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Function evaluation is a process that involves finding the output value of a function for specific input values. In multivariable calculus, you deal with functions of more than one variable. Here, the function \( f(x, y) = \frac{y}{x+y} \) needs to be evaluated at three different points: \((2, 1)\), \((-1, 2)\), and \((-2, -1)\). To evaluate the function, you substitute the values of \(x\) and \(y\) from each point into the function. This helps determine the value that the function takes for these specific inputs.

• For \((2, 1)\), substitute \(x = 2\) and \(y = 1\) to get \( f(2, 1) = \frac{1}{3} \).
• For \((-1, 2)\), substitute \(x = -1\) and \(y = 2\) to get \( f(-1, 2) = 2 \).
• For \((-2, -1)\), substitute \(x = -2\) and \(y = -1\) to get \( f(-2, -1) = \frac{1}{3} \).

This substitution helps in identifying how the function behaves in different regions of the coordinate plane.

Coordinate substitution is a key method for evaluating functions at given points. By substituting different coordinate values into a function, you discover the specific output based on those inputs. This method is straightforward:- Replace the variable \(x\) with its specified value.- Replace the variable \(y\) with its specified value.- Simplify to find the function's value at that point.For example, let’s apply this to the point \( (2, 1) \) for the function \( f(x, y) = \frac{y}{x+y} \):1. Substitute \(x = 2\) and \(y = 1\).2. Simplify to get \( f(2, 1) = \frac{1}{2 + 1} = \frac{1}{3} \).This simple procedure is repeated for each point, ensuring you maintain accuracy in each substitution step. That's essential because one minor mistake in substitution can change the function's output entirely.

Fraction simplification involves reducing a fraction to its simplest form, making it easier to understand and compare. In function evaluation, fractions often appear after substituting coordinates into the function. Simplifying these fractions is critical to arrive at a meaningful answer.When finding the function values:- Simplify \( \frac{y}{x + y} \) by combining terms in the denominator to get a straightforward fraction.- For example, for the point \((-2, -1)\), after substitution, the expression becomes \( \frac{-1}{-3} \). - Simplify this fraction to \( \frac{1}{3} \), as the negatives in both numerator and denominator cancel each other.This process of simplifying ensures that all answers are as reduced as possible, which aids in clarity and further mathematical operations. Always double-check simplifications to ensure accuracy, as these can influence the correctness of your overall results.