[bahjat93]

Hello my fellow pirates,
I am stuck on a question
http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN10.PDF
The question is 6b),
It says state the coordinates of the stationary point and yet when you look at the mark scheme it only gives you the X coordinates!
Also if you then plug in the X coordinates into your original equation is gives you a bizzare number :/.

Can some1 explain to me why this is my fellow mutineers

[bahjat93]

I need your help people ASAP, i have the exam in less then 8 hours

[Blutooth]

(Original post by bahjat93)
I need your help people ASAP, i have the exam in less then 8 hours

What did you get when you 1) differentiated y (lnx/x) 2)set y to 0 and tried to solve for x.

[bahjat93]

(Original post by Blutooth)
What did you get when you 1) differentiated y (lnx/x) 2)set y to 0 and tried to solve for x.

I've checked my answer and it is correct,
x=e^-1

[Blutooth]

(Original post by bahjat93)
I've checked my answer and it is correct,
x=e^-1

You're wrong.
X=e^1 not e^-1
y=e^-1

[Rennit]

(Original post by bahjat93)
Hello my fellow pirates,
I am stuck on a question
http://store.aqa.org.uk/qual/gce/pdf...W-QP-JUN10.PDF
The question is 6b),
It says state the coordinates of the stationary point and yet when you look at the mark scheme it only gives you the X coordinates!
Also if you then plug in the X coordinates into your original equation is gives you a bizzare number :/.

Can some1 explain to me why this is my fellow mutineers

If you've got the exam in 8 hours then my advice to you is GET SOME DAMN SLEEP

[Blutooth]

(Original post by bahjat93)
I've checked my answer and it is correct,
x=e^-1

Trivially subbing e^-1 into ln(x)/x yields -1*e which is clearly negative, so by the shape of the graph it can't be the stationary point.

[Blutooth]

(Original post by bahjat93)
I've checked my answer and it is correct,
x=e^-1

y'=(1-ln(x))/x^2

Did you get this?

[bahjat93]

(Original post by Rennit)
If you've got the exam in 8 hours then my advice to you is GET SOME DAMN SLEEP

I can sleep when i'm dead . Thats all i will be doing then

(Original post by Blutooth)
Trivially subbing e^-1 into ln(x)/x yields -1*e which is clearly negative, so by the shape of the graph it can't be the stationary point.

(Original post by Blutooth)
y'=(1-ln(x))/x

Did you get this?

Yep, i see the error of my ways ty dude