Problem 21 - Calculus Early Transcendentals: Pearson New International Edition

Question

Describe the largest set \(S\) on which it is correct to say that \(f\) is continuous. \(f(x, y)=\frac{x^{2}+3 x y+y^{2}}{y-x^{2}}\)

Step-by-Step Solution

Answer

The largest set is \(\{(x, y) \mid y \neq x^2\}\).

Step 1: Identify Points of Discontinuity

For the function \[f(x, y) = \frac{x^2 + 3xy + y^2}{y - x^2},\]the denominator is zero whenever \[y = x^2.\]Therefore, the function may be discontinuous at these points. The set of such points can be expressed as \[\{(x, y) \mid y = x^2\}.\]

Step 2: Define the Complement of the Discontinuity Set

The complement of the points where the function is potentially discontinuous (where it remains continuous) is given by all points that do not satisfy \[y = x^2.\]This is \[\{(x, y) \mid y eq x^2\}.\]

Step 3: Largest Set for Continuity

The largest set on which the function \(f(x, y)\) is continuous is thus composed of all points where \(y eq x^2\), as this avoids the division by zero issue in the function's formula.

Key Concepts

Discontinuity Points

In multivariable calculus, identifying discontinuity points is crucial for understanding where a function behaves irregularly. A function becomes discontinuous when certain conditions make it undefined. For our function \(f(x, y) = \frac{x^2 + 3xy + y^2}{y - x^2}\), these are the points where the denominator is zero.

In simpler terms, a discontinuity occurs wherever \(y = x^2\). At these locations, the denominator \(y - x^2\) becomes zero, causing the function to "blow up," or become undefined. This leads to a gap or "disruption" in the function's graph.

Finding these points is essential because they help us understand the behavior of the function and identify where it fails to be smooth and continuous.

Domain of a Function

The domain of a multivariable function includes all possible values for which the function is defined. To determine the domain of \(f(x, y) = \frac{x^2 + 3xy + y^2}{y - x^2}\), we need to look at where this expression makes sense.

For this function, the domain includes all \((x, y)\) pairs except where \(y = x^2\). This is because when \(y = x^2\), the function becomes undefined due to division by zero. By excluding these points, we ensure that the function can be evaluated without encountering any mathematical errors.

To express this formally, the domain is:

• \((x, y)\) such that \(y eq x^2\)

Understanding the domain clarifies which inputs we can use, ensuring the function remains valid and calculable.

Limits and Continuity

Limits and continuity are foundational concepts in calculus that help determine whether a function behaves predictably. When examining continuity in multivariable functions, we must check if limits exist as we approach a potential point of discontinuity. A function is continuous if the limit exists and matches the function's value at the point.

For \(f(x, y)\), when approaching a point where \(y = x^2\), the limit does not exist uniformly. The values of the function grow infinitely large as you approach \((x, x^2)\), signaling a discontinuity.

To ensure continuity, we work where \(y eq x^2\). In these regions, the limits approach a finite and defined value, reflecting a smooth transition without jumps or disruptions.

• The function continues as you'd intuitively expect.
• There's no unexpected behavior or undefined leaps in value.

By understanding limits and continuity, we assess where a function like \(f(x, y)\) behaves consistently and predictably across its domain.