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When facing a multivariable function like \(f(x, y, z) = x^2 - 3y + z\), function evaluation entails finding the output by substituting specific values for each variable. This makes it possible to know what the function equals at a given point. For such exercises, you'll often encounter tuple points that indicate exact values to replace variables, like \((3, -1, 1)\) in our example.
To evaluate a function:

• Identify each variable present.
• Substitute the designated value as given in the problem.
• Solve any arithmetic operations involved.

Doing these steps helps ascertain the function's behavior or result at specified coordinates. Evaluating functions is a foundational skill in calculus and crucial for deeper comprehension of graph behavior and system dynamics.

The substitution method is integral to evaluating multivariable functions. It involves replacing each variable with a known value. This method simplifies complex functions into manageable arithmetic operations. Take \(f(x, y, z) = x^2 - 3y + z\) and the point \((3, -1, 1)\). Here’s how substitution unfolds:

• First, substitute \(x = 3\) in place of \(x\).
• Then, replace \(y = -1\) in the expression with \(y\).
• Lastly, use \(z = 1\) in lieu of \(z\).

With each variable swapped, the function simplifies from an abstract expression into an arithmetic calculation, making it easier to derive the desired answer.

Once values substitute the variables in a function, various algebraic operations come into play. Take step-by-step computations:In the given function \(f(x, y, z) = x^2 - 3y + z\): 1. Squaring \(x\) involves simple multiplication, \(3^2 = 9\). This squaring step captures potential changes to the function as \(x\) varies.
2. The next operation, \(-3y\), entails multiplying \(-3\) by \(-1\) to get 3. Multiplying negatives often results in a positive outcome.
3. Finally, adding \(z\) (which is 1) to the sum brings about \(9 + 3 + 1 = 13\).Algebraic operations simplify once substitutions are in place. This clear path from abstract equations to numerical solutions bolsters both problem-solving and analytical abilities in calculus.