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# Fp1

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1. http://mathsathawthorn.pbworks.com/f/FP1Jan07.pdf
In q6a we have to find the coordinates of the stationary point on the graph of f so so far ive worked out dy/dx which is (-2ylnx)/x and i know at a stationary point dy/dx=0 , but what do i do next?
2. (Original post by Ayaz789)
http://mathsathawthorn.pbworks.com/f/FP1Jan07.pdf
In q6a we have to find the coordinates of the stationary point on the graph of f so so far ive worked out dy/dx which is (-2ylnx)/x and i know at a stationary point dy/dx=0 , but what do i do next?
Substitute in the value of y, and look for where the numerator is zero, i.e. one of the multiplicands is zero. Note that the domain is restricted to x > 0.
3. (Original post by ghostwalker)
Substitute in the value of y, and look for where the numerator is zero, i.e. one of the multiplicands is zero. Note that the domain is restricted to x > 0.
So if -2ylnx=0 How do i know the value of y and x?
4. (Original post by Ayaz789)
So if -2ylnx=0 How do i know the value of y and x?
You're told y as a function of x at the start, so sub that in, and you now have a product of functions of x. For that to equal zero, one of the multiplicands must be zero. Check each one to find what value of x - if any - will make it zero.
5. (Original post by ghostwalker)
You're told y as a function of x at the start, so sub that in, and you now have a product of functions of x. For that to equal zero, one of the multiplicands must be zero. Check each one to find what value of x - if any - will make it zero.
Step at a time please Okay so i have put they y value into x to get 0=-2^-lnx.lnx as the x cancels out?
6. (Original post by Ayaz789)
Step at a time please Okay so i have put they y value into x to get 0=-2^-lnx.lnx as the x cancels out?
EEK!!!! The x does not cancel.

dy/dx = -2x^{-ln x}{ln x}/x

= -2x^{-(ln x)-1}{ln x} = 0

So, either

ln x = 0, or

-2x^{-(ln x)-1} = 0

and recall that x>0, so can that second part ever be zero?
7. (Original post by ghostwalker)
EEK!!!! The x does not cancel.

dy/dx = -2x^{-ln x}{ln x}/x

= -2x^{-(ln x)-1}{ln x} = 0

So, either

ln x = 0, or

-2x^{-(ln x)-1} = 0

and recall that x>0, so can that second part ever be zero?
-2x^{-(ln x)-1}{ln x} = 0 where did the -1 come from? Im really sorry but idk :/
8. (Original post by Ayaz789)
-2x^{-(ln x)-1}{ln x} = 0 where did the -1 come from? Im really sorry but idk :/
Rules of indices.

In essence,

x^a/ x = x^a / x^1 = x^(a-1)
9. (Original post by ghostwalker)
Rules of indices.

In essence,

x^a/ x = x^a / x^1 = x^(a-1)
Ohh yhhhh
10. (Original post by ghostwalker)
EEK!!!! The x does not cancel.

dy/dx = -2x^{-ln x}{ln x}/x

= -2x^{-(ln x)-1}{ln x} = 0

So, either

ln x = 0, or

-2x^{-(ln x)-1} = 0

and recall that x>0, so can that second part ever be zero?
Thank you!
11. (Original post by ghostwalker)
Rules of indices.

In essence,

x^a/ x = x^a / x^1 = x^(a-1)
I need help!
12. (Original post by Ayaz789)
I need help!
Focus - be specific.

Or is that just a general exclamation, for which you're not expecting a reply?

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