# How do you find the linearization at (-2,1) of #f(x,y) = x^2y^3 - 4sin(x+2y) #?

Let us consider the surface

the surface normal vector is given by

or

So

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To find the linearization of ( f(x, y) ) at the point ( (-2, 1) ), we first need to find the partial derivatives of ( f ) with respect to ( x ) and ( y ) at that point. Then, we use these derivatives to construct the linear approximation formula:

[ L(x, y) = f(-2, 1) + f_x(-2, 1)(x + 2) + f_y(-2, 1)(y - 1) ]

Where:

- ( f_x(-2, 1) ) is the partial derivative of ( f ) with respect to ( x ) evaluated at ( (-2, 1) ).
- ( f_y(-2, 1) ) is the partial derivative of ( f ) with respect to ( y ) evaluated at ( (-2, 1) ).

Let's find these derivatives:

[ f_x = 2xy^3 - 4\cos(x + 2y) ] [ f_y = 3x^2y^2 - 8\cos(x + 2y) ]

Now, we evaluate these derivatives at ( (-2, 1) ):

[ f_x(-2, 1) = 2(-2)(1)^3 - 4\cos((-2) + 2(1)) ] [ f_y(-2, 1) = 3(-2)^2(1)^2 - 8\cos((-2) + 2(1)) ]

After calculating these, we substitute them into the linear approximation formula to find ( L(x, y) ).

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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.

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