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The square root function is a mathematical function denoted by the symbol \( \sqrt{} \). This function essentially finds a number which, when multiplied by itself, gives the original number beneath the square root symbol.
In our exercise, we have a square root function dealing with variables, specifically two variables, \( x \) and \( y \). Here, the function \( f(x, y) = \sqrt{75 - x^{2} - y^{2}} \) indicates that for any given values of \( x \) and \( y \), we find what under the square root results in a non-negative number.
This particular form where we subtract squares of variables from a constant under the square root creates a sphere if set equal to zero in three-dimensional space. This characteristic is important in both calculus and geometry.

Evaluating a function means finding its output value for specific inputs. It's like feeding numbers into a machine and getting results.
For this exercise, we evaluate \( f(x, y) \) at the point \((5, -1)\).

• Identify the given point: Here, it's \((5, -1)\).
• Substitute these values into the function: Replace \( x \) with 5 and \( y \) with -1.
• Simplify the result: Follow the order of operations – exponentiation first, then subtraction, and square root at the end.

By substituting and simplifying the function, you find \( f(5, -1) = 7 \), meaning when \( x = 5 \) and \( y = -1 \), the function evaluates to 7.

Two-variable functions do not depend on a single input but rather on two inputs, usually denoted as \( x \) and \( y \). This makes them a step up in complexity from single-variable functions since the output depends on varying two different inputs.
In our function \( f(x, y) = \sqrt{75 - x^{2} - y^{2}} \), the two variables \( x \) and \( y \) influence the value under the square root.

• Two-variable functions are often visualized in three-dimensional space, with the input forming a plane, and the output being a surface.
• These types of functions are crucial in fields like physics and engineering for modeling phenomena that depend on multiple factors.

Function evaluation in multivariable calculus often involves substituting specific values for both variables. As illustrated, this makes solving these functions systematic and straightforward, leading to a clear numerical answer.