limitlesspuffy
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find the stationary points of :
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Pangol
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(Original post by limitlesspuffy)
find the stationary points of :
What happens when you try the usual approach? Is there a particular point that you are stuck at?
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limitlesspuffy
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(Original post by Pangol)
What happens when you try the usual approach? Is there a particular point that you are stuck at?
so dy/dx i've worked out

It is 15x^1/4 -5/9x

Then you set that to 0 and solve.

I've got 135x^1/4 -5x=0 so far

and idk how to solve after that
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Pangol
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(Original post by limitlesspuffy)
so dy/dx i've worked out

It is 15x^1/4 -5/9x

Then you set that to 0 and solve.

I've got 135x^1/4 -5x=0 so far

and idk how to solve after that
Factorise out the x1/4 and see where that takes you.
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limitlesspuffy
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(Original post by Pangol)
Factorise out the x1/4 and see where that takes you.
x1/4(135-5x4)=0
So x=0 and 135-5x4=0
so x4=27
x=2.2795 (4 d.p)
I feel like this is wrong.
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Pangol
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(Original post by limitlesspuffy)
x1/4(135-5x4)=0
So x=0 and 135-5x4=0
so x4=27
x=2.2795 (4 d.p)
I feel like this is wrong.
I wouldn't bother multiplying by 9, that can come later. It is the power that is attached to the x in the bracket that needs looking at. If you multiply out your factorisation, do you get back where you started? What is (x1/4)(x4)?
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limitlesspuffy
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(Original post by Pangol)
I wouldn't bother multiplying by 9, that can come later. It is the power that is attached to the x in the bracket that needs looking at. If you multiply out your factorisation, do you get back where you started? What is (x1/4)(x4)?
(x1/4)(x4)=x1 so it is what i started with
so if i don't multiply by 9, I get 15 - 5/9x4 =0
Then I still get x4=27

oh wait in (x1/4)(x4), do you add 1/4 and 4
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Pangol
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(Original post by limitlesspuffy)
(x1/4)(x4)=x1 so it is what i started with
so if i don't multiply by 9, I get 15 - 5/9x4 =0
Then I still get x4=27

oh wait in (x1/4)(x4), do you add 1/4 and 4
You do indeed add them.
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limitlesspuffy
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(Original post by Pangol)
You do indeed add them.
so I got x=81
Is x=0 also a solution?
Which means can x1/4=0 since 01/4=0
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Pangol
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(Original post by limitlesspuffy)
so I got x=81
Is x=0 also a solution?
x = 81 is correct. Note that the question asks for the stationary points, so you will have to provide a coordinate, not just an x-value.

x = 0 is a solution of dy/dx = 0, but look again at the equation as given to you in the question to see if this counts.
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limitlesspuffy
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(Original post by Pangol)
x = 81 is correct. Note that the question asks for the stationary points, so you will have to provide a coordinate, not just an x-value.

x = 0 is a solution of dy/dx = 0, but look again at the equation as given to you in the question to see if this counts.
OHHhh, thank you so much for your help
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NMB27
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if y=12x^5/4-5/18x^2-1000 then dy/dx = 15x^1/4-5/9x
we then know that 15x^1/4 = 5/9x
then divide both sides by x so 15x^-3/4 = 5/9
multiply both sides by 15 so x^-3/4 = 1/27
therefore 1/x^3/4 = 1/27 so x^3/4 must be equal to 27
x^3/4 = the fourth root of x^3 so that = 27
cube root both sides: fourth root of x = 3
both sides to the fourth power and x = 81 which is the only value for x
then substitute that into the original equation for y to get the point
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Pangol
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(Original post by NMB27)
if y=12x^5/4-5/18x^2-1000 then dy/dx = 15x^1/4-5/9x
we then know that 15x^1/4 = 5/9x
then divide both sides by x so 15x^-3/4 = 5/9
multiply both sides by 15 so x^-3/4 = 1/27
therefore 1/x^3/4 = 1/27 so x^3/4 must be equal to 27
x^3/4 = the fourth root of x^3 so that = 27
cube root both sides: fourth root of x = 3
both sides to the fourth power and x = 81 which is the only value for x
then substitute that into the original equation for y to get the point
The forum guidelines say not to post fully worked solutions. And this has already been finished anyway!
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