To find the linearization $L(x, y)$ of the function $f(x, y) = \frac{1 + y}{1 + x}$ at the point $(1, 3)$ and use it to approximate $f(0.9, 3.2)$, let’s follow these steps:

Step 1: Calculate $f(1, 3)$

Evaluate $f$ at the point $(1, 3)$: $f(1, 3) = \frac{1 + 3}{1 + 1} = \frac{4}{2} = 2$

Step 2: Find Partial Derivatives $f_x$ and $f_y$

1. Partial derivative with respect to $x$: $f_x(x, y) = \frac{\partial}{\partial x} \left( \frac{1 + y}{1 + x} \right) = -\frac{1 + y}{(1 + x)^2}$ Evaluating at $(1, 3)$: $f_x(1, 3) = -\frac{1 + 3}{(1 + 1)^2} = -\frac{4}{4} = -1$

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

Step 3: Form the Linearization $L(x, y)$

The linearization formula is: $L(x, y) = f(a, b) + f_x(a, b)(x - a) + f_y(a, b)(y - b)$ where $(a, b) = (1, 3)$.

Substitute the values: $L(x, y) = 2 - 1(x - 1) + \frac{1}{2}(y - 3)$ Simplify: $L(x, y) = 2 - (x - 1) + \frac{1}{2}(y - 3) = 3 - x + \frac{y}{2}$

Step 4: Approximate $f(0.9, 3.2)$

Now, substitute $x = 0.9$ and $y = 3.2$ into the linearization: $L(0.9, 3.2) = 3 - 0.9 + \frac{3.2}{2}$ Calculate each term: $= 3 - 0.9 + 1.6 = 3.7$

Final Answer

The approximate value of $f(0.9, 3.2)$ is 3.7.

Would you like further details on any step, or have any questions?

Here are some related questions:

1. How do partial derivatives relate to linearization?
2. What is the geometric interpretation of a linearization?
3. How accurate is linearization for points close to $(a, b)$?
4. Can linearization be used for functions with three or more variables?
5. How does linearization relate to the concept of tangent planes?

Tip: When linearizing, always ensure that the chosen point $(a, b)$ is within the domain of differentiability for accurate approximation.