Problem 30 - Calculus for Biology and Medicine

Question

a) Write $$ h(x, y)=\cos (y-x) $$ as a composition of two functions. (b) For which values of \((x, y)\) is \(h(x, y)\) continuous?

Step-by-Step Solution

Answer

a) \( h(x, y) = f(g(x, y)) \) with \( g(x, y) = y-x \) and \( f(u) = \cos(u) \). b) Continuous for all \( (x, y) \in \mathbb{R}^2 \).

Step 1: Recognizing Function Composition

To express the given function \( h(x, y) = \cos(y - x) \) as a composition of two functions, recognize that it involves both an addition operation and the cosine function. Thus, consider breaking it down into these fundamental operations.

Step 2: Defining the Inner Function

Define the inner function \( g(x, y) \) which handles the subtraction inside the cosine function. Specifically, set \( g(x, y) = y - x \). This completes one part of the composition where the expression inside the cosine function is calculated separately.

Step 3: Defining the Outer Function

Define the outer function \( f(u) \), where \( u \) represents the result of the inner function. Set \( f(u) = \cos(u) \). This function applies the cosine operation to the result of the inner function \( g(x, y) \).

Step 4: Express the Composition

With the two functions defined, express \( h(x, y) \) as a composition: \( h(x, y) = (f \circ g)(x, y) = f(g(x, y)) = f(y - x) = \cos(y - x) \). This confirms that \( h(x, y) \) has been decomposed into \( f(u) \) and \( g(x, y) \).

Step 5: Continuity Investigation

For part (b), we need to determine where \( h(x, y) = \cos(y - x) \) is continuous. Since both \( f(u) = \cos(u) \) and \( g(x, y) = y - x \) are continuous for all real numbers, it follows that the composition \( f(g(x, y)) = \cos(y - x) \) is continuous for all \((x, y)\) values in \( \mathbb{R}^2 \).

Key Concepts

Continuity of Functions

The concept of continuity in functions is fundamental in mathematics and engineering. A function is said to be continuous if you can draw it without lifting your pencil from the paper. In more precise terms, a function is continuous at a point if three conditions are met:

• The function is defined at the point.
• The limit as the variable approaches the point exists.
• The limit of the function as it approaches the point is equal to the function’s value at that point.

For the function \(h(x, y) = \cos(y - x)\), continuity means it operates smoothly without any interruptions over its domain. Since the cosine function \(f(u) = \cos(u)\) is continuous for all real numbers and the subtraction operation \(g(x, y) = y - x\) also has no discontinuities over \(\mathbb{R}^2\), combined they ensure that \(h(x, y)\) is continuous everywhere in \(\mathbb{R}^2\). This means whether we examine points near 0, or explore infinity, \(h(x, y)\) will smoothly attain values without abrupt jumps or undefined spots.

Inner Function

The "inner function" in function composition refers to the initial operation or transformation applied within the chain of functions. Think of it as the first step in the recipe before the outer function completes the dish. In the given problem, our inner function is defined as \(g(x, y) = y - x\). This function takes two parameters, \(x\) and \(y\), and outputs their difference. This step is crucial as it sets the stage for further transformations by the outer function. Despite its simplicity, the inner function processes the input values allowing for an intermediate result which can then be transformed by other operations. In our expression, the subtraction of \(x\) from \(y\) within \(g(x, y)\) essentially adjusts the phase input for the cosine function to handle, ensuring future transformations maintain the relationship between \(x\) and \(y\).

Outer Function

The outer function in a composition framework is the final layer that applies a transformation to the result provided by the inner function. In our scenario, the outer function is \(f(u) = \cos(u)\). It takes the result of the inner function \(g(x, y) = y - x\) and applies the cosine operation to it.The purpose of the outer function is to process the output from the inner stage and transform it into the final desired form. Here, the role of \(f(u)\) highlights the whole aim of composition: to build complex functions from simpler ones and achieve desired transformations. By implementing the cosine operation after determining \(y - x\) with \(g(x, y)\), the outer function leverages trigonometric properties like periodicity and boundedness, ensuring that \(h(x, y)\) provides a continuous and oscillating output that fits smoothly within the real number system.