What are the values and types of the critical points, if any, of #f(x,y)=2x + 4y + 4#?

Answer 1

That function has no critical points. (It is a plane.)

#f_x(x,y) = 2# and #f_y(x,y) = 4#.

It is not possible to make the first derivatives undefined or zero. So there are no critical points.

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Answer 2

To find the critical points of ( f(x, y) = 2x + 4y + 4 ), we need to calculate its partial derivatives with respect to ( x ) and ( y ), set them equal to zero, and solve for ( x ) and ( y ).

The partial derivatives are:

[ \frac{\partial f}{\partial x} = 2 ] [ \frac{\partial f}{\partial y} = 4 ]

Setting these partial derivatives equal to zero, we get:

[ 2 = 0 ] [ 4 = 0 ]

Since these equations have no solutions, there are no critical points for the function ( f(x, y) = 2x + 4y + 4 ). Therefore, there are no values or types of critical points to determine.

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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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