91Ó°ÊÓ

Understanding domain restrictions is crucial when working with functions, especially rational ones. A domain refers to all possible input values for a function. To identify domain restrictions, you must look at conditions where a function is not defined. In the given function, \( H(x, y) = \frac{e^x + e^y}{e^{xy} - 1} \), we must prevent the denominator from being zero because division by zero is undefined.

Hence, you set \( e^{xy} - 1 eq 0 \), which leads to \( e^{xy} eq 1 \). The expression \( e^{xy} = 1 \) only occurs when \( xy = 0 \). Therefore, \( x \) and \( y \) should not both be zero simultaneously.

In essence, the domain restriction in this situation means the points \( (x, y) \) should satisfy \( xy eq 0 \). This allows either \( x \) or \( y \) to be zero, but not both, thereby preventing undefined calculations.

Rational functions are a special class of functions that are expressed as the ratio of two polynomials. In the function \( H(x, y) = \frac{e^x + e^y}{e^{xy} - 1} \), although the numerator and denominator involve exponential expressions rather than simple polynomials, the principles remain the same.

A rational function is defined wherever its denominator is not zero. This makes rational functions fascinating because their domains often highlight interesting behavior. When analyzing such functions, identifying where the denominator vanishes is key to finding potential discontinuities or undefined points.

In this particular example, the denominator \( e^{xy} - 1 \) imposes a restriction on the continuity of the function. Understanding rational functions involves recognizing the conditions under which the different parts of the function may fail to produce valid outputs.

Points of discontinuity are specific values where a function breaks or jumps, meaning it is not continuous at those points. For a function to be continuous at a point, its value as well as its limit, as you approach from any direction, must exist and be equal.

For the function \( H(x, y) = \frac{e^x + e^y}{e^{xy} - 1} \), discontinuity occurs wherever \( xy = 0 \), which happens along the lines \( x = 0 \) and \( y = 0 \). On these lines, the denominator becomes zero, making the function undefined.

Thus, points at or on these lines are not included in the set where the function is continuous. Practically, this means the function "breaks" along these axes. To describe where \( H(x, y) \) remains continuous, you exclude those discontinuous points, encapsulating all inputs where the function operates smoothly without any interruptions.