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differentiation

find the stationary points of :
Reply 1
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?
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
Reply 3
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.
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.
Reply 5
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)?
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
(edited 3 years ago)
Reply 7
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.
(edited 3 years ago)
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
(edited 3 years ago)
Reply 9
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.
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 :smile:
Reply 11
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
Reply 12
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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