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Stationary points

I need help here.
I have a question worded exactly like that:
"Deduce that the stationary points of the curve y = x sqrt(x) - 2 sqrt(x) occurs when x = 2/3"
It sounds simple enough, substitute x for 2/3 in the equation and you found the stationary points. Except the question is on 7 marks and I doubt and it would that simple that answer.
Am I getting something wrong?
Reply 1
Original post by Mlopez14
I need help here.
I have a question worded exactly like that:
"Deduce that the stationary points of the curve y = x sqrt(x) - 2 sqrt(x) occurs when x = 2/3"
It sounds simple enough, substitute x for 2/3 in the equation and you found the stationary points. Except the question is on 7 marks and I doubt and it would that simple that answer.
Am I getting something wrong?

That is not how you find stationary points of a graph.

Seeking stationary points usually involves differentiating a curve (assuming differentiatble) and setting the derivative to 0.
Can you carry on from here?
Reply 2
Original post by simon0
That is not how you find stationary points of a graph.

Seeking stationary points usually involves differentiating a curve (assuming differentiatble) and setting the derivative to 0.
Can you carry on from here?

That is what I tried doing at first, I differentiated it which got me (3sqrtx)/2 - 1/sqrtx. I got stuck at that point which is how I went back to the substitution.
I was trying to get a quadratic with dy/dx but I could not get that.
Reply 3
We could try removing x^(1/2) from the denominator of the second term by multiplying the derivative by x^(1/2).

Also noting only one solution exists for derivative equal to 0 may be required.
(edited 4 years ago)
Original post by Mlopez14
That is what I tried doing at first, I differentiated it which got me (3sqrtx)/2 - 1/sqrtx.

Set it equal to zero and solve for x.
Original post by Mlopez14
That is what I tried doing at first, I differentiated it which got me (3sqrtx)/2 - 1/sqrtx. I got stuck at that point which is how I went back to the substitution.
I was trying to get a quadratic with dy/dx but I could not get that.


since (3sqrtx)/2 - 1/sqrtx is equal to zero. factorize the denominator 2sqrtx
you should get 3x-2= 0
therefore x= 2/3
(edited 4 years ago)
Reply 6
Original post by simon0
We could try removing x^(1/2) from the denominator of the second term by multiplying the derivative by x^(1/2).

So I rationalised it which got me x^(1/2)/x. Do I then put them on common denominator (2x) which gets me 3x-2/ 2sqrtx. I suppose I could then rationalise it again and then set it equal to 0 which should be quite easy to solve?
Reply 7
Okay, we have:

f(x)=(32)x1x=0. f^{'} (x) = \big( \frac{3}{2} \big) \sqrt{x} - \frac{1}{\sqrt{x}} = 0.

Rationalising would give:

3x22x1/2=0. \displaystyle \frac{3x - 2}{2x^{1/2}} = 0.

What values of x satisfies this equation?
(edited 4 years ago)
Reply 8
Original post by simon0
Okay, we have:

f(x)=(32)x1x=0. f^{'} (x) = \big( \frac{3}{2} \big) \sqrt{x} - \frac{1}{\sqrt{x}} = 0.

Rationalising would give:

3x22x1/2=0. \displaystyle \frac{3x - 2}{2x^{1/2}} = 0.

What values of x satisfies this equation?

Yes, I can. It is just a matter of solving the equation.
Thank you very much

PS: How did you manage to display the equation in this form? I did not know you could do that in TSR .
Reply 9
Original post by Mlopez14
Yes, I can. It is just a matter of solving the equation.
Thank you very much

PS: How did you manage to display the equation in this form? I did not know you could do that in TSR .

For the equations, it is a using a typesetting language called "LaTeX" (pronounced "Lay-Tech"), which can be used on this forum.
I tend to think of it as a markup language (so has its own syntax) which is is useful so to display equations but can take some time to get used to.

Student Room details of LaTeX:
https://www.thestudentroom.co.uk/help/latex

Otherwise typing in equations but using brackets so to make your equations unambiguous is fine in my eyes.

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