}

up into 8x31−2x3and 11−2x3\dfrac{8x^3}{1-2x^3} and\ \dfrac{1}{1-2x^3}

can i just make the first bit into 24x21−6x2\dfrac{24x^2}{1-6x^2} or am i killing people with my "maths" here?

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Avatar for JLegion
JLegion
2
Original post by thegreatwhale
http://www.examsolutions.net/a-level-maths-papers/Edexcel/Core-Maths/Core-Maths-C3/2008-January/paper.php#Q8

how would i go around doing part d of this?

if i split 8x3−11−2x3\dfrac{8x^3 -1}{1-2x^3}

up into 8x31−2x3and 11−2x3\dfrac{8x^3}{1-2x^3} and\ \dfrac{1}{1-2x^3}

can i just make the first bit into 24x21−6x2\dfrac{24x^2}{1-6x^2} or am i killing people with my "maths" here?


That would be illegal I'm afraid. You need to multiply both top and bottom by 3, giving you the result (24x^3)/(3-6x^2).

I'd recommend differentiating gf(x) (using quotient rule; maybe the product rule) to get yourself another fraction. Put that equal to zero and solve.
Original post by JLegion
That would be illegal I'm afraid. You need to multiply both top and bottom by 3, giving you the result (24x^3)/(3-6x^2).

I'd recommend differentiating gf(x) (using quotient rule; maybe the product rule) to get yourself another fraction. Put that equal to zero and solve.


couldn't i just use quotient rule as it is?(just learnt it xD)

and say that

(1−2x3)(24x2)−(8x3−1)(−6x2)(1−2x3)2\dfrac{(1-2x^3)(24x^2)-(8x^3 -1)(-6x^2)}{(1-2x^3)^2}

24x2−48x5+48x5−6x2(1−2x3)2\dfrac{24x^2 -48x^5 +48x^5 -6x^2}{(1-2x^3)^2}

18x2(1−2x3)2\dfrac{18x^2}{(1-2x^3)^2}
Original post by thegreatwhale
couldn't i just use quotient rule as it is?(just learnt it xD)

and say that

(1−2x3)(24x2)−(8x3−1)(−6x2)(1−2x3)2\dfrac{(1-2x^3)(24x^2)-(8x^3 -1)(-6x^2)}{(1-2x^3)^2}

24x2−48x5+48x5−6x2(1−2x3)2\dfrac{24x^2 -48x^5 +48x^5 -6x^2}{(1-2x^3)^2}

18x2(1−2x3)2\dfrac{18x^2}{(1-2x^3)^2}


Yes that works, and hence you can deduce that the stationary point is at x=0. Now just substitute into the expression for gf(x) to get the y-coordinate.
Original post by HapaxOromenon3
Yes that works, and hence you can deduce that the stationary point is at x=0. Now just substitute into the expression for gf(x) to get the y-coordinate.


How do i know it's a quadratic? it doesn't look like a normal one :/
Original post by thegreatwhale
How do i know it's a quadratic? it doesn't look like a normal one :/


At stationary points, dy/dx = 0.
Thus 18x^2/(1-2x^3)^2 = 0.
Multiply both sides by (1-2x^3)^2 to give 18x^2 = 0, then divide by 18 to give x^2 = 0, so x = 0.
Notice that in general, if a fraction equals 0, we can automatically write down that the numerator must be 0, and it doesn't matter what the denominator is (as long as we don't have a division by zero).
Original post by HapaxOromenon3
At stationary points, dy/dx = 0.
Thus 18x^2/(1-2x^3)^2 = 0.
Multiply both sides by (1-2x^3)^2 to give 18x^2 = 0, then divide by 18 to give x^2 = 0, so x = 0.
Notice that in general, if a fraction equals 0, we can automatically write down that the numerator must be 0, and it doesn't matter what the denominator is (as long as we don't have a division by zero).


i understand dydx=0\dfrac{\mathrm d y}{\mathrm d x}=0 at a stationary point

but how u know it's a quadratic? because quadratic generally have u shape or n shape and that way it has only 1 stationary point or am i wrong and does the graph of it look totally different?
Reply 7
Avatar for fefssdf
fefssdf
20
Original post by JLegion
That would be illegal I'm afraid. You need to multiply both top and bottom by 3, giving you the result (24x^3)/(3-6x^2).

I'd recommend differentiating gf(x) (using quotient rule; maybe the product rule) to get yourself another fraction. Put that equal to zero and solve.


Thanks for pointing out the OP's illegal move , do you have the number for the maths police ?
Reply 8
Avatar for JLegion
JLegion
2
Original post by thegreatwhale
couldn't i just use quotient rule as it is?(just learnt it xD)

and say that

(1−2x3)(24x2)−(8x3−1)(−6x2)(1−2x3)2\dfrac{(1-2x^3)(24x^2)-(8x^3 -1)(-6x^2)}{(1-2x^3)^2}

24x2−48x5+48x5−6x2(1−2x3)2\dfrac{24x^2 -48x^5 +48x^5 -6x^2}{(1-2x^3)^2}

18x2(1−2x3)2\dfrac{18x^2}{(1-2x^3)^2}


Yes you can. I was just letting you know that when you split the fraction and multiplied by 3, it was done incorrectly.

Now you know the derivative. Equate it to 0, and find the value of gf(x) for this value of x to find the coordinates of the stationary point :smile:
Original post by JLegion
Yes you can. I was just letting you know that when you split the fraction and multiplied by 3, it was done incorrectly.

Now you know the derivative. Equate it to 0, and find the value of gf(x) for this value of x to find the coordinates of the stationary point :smile:

ah right oh i understand the question told me lol >.>
thanks got the answer now
Original post by fefssdf
Thanks for pointing out the OP's illegal move , do you have the number for the maths police ?


this is what happens when you go through the textbook up to chapter 3 and you aint done the chapter on differentiation but you still wanna do dem big mark questions an finish the rest of the question.
Original post by thegreatwhale
ah right oh i understand the question told me lol >.>
thanks got the answer now


this is what happens when you go through the textbook up to chapter 3 and you aint done the chapter on differentiation but you still wanna do dem big mark questions an finish the rest of the question.


Aha oh right Yh I know what you mean ; have you started c3 early then ? I started c3 last summer but didn't really remember much of it lol . I wish I had actually learnt it properly so I had more time for other stuff hmmm
Original post by fefssdf
Aha oh right Yh I know what you mean ; have you started c3 early then ? I started c3 last summer but didn't really remember much of it lol . I wish I had actually learnt it properly so I had more time for other stuff hmmm


yes, i'm still tryin to learn it now myself so i can get a head start and get things going back when i go back to school
Original post by thegreatwhale
yes, i'm still tryin to learn it now myself so i can get a head start and get things going back when i go back to school


Brilliant haha ; do all the exercises and keep them in a folder and mark them and then when your teacher sets homework from the book then you would've done it already aha :smile:
Original post by fefssdf
Brilliant haha ; do all the exercises and keep them in a folder and mark them and then when your teacher sets homework from the book then you would've done it already aha :smile:


ah thing is though my teacher doesn't do that he rushes through and tries to get us to do all the past papers so we have as much experience of papers as possible
Original post by thegreatwhale
ah thing is though my teacher doesn't do that he rushes through and tries to get us to do all the past papers so we have as much experience of papers as possible


Oh ok well just do all the exercises in the book then so when you come to do papers you'll be confident with the content
When I did the problem initially :moon: I noticed that 8x3−11−2x3=3(1−2x3)−1−4\dfrac{8x^3-1}{1-2x^3}=3(1-2x^3)^{-1}-4 and I found the derivative using the chain rule leading to 18x2(1−2x3)2\dfrac{18x^2}{(1-2x^3)^2} as you have.
(edited 7 years ago)
Original post by fefssdf
Oh ok well just do all the exercises in the book then so when you come to do papers you'll be confident with the content


Original post by Kvothe the arcane
When I did the problem initially :moon: I noticed that 8x3−11−2x3=3(1−2x3)−1−4\dfrac{8x^3-1}{1-2x^3}=3(1-2x^3)^{-1}-4 and I found the derivative using the chain rule leading to 18x2(1−2x3)2\dfrac{18x^2}{(1-2x^3)^2} as you have.


woah where did that come from?? o..o
i understand 11−2x3=(1−2x3)−1\dfrac{1}{1-2x^3} =(1-2x^3)^{-1}

but i'm a little unsure of where the 3 comes from ad i'm assuming the -4 is to balance things out
Original post by thegreatwhale
woah where did that come from?? o..o
i understand 11−2x3=(1−2x3)−1\dfrac{1}{1-2x^3} =(1-2x^3)^{-1}

but i'm a little unsure of where the 3 comes from ad i'm assuming the -4 is to balance things out


-4 is the remainder I think
Original post by fefssdf
-4 is the remainder I think


lol still don't know where the 3 comes from and where the 8x3−18x^3 -1 went
Original post by thegreatwhale
lol still don't know where the 3 comes from and where the 8x3−18x^3 -1 went


I'm actually so confused right now

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