To approximate $f(-1.6, -1.9)$ using the linear approximation of the function $f(x, y) = x^3 + 4y^2 - 3xy$ at the point $(-2, -2)$, we follow these steps:

Step 1: Find the linear approximation formula

The linear approximation of a function $f(x, y)$ at a point $(a, b)$ is given by: $L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)$ Where:

• $f_x(a, b)$ is the partial derivative of $f$ with respect to $x$ evaluated at $(a, b)$
• $f_y(a, b)$ is the partial derivative of $f$ with respect to $y$ evaluated at $(a, b)$

Step 2: Calculate the necessary derivatives

First, we compute the partial derivatives $f_x(x, y)$ and $f_y(x, y)$.

$f_x(x, y)$:

$f_x(x, y) = \frac{\partial}{\partial x} (x^3 + 4y^2 - 3xy) = 3x^2 - 3y$

$f_y(x, y)$:

$f_y(x, y) = \frac{\partial}{\partial y} (x^3 + 4y^2 - 3xy) = 8y - 3x$

Step 3: Evaluate the function and its partial derivatives at $(-2, -2)$

Evaluate $f(-2, -2)$:

$f(-2, -2) = (-2)^3 + 4(-2)^2 - 3(-2)(-2)$ $= -8 + 16 - 12 = -4$

Evaluate $f_x(-2, -2)$:

$f_x(-2, -2) = 3(-2)^2 - 3(-2) = 3(4) + 6 = 12 + 6 = 18$

Evaluate $f_y(-2, -2)$:

$f_y(-2, -2) = 8(-2) - 3(-2) = -16 + 6 = -10$

Step 4: Write the linear approximation

Using the values from above, we can now write the linear approximation $L(x, y)$: $L(x, y) = f(-2, -2) + f_x(-2, -2)(x + 2) + f_y(-2, -2)(y + 2)$ $L(x, y) = -4 + 18(x + 2) - 10(y + 2)$

Step 5: Approximate $f(-1.6, -1.9)$

Now, plug in $x = -1.6$ and $y = -1.9$ into the linear approximation formula: $L(-1.6, -1.9) = -4 + 18(-1.6 + 2) - 10(-1.9 + 2)$ $= -4 + 18(0.4) - 10(0.1)$ $= -4 + 7.2 - 1$ $= 2.2$

Final Answer:

The linear approximation of $f(-1.6, -1.9)$ is $\boxed{2.2}$.

---

Would you like more details or further explanation? Here are some related questions for you:

1. How do you derive the linear approximation formula for functions of two variables?
2. What does the linear approximation represent in terms of the geometry of the function's graph?
3. How does the accuracy of the linear approximation change when the point $(a, b)$ is closer to or farther from the point of interest?
4. How do you interpret the partial derivatives $f_x(x, y)$ and $f_y(x, y)$ in the context of this problem?
5. What is the difference between linear approximation and higher-order approximations (like quadratic approximations)?

Tip: The closer the point of approximation is to the target point, the more accurate the linear approximation will be.