To find the linearization of the function $f(x, y) = x^2 - xy + \frac{1}{2}y^2 + 3$ at the point $(3, 2)$, we need to calculate the partial derivatives with respect to $x$ and $y$ and then use the formula for the linear approximation.

Step 1: Compute the partial derivatives

1. Partial derivative with respect to $x$: $f_x(x, y) = \frac{\partial}{\partial x}\left( x^2 - xy + \frac{1}{2}y^2 + 3 \right) = 2x - y$

2. Partial derivative with respect to $y$: $f_y(x, y) = \frac{\partial}{\partial y}\left( x^2 - xy + \frac{1}{2}y^2 + 3 \right) = -x + y$

Step 2: Evaluate the partial derivatives at the point $(3, 2)$

1. $f_x(3, 2) = 2(3) - 2 = 6 - 2 = 4$
2. $f_y(3, 2) = -(3) + 2 = -3 + 2 = -1$

Step 3: Write the linearization formula

The linearization $L(x, y)$ of $f(x, y)$ at the point $(3, 2)$ is given by: $L(x, y) = f(3, 2) + f_x(3, 2)(x - 3) + f_y(3, 2)(y - 2)$

Step 4: Evaluate $f(3, 2)$

$f(3, 2) = 3^2 - 3(2) + \frac{1}{2}(2)^2 + 3 = 9 - 6 + 2 + 3 = 8$

Step 5: Substitute the values into the linearization formula

$L(x, y) = 8 + 4(x - 3) - 1(y - 2)$

Simplify the expression:

$L(x, y) = 8 + 4x - 12 - y + 2$ $L(x, y) = 4x - y - 2$

Conclusion:

The linearization of the function at $(3, 2)$ is $L(x, y) = 4x - y - 2$.

Therefore, the correct answer is: c. $4x - y - 2$

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Would you like further details or have any questions?

Here are 5 related questions you might be interested in:

1. How do you find the linearization of a function at a different point?
2. What is the geometric interpretation of a linearization?
3. How do partial derivatives help in linear approximations?
4. Can linearization be used for functions of more than two variables?
5. What are some practical applications of linearization in engineering?

Tip: Linearization is especially useful in approximating the value of a function near a given point.