Finding the critical point and its nature. With solid attempt

[tamintl]

Finding the critical point and its nature. With solid attempt!

Homework Statement

Find all critical points of the function

f(x, y) = xy² - 2xy - 2x² - 3x +7

and determine their nature.

Homework Equations

none

The Attempt at a Solution

I know that to find the critical points you must set f_x = 0 and f_y=0

Doing this I get:


f_x = -4x + y² - 2y - 3 = 0

and

f_y = 2xy - 2x = 0

Thus, f_x = f_y

-4x + y² -2y - 3 = 0

-4x + (y+1)(y-3) = 0

I don't really know how to proceed? Any help would be great.

 

[Dick]

Factor the fy equation first. What conclusions can you draw from that?

 

[tamintl]

Factor the fy equation first. What conclusions can you draw from that?

f_y = 2xy - 2x = 2x(y-1)

Thus, x=0 and y=1

SO, critical point is (0,1)...

So is that the only critical point or do I sub this back into f_x to get another?

Thanks Dick!

 

[Dick]

f_y = 2xy - 2x = 2x(y-1)

Thus, x=0 and y=1

SO, critical point is (0,1)...

So is that the only critical point or do I sub this back into f_x to get another?

Thanks Dick!

Careful! 2x(y-1)=0 if x=0 OR y=1. Not necessarily both. Put those two possibilities back into fx and see what happens.

 

[tamintl]

Careful! 2x(y-1)=0 if x=0 OR y=1. Not necessarily both. Put those two possibilities back into fx and see what happens.

Silly me!

Okay subbing back into f_x gives me: (0,-1), (0,3), (-1,1)

Assuming these are correct do I now find f_xx, f_yy, and f_xy
to determine the nature (using theorem) ??

On a side note how do you find f_xy? I'm confused about that particular instance.

Thanks Dick! I really appreciate your help

 

[Dick]

Silly me!

Okay subbing back into f_x gives me: (0,-1), (0,3), (-1,1)

Assuming these are correct do I now find f_xx, f_yy, and f_xy
to determine the nature (using theorem) ??

On a side note how do you find f_xy? I'm confused about that particular instance.

Thanks Dick! I really appreciate your help

Sure, now use the second derivative test on those three critical points. fxy isn't hard to to find, just take your fx and differentiate with respect to y, OR take fy and differentiate with respect to x. You'll get the same thing.

 

[tamintl]

Sure, now use the second derivative test on those three critical points. fxy isn't hard to to find, just take your fx and differentiate with respect to y, OR take fy and differentiate with respect to x. You'll get the same thing.

Got it thanks!

Now I have anther question:

Find all critical points of the function: f(x, y) = y sin x + cos x and determine their nature.

So we have f_x = ycosx - sinx and f_y = sinx

hence, for

f_x we have y = sinx/cosx = tanx

and

f_y we have sinx = 0

Therefore we have x=0... when we sub this into y=tanx we get y=0

Hence critical point is (0,0)

Would this be reasonable?

Regards

 

[Dick]

Got it thanks!

Now I have anther question:

Find all critical points of the function: f(x, y) = y sin x + cos x and determine their nature.

So we have f_x = ycosx - sinx and f_y = sinx

hence, for

f_x we have y = sinx/cosx = tanx

and

f_y we have sinx = 0

Therefore we have x=0... when we sub this into y=tanx we get y=0

Hence critical point is (0,0)

Would this be reasonable?

Regards

x=0 isn't the only solution to sin(x)=0. What are the others?

 

[tamintl]

∏, 2∏

Now what :/

edit: x=∏(n) for all n ε Z

 

[Dick]

∏, 2∏

Now what :/

edit: x=∏(n) for all n ε Z

Put x=pi*n into fx. What does that tell you about y?

 

[tamintl]

Put x=pi*n into fx. What does that tell you about y?

As you put it into fx, sinx = 0... therefore you have fx = -y for odd n and fx = y for even n..

Hence y=0 again for x=n*∏

?

 

[Dick]

As you put it into fx, sinx = 0... therefore you have fx = -y for odd n and fx = y for even n..

Hence y=0 again for x=n*∏

?

Fine. So your critical points are (-pi,0), (0,0), (pi,0), (2pi,0)... right? (n*pi,0) where n is any integer.