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1. Please can you list the steps to work out the following question please:

"given that , find the value of x for which A has a stationary
point. Use the second derivative to determine whether A is a maximum or minimum point."
2. First of all rewrite it in a form you can differentiate, and then differntiate and set it eqaul to 0 and solve for x, then sub the x value to gain a value for a use second dirriv to see if its a max or min
3. Differentiate the equation. Set dA/dx as zero and solve to find x. Differentiate again to get d2A/dx2 and enter in your x value. If it comes out negative it's a maximum, if it comes out positive it's a minimum.
4. (Original post by berwick53)
Please can you list the steps to work out the following question please:

"given that , find the value of x for which A has a stationary
point. Use the second derivative to determine whether A is a maximum or minimum point."
2000/x becomes 2000x^-1

and differentiate as normal.

equate that to 0 and solve for x.

that will give you the stationary point(s)

find the second derivative and for that vaule of x, if the second derivative is more than zero, it is a minimum point, and vice versa.
5. It's tradition to not spoon feed answers on TSR. So I'll give you some point to consider, what do you need to do to get a second dervative? What do you need to do to the 2000/x to make it easier to work with. Since it is a stationary point what would the gradient (and hence dy/dx) be?
6. so dy/dx = 2x + -2000x^-2
7. (Original post by sicarius1992)
It's tradition to not spoon feed answers on TSR. So I'll give you some point to consider, what do you need to do to get a second dervative? What do you need to do to the 2000/x to make it easier to work with. Since it is a stationary point what would the gradient (and hence dy/dx) be?
It was a nice thought but unfortunately many others have decided to prove their maths skills on this one
8. (Original post by jonasdb)
It was a nice thought but unfortunately many others have decided to prove their maths skills on this one
Oh well, they must be used to being spoon fed themselves Modern education is like that, no one learns anymore, they just memorise
9. (Original post by sicarius1992)
It's tradition to not spoon feed answers on TSR. So I'll give you some point to consider, what do you need to do to get a second dervative? What do you need to do to the 2000/x to make it easier to work with. Since it is a stationary point what would the gradient (and hence dy/dx) be?
dy/dx, strictly that should be da(x)/dx.

Don't confuse the fact that functions can be expressed as any letter not just y.
10. (Original post by titanlux)
dy/dx, strictly that should be da(x)/dx.

Don't confuse the fact that functions can be expressed as any letter not just y.
Yeah you're right, I just didn't read the OP properly haha, just skimmed over the main part of the equation hehe

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Updated: December 6, 2010
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