Problem 33 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let $$ f(x, y)=\left\\{\begin{array}{ll} \frac{x^{2}-4 y^{2}}{x-2 y}, & \text { if } x \neq 2 y \\ g(x), & \text { if } x=2 y \end{array}\right. $$ If \(f\) is continuous in the whole plane, find a formula for \(g(x)\).

Step-by-Step Solution

Answer

The function \(g(x)\) is \(x\).

Step 1: Identify when the problematic points occur

The problematic points are where the function might not be continuous due to the definition, which is at \(x = 2y\). For these points, the function is defined differently, using \(g(x)\).

Step 2: Simplify the expression

For \(x eq 2y\), the function \(f(x, y)\) is \( \frac{x^{2} - 4y^{2}}{x - 2y} \). This can be simplified using the identity \(a^2 - b^2 = (a-b)(a+b)\), so \(x^2 - 4y^2\) becomes \((x - 2y)(x + 2y)\). Thus, the function simplifies to \(x + 2y\) when \(x eq 2y\).

Step 3: Evaluate the limit as \(x \to 2y\)

We need to ensure \(f(x, y) = g(x)\) when \(x = 2y\) to maintain continuity. Take the limit of \(x + 2y\) as \(x \to 2y\). Substitute \(x = 2y\) into \(x + 2y\) to get \(2y + 2y = 4y\). Therefore, the limit is \(2y\). Correspondingly, when \(x = 2y\), then \(y = x/2\), and so it becomes \(2(x/2) = x\).

Step 4: Determine the value of g(x)

For \(f\) to be continuous at \(x = 2y\), set \(g(x)\) equal to the limit found, so \(g(x) = x\). This way, the function remains consistent and continuous across the plane.

Key Concepts

Limits

In calculus, limits help us understand the behavior of functions as inputs approach a particular point. When dealing with multivariable functions, like our piecewise function, limits assure us that there is a seamless continuation at every point—no abrupt changes or jumps. In our problem, the function is defined differently when \( x = 2y \), which means the limit as \( x \) approaches \( 2y \) should be carefully evaluated to ensure continuity. Let's break down the process:

• Consider the expression \( \frac{x^{2}-4y^{2}}{x-2y} \). For points where \( x eq 2y \), this expression simplifies, but at \( x = 2y \), it's not directly computable, leading to a problem if we don't analyze the limit.
• Simplifying high-degree expressions often includes factoring, and here we utilize the identity \( a^2 - b^2 = (a-b)(a+b) \).

In our solution, by doing the limit evaluation, we ensure that when \( x \to 2y \), the resulting function value matches the piece defined for this exact situation. This ensures the function continuously covers the whole plane with no points of discontinuity.

Piecewise Functions

Piecewise functions, like the one we're analyzing, assign different expressions to different parts of the domain. Understanding them involves grasping how these expressions change depending on given conditions. Here, our function \( f(x, y) \) involves:

• A standard form \( \frac{x^{2}-4y^{2}}{x-2y} \) for \( x eq 2y \).
• An undetermined form \( g(x) \) for \( x = 2y \)—until we define it for continuity.

The key aspect of piecewise functions in calculus is ensuring that all the "pieces" fit together smoothly. This means checking for continuity at the boundaries where conditions change. In our problem, continuity at \( x = 2y \) ensures no gaps or jumps occur, which is often the primary challenge with piecewise functions. Through setting \( g(x) \) equal to the calculated limit, the transition between the distinct parts is left seamless, assuring a continuous function behavior across its entire domain.

Simplification of Algebraic Expressions

Simplification is crucial in calculus to make complex expressions manageable and to accurately compute limits, derivatives, or integrals. Our function \( f(x, y) = \frac{x^{2}-4y^{2}}{x-2y} \) initially appears complicated for usual methods of evaluation but simplifies nicely.The process can include:

• Recognizing algebraic identities like \( a^2 - b^2 = (a-b)(a+b) \), which breaks down expressions into simpler, factorable components.
• Eliminating common factors in numerators and denominators, for instance, \( (x - 2y) \) in both the numerator and denominator.

When we simplify \( x^2 - 4y^2 \) to \( (x - 2y)(x + 2y) \), the function itself turns into \( x + 2y \), making it much easier to evaluate limits and check for continuity. Thus, simplification isn't just a matter of convenience; it's necessary for assessing function behavior, especially in variable-heavy settings like multivariable calculus.