Problem 11 - APEX Calculus

Question

Give the domain and range of the multivariable function. $$ f(x, y)=\frac{1}{x^{2}+y^{2}+1} $$

Step-by-Step Solution

Answer

Domain: \((- ablafty, + ablafty) \times (- ablafty, + ablafty)\); Range: \((0, 1]\).

Step 1: Understand the Function Type

The given function is a multivariable function, which means it takes more than one input—in this case, two inputs: \(x\) and \(y\). The function is defined as \( f(x, y) = \frac{1}{x^2 + y^2 + 1} \).

Step 2: Determine the Domain

To find the domain of the function, we need to identify the set of all possible input values \((x, y)\) for which the function is defined. The function \( f(x, y) = \frac{1}{x^2 + y^2 + 1} \) is a rational function where the denominator \(x^2 + y^2 + 1\) must not be zero. However, \(x^2 + y^2 + 1\) can never be zero because \(x^2 \geq 0\) and \(y^2 \geq 0\), so \(x^2 + y^2 + 1 \geq 1\). Therefore, the function is defined for all real numbers \(x\) and \(y\). Thus, the domain is all real numbers, \((-\infty, +\infty) \times (-\infty, +\infty)\).

Step 3: Determine the Range

To find the range of the function, we need to look at the output values that \( f(x, y) \) can take. The denominator \(x^2 + y^2 + 1\) has a minimum value of 1 and increases without bound. Thus, \( \frac{1}{x^2 + y^2 + 1}\) has a maximum value of \(\frac{1}{1} = 1\) and a minimum value approaching \(0\) as \(x^2 + y^2\) approaches infinity. Therefore, the range of \(f(x, y)\) is \((0, 1]\).

Key Concepts

Domain of a Function

The domain of a function refers to the set of all possible inputs that the function can accept. In the context of a multivariable function like \(f(x,y) = \frac{1}{x^2 + y^2 + 1}\), the domain concerns pairs of inputs \((x, y)\). For this specific function, we must ensure that the denominator \(x^2 + y^2 + 1\) cannot be zero, as division by zero is undefined in mathematics.
The key here is to analyze the expression \(x^2 + y^2 + 1\). Here, \(x^2\) and \(y^2\) are always non-negative (i.e., \(x^2 \geq 0\) and \(y^2 \geq 0\). As a result, \(x^2 + y^2\) is also non-negative. However, with the addition of 1, the entire expression \(x^2 + y^2 + 1\) will always be positive, and specifically, it is always greater than or equal to 1.

• Since \(x^2 + y^2 + 1\) is always greater than zero, it can never be zero.
• This means that \(f(x,y)\) is defined for all real numbers \(x\) and \(y\).
• Hence, the domain of the function is the entire two-dimensional real plane: \((-\infty, +\infty) \times (-\infty, +\infty)\).

Range of a Function

The range of a function represents the set of all possible outputs of the function. For the multivariable function \(f(x,y) = \frac{1}{x^2 + y^2 + 1}\), determining the range involves analyzing the values that \(f(x, y)\) can actually take when we vary \(x\) and \(y\) over their entire domain.
The denominator in this function, \(x^2 + y^2 + 1\), has a minimum value of 1 (when both \(x\) and \(y\) are zero) and increases without bound as either \(x\) or \(y\) (or both) increase. This means the expression \(\frac{1}{x^2 + y^2 + 1}\):

• Reaches its maximum value of 1 when \(x = 0\) and \(y = 0\).
• Approaches a minimum value of 0 as \(x^2 + y^2\) becomes increasingly large.

Therefore, the range of this function is the interval \((0, 1]\), which includes all real numbers greater than zero and up to 1 (with 1 included).

Rational Functions

A rational function is a type of function represented by the ratio of two polynomials. The numerator and the denominator both need to be polynomial expressions. For the function \(f(x, y) = \frac{1}{x^2 + y^2 + 1}\), the numerator is simply 1, which is a constant polynomial, and the denominator is a polynomial given by \(x^2 + y^2 + 1\).

This particular form emphasizes that the function is well-behaved unless the denominator is zero. However, as discussed:

• The denominator \(x^2 + y^2 + 1\) can never be zero.
• This makes the function well-defined over the entire set of real numbers \((x, y)\).
• Understanding that the denominator is never zero simplifies analyzing such functions significantly in multivariable contexts.

Multivariable Functions

Multivariable functions depend on more than one input variable. The function \(f(x,y) = \frac{1}{x^2 + y^2 + 1}\) is a clear example, as it uses two variables: \(x\) and \(y\). This function outputs a single value based on the combination of \(x\) and \(y\). Understanding these functions involves considering:

• How the combination of multiple inputs influences the output.
• Visualization, often requiring graphs in 3D space to understand fully.
• Handling more complex constraints such as identifying what combinations of parameters don’t work.

Multivariable functions are a step beyond simple one-variable functions, introducing students to more complex and real-world scenarios where outcomes depend on multiple factors. In this exercise, we see the introduction of new dimensions which reveal the richness and complexity of calculus beyond single-variable cases.