stationary points

[Mel Mitch]

Hello ,
Please help.

Find the coordinates of the stationary points of the curve:

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


How do i prove this ????
Answer
the coordinates are (0,0) and (4,8)

 

[BigGlenntheHeavy]

x must be greater than 2. Why?

Hence domain is (2,infinity) and range is [(32?6)/9, infinity).

 

[Mel Mitch]

it's a calculus III question under (function of a function) but i don't how to solve it.....i'm sure some dy/dx must be used

 

[Deleted member 4993]

Hello ,
Please help.

Find the coordinates of the stationary points of the curve:

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

Is your function

\(\displaystyle y \, = \, x^2\cdot (x-2)^{-\frac{1}{2}}\)

If it is, then Glen is pointing out that the answer in your book is probably wrong.

To calculate stationary points - first you got to know the definition of stationary point.

What is it?
How do i prove this ????
Answer
the coordinates are (0,0) and (4,8)

 

[BigGlenntheHeavy]

Mel, whose homework are you doing? Are you trying to impress a girl by showing her how smart you are by correcting her homework vis-a-vis this message board?

 

[Mel Mitch]

sorry i published they wrong it wrong thing.....sorry

Find the coordinates of the stationary points of the curve:

y= x^2.(x-2)^-1




Listen as i published some time earlier i'm studying for an exam and my last one i got a C.....so i don't see my self as impressing anyone other than upping my grades come this september .

I'm struggling and i needed some help( as the site claims above).....and my lec told us to studied the examination questions from the book "Understanding Pure Mathematics" Sadler, and Thorning.....

because the general exercise question i could read and follow the direction and solve the problem but when it comes to the examination questions i'm on a sailing boat.

The persons i use to study with are no longer doing this course ..... so i'm on my own....and my lec is pretty tough.

And Big G u'r talking to a Melissa....hence someone without a dangling genitalia :wink:

 

[Deleted member 4993]

sorry i published they wrong it wrong thing.....sorry

Find the coordinates of the stationary points of the curve:

y= x^2.(x-2)^-1

Definition of stationary point ? where dy/dx = 0

dy/dx = [2x(x-2) - x[sup:g6ev93cg]2[/sup:g6ev93cg]]/(x-2)[sup:g6ev93cg]2[/sup:g6ev93cg]

Now find the points where the function above goes to "zero".

[2x(x-2) - x[sup:g6ev93cg]2[/sup:g6ev93cg]]/(x-2)[sup:g6ev93cg]2[/sup:g6ev93cg] = 0 ? [2x(x-2) - x[sup:g6ev93cg]2[/sup:g6ev93cg]] = 0

x(x-4) = 0

So stationary points are at x = 0 and x = 4

y(0) = 0

and

y(4) = 4[sup:g6ev93cg]2[/sup:g6ev93cg]/(4-2) = 8

so the stationary points are (0,0) and (4,8)

 

[Mel Mitch]

thanks khan