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

• tamintl
So what's your conclusion?In summary, the critical points of the function f(x, y) = y sin x + cos x are (-pi,0), (0,0), (pi,0), and (2pi,0). These points correspond to the x-value being any integer multiple of pi, and the y-value being 0. The nature of these critical points can be determined by finding the second derivatives, fxx, fyy, and fxy, and using the second derivative test.

#### 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.

none

## The Attempt at a Solution

I know that to find the critical points you must set fx = 0 and fy=0

Doing this I get:

fx = -4x + y2 - 2y - 3 = 0

and

fy = 2xy - 2x = 0

Thus, fx = fy

-4x + y2 -2y - 3 = 0

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

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

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

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

fy = 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 fx to get another?

Thanks Dick!

tamintl said:
fy = 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 fx 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.

Dick said:
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 fx gives me: (0,-1), (0,3), (-1,1)

Assuming these are correct do I now find fxx, fyy, and fxy
to determine the nature (using theorem) ??

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

Thanks Dick! I really appreciate your help

tamintl said:
Silly me!

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

Assuming these are correct do I now find fxx, fyy, and fxy
to determine the nature (using theorem) ??

On a side note how do you find fxy? 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.

Dick said:
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 fx = ycosx - sinx and fy = sinx

hence, for

fx we have y = sinx/cosx = tanx

and

fy 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

tamintl said:
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 fx = ycosx - sinx and fy = sinx

hence, for

fx we have y = sinx/cosx = tanx

and

fy 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?

∏, 2∏

Now what :/

edit: x=∏(n) for all n ε Z

tamintl said:
∏, 2∏

Now what :/

edit: x=∏(n) for all n ε Z

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

Dick said:
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*∏

?

tamintl said:
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.

Last edited: