How do you find the critical points of the function #f(x,y)=x^2+y^2+(x^2)(y)+4#?
(0,0), (
corresponding values of y =0,-1,-1.
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To find the critical points of the function ( f(x, y) = x^2 + y^2 + x^2y + 4 ), we need to find where the partial derivatives with respect to ( x ) and ( y ) are both equal to zero.
The partial derivative with respect to ( x ) is ( \frac{\partial f}{\partial x} = 2x + 2xy ).
Setting ( \frac{\partial f}{\partial x} = 0 ) gives us the equation ( 2x + 2xy = 0 ), which simplifies to ( x(1 + y) = 0 ).
This gives us two possibilities: either ( x = 0 ) or ( y = -1 ).
Similarly, the partial derivative with respect to ( y ) is ( \frac{\partial f}{\partial y} = 2y + x^2 ).
Setting ( \frac{\partial f}{\partial y} = 0 ) gives us the equation ( 2y + x^2 = 0 ).
If we substitute ( x = 0 ) into this equation, we get ( 2y = 0 ), so ( y = 0 ).
Therefore, the critical points are at ( (0, 0) ) and ( (0, -1) ).
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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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