What are the extrema and saddle points of #f(x,y) = e^y(y^2-x^2)#?

Answer 1

#{0,0}# saddle point
#{0,-2}# local maximum

#f(x,y) = e^y(y^2-x^2)#

so the sationary points are determined by solving

#grad f(x,y) = vec 0#

or

#{ (-2 e^y x = 0), (2 e^y y + e^y (-x^2 + y^2) = 0) :}#

giving two solutions

#((x=0,y=0),(x=0,y=-2))#

Those points are qualified using

#H = grad(grad f(x,y))#

or

#H =((-2 e^y, -2 e^y x),(-2 e^y x, 2 e^y + 4 e^y y + e^y (-x^2 + y^2))) #

so

#H(0,0) = ((-2, 0),(0, 2))# has eigenvalues #{-2,2}#. This result qualifies point #{0,0}# as a saddle point.

#H(0,-2)=((-2/e^2, 0),(0, -2/e^2))# has eigenvalues #{-2/e^2, -2/e^2}#. This result qualifies point #{0,-2}# as a local maximum.

Attached the #f(x,y)# contour map near the points of interest

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Answer 2

To find the critical points of ( f(x, y) = e^y(y^2 - x^2) ), we first need to find the partial derivatives with respect to ( x ) and ( y ) and then solve for where both partial derivatives are equal to zero. Then, we can use the second partial derivative test to classify the critical points as extrema or saddle points.

The partial derivatives are: [ \frac{\partial f}{\partial x} = -2xe^yy^2 ] [ \frac{\partial f}{\partial y} = e^y(2y^2 - x^2 + 1) ]

Setting both partial derivatives equal to zero, we get: [ -2xe^yy^2 = 0 ] [ e^y(2y^2 - x^2 + 1) = 0 ]

From the first equation, we have ( x = 0 ) or ( y = 0 ). Plugging these into the second equation:

If ( x = 0 ): [ e^y(2y^2 + 1) = 0 ] This implies ( y = 0 ) (no real solutions for ( y ) when ( 2y^2 + 1 = 0 )).

If ( y = 0 ): [ e^0(0 - x^2 + 1) = 0 ] This implies ( x^2 = 1 ), so ( x = \pm 1 ).

Therefore, the critical points are at ( (0, 0) ) and ( (\pm 1, 0) ).

Now, to classify these critical points, we need to use the second partial derivative test. We compute the second partial derivatives: [ \frac{\partial^2 f}{\partial x^2} = -2e^yy^2 ] [ \frac{\partial^2 f}{\partial y^2} = e^y(4y^2 - x^2 + 1) ] [ \frac{\partial^2 f}{\partial x \partial y} = -4xe^yy ]

For ( (0, 0) ): [ D = \frac{\partial^2 f}{\partial x^2} \frac{\partial^2 f}{\partial y^2} - \left(\frac{\partial^2 f}{\partial x \partial y}\right)^2 = 0 ] Since ( D = 0 ), the second derivative test is inconclusive.

For ( (1, 0) ) and ( (-1, 0) ): [ D = -2e^0 \cdot 0(4 \cdot 0^2 - 1 + 1) - \left(-4 \cdot 0 \cdot e^0 \cdot 0\right)^2 = 0 ] Since ( D = 0 ), the second derivative test is inconclusive.

Therefore, the critical points ( (0, 0) ), ( (1, 0) ), and ( (-1, 0) ) may be saddle points.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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