The function \(f(x, y) = x^2 + y^2 + 2\) involves square terms of \(x\) and \(y\), which are defined for all real numbers. Therefore, the domain of \(f\) is all ordered pairs of real numbers: \[ \text{Domain} = \mathbb{R}^2 \]
Since \(x^2\) and \(y^2\) are both non-negative, the smallest value of \(x^2 + y^2\) is 0 (occurring when both \(x\) and \(y\) are 0). Thus, the minimum value of \(f(x, y)\) is 2. There is no upper limit to the values of \(x^2 + y^2\), so: \[ \text{Range} = [2, \infty) \]
The function \(f(x, y) = x + 2y\) is a linear function of \(x\) and \(y\), and there are no restrictions on the values of \(x\) and \(y\). Hence: \[ \text{Domain} = \mathbb{R}^2 \]
As a linear function, \(f(x, y)\) can take any real number depending on the values of \(x\) and \(y\). Therefore, \[ \text{Range} = \mathbb{R} \]
The function \(f(x, y) = \frac{1}{x + 2y}\) is defined as long as the denominator is not zero. Therefore, the domain excludes points where \(x + 2y = 0\). Solving for \(y\) gives \(y = -\frac{x}{2}\), which is a line in the xy-plane. Thus, the domain of \(f\) is: \[ \text{Domain} = \mathbb{R}^2 \setminus \{(x, y) \mid y = -\frac{x}{2}\} \]
Since \(f(x, y)\) can approach any non-zero real number as \(x + 2y\) approaches any non-zero value, and it can approach both positive and negative infinity depending on whether \(x + 2y\) approaches zero from positive or negative directions respectively, the range of \(f\) is all real numbers except for zero: \[ \text{Range} = \mathbb{R} \setminus \{0\} \]
We have analyzed the domain and range for three multi-variable functions:
\(f(x, y) = x^2 + y^2 + 2\) with domain \(\mathbb{R}^2\) and range \([2, \infty)\).
\(f(x, y) = x + 2y\) with domain and range both being \(\mathbb{R}^2\) and \(\mathbb{R}\) respectively.
\(f(x, y) = \frac{1}{x + 2y}\) with domain \(\mathbb{R}^2 \setminus \{(x, y) \mid y = -\frac{x}{2}\}\) and range \(\mathbb{R} \setminus \{0\}\).