Show that f has a stationary point at (0, 0) for every k ∈ R

In summary, the conversation was about finding the stationary point of the function f(x,y) = x^2 + kxy + y^2, and it was determined that this point is (0,0) for every value of k in the set of real numbers. The solution involved taking derivatives of the function and setting them equal to 0, without actually solving for any values.
  • #1
cathal84
15
0

Homework Statement


Let f(x, y) = x^2 + kxy + y^2 , where k is some constant in R. i. Show that f has a stationary point at (0, 0) for every k ∈ R

Homework Equations


...

The Attempt at a Solution


I may have the solution or i may have gone completely wrong I am not entirely sure.
i first found the derivative of f(x,y) with respect to x it was 2x+ky
then found the derivative of f(x,y) with respect to y it was 2y+kx
i then let both of them equal 0

then i solved 2x+ky=0 looking for a value of x and i got x=-ky/2
i then put this value for x back into 2y+kx=0 initially looking for a value of y but then i got y to cancel and i got a value for k instead. i got k=2.

so then i rewrote my derivatives as 2x+2y=0
and 2y+2x=0
which i have just realized are the same equation. anyway,
i then went ahead and tried to solve one of them for a value of x so i could sub it back
and i got x=-y
then i went and subbed this new value for x back into the equation and I am getting 0=0
have i answered the equation at all?
would greatly appreciate any input.
if i am completely wrong it would help greatly if someone pointed me in the right direction
thanks



 
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  • #2
cathal84 said:

Homework Statement


Let f(x, y) = x^2 + kxy + y^2 , where k is some constant in R. i. Show that f has a stationary point at (0, 0) for every k ∈ R

Homework Equations


...

The Attempt at a Solution


I may have the solution or i may have gone completely wrong I am not entirely sure.
i first found the derivative of f(x,y) with respect to x it was 2x+ky
then found the derivative of f(x,y) with respect to y it was 2y+kx
i then let both of them equal 0

then i solved 2x+ky=0 looking for a value of x and i got x=-ky/2
i then put this value for x back into 2y+kx=0 initially looking for a value of y but then i got y to cancel and i got a value for k instead. i got k=2.

so then i rewrote my derivatives as 2x+2y=0
and 2y+2x=0
which i have just realized are the same equation. anyway,
i then went ahead and tried to solve one of them for a value of x so i could sub it back
and i got x=-y
then i went and subbed this new value for x back into the equation and I am getting 0=0
have i answered the equation at all?
would greatly appreciate any input.
if i am completely wrong it would help greatly if someone pointed me in the right direction
thanks



You have not dealt with the question asked: it asked you to show that ##(x,y) = (0,0)## is a stationary point, no matter what the value of ##k## may be. No equation-solving is involved at all!

Of course, you may need to try solving the equations if you are asked to determine all the stationary points, but that was not the question.
 
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  • #3
Ray Vickson said:
You have not dealt with the question asked: it asked you to show that ##(x,y) = (0,0)## is a stationary point, no matter what the value of ##k## may be. No equation-solving is involved at all!

Of course, you may need to try solving the equations if you are asked to determine all the stationary points, but that was not the question.
thanks for your reply,
Ray Vickson said:
You have not dealt with the question asked: it asked you to show that ##(x,y) = (0,0)## is a stationary point, no matter what the value of ##k## may be. No equation-solving is involved at all!

Of course, you may need to try solving the equations if you are asked to determine all the stationary points, but that was not the question.
thank you this has answered my question
 

Related to Show that f has a stationary point at (0, 0) for every k ∈ R

1. What does it mean for a function to have a stationary point?

A stationary point on a function is a point where the slope or derivative is equal to zero. This means that at this point, the function is neither increasing nor decreasing.

2. How can you prove that a function has a stationary point at (0, 0)?

To prove that a function has a stationary point at (0, 0), we need to show that the derivative of the function at this point is equal to zero. This can be done by taking the derivative of the function and setting it equal to zero, then solving for the values of x and y at the stationary point.

3. What is the significance of the value of k in this statement?

The value of k in this statement represents a constant in the function. It is included to show that the stationary point at (0, 0) exists for all real values of k, not just a specific value.

4. Can a function have multiple stationary points?

Yes, a function can have multiple stationary points. These are points where the derivative is equal to zero, and can occur at different values of x and y.

5. How does knowing that a function has a stationary point at (0, 0) help in understanding the behavior of the function?

Knowing that a function has a stationary point at (0, 0) can help in understanding the behavior of the function, as it tells us that at this point the function is neither increasing nor decreasing. This can provide insight into the overall shape and characteristics of the function.

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