Domain and Range of Multi-variable Functions
In the exercises below, we determine the domain and range of each
multi-variable function. Each function’s domain and range were
determined based on their algebraic properties and the behavior of
functions like squares, square roots, and trigonometric functions.
Problem 7: \(f(x, y) = x^2 + y^2 +
2\)
- Domain: The domain of \(f(x, y)\) is all pairs \((x, y)\) in \(\mathbb{R}^2\) because there are no
restrictions on \(x\) and \(y\) for the function \(x^2 + y^2 + 2\).
- Range: The minimum value is when \(x = 0\) and \(y =
0\), so the minimum of \(f\) is
\(2\). Since \(x^2 + y^2\) can take any non-negative
value, the range is \([2,
\infty)\).
Problem 8: \(f(x, y) = x +
2y\)
- Domain: The domain is \(\mathbb{R}^2\), as there are no
restrictions on the values that \(x\)
and \(y\) can take.
- Range: The range is \(\mathbb{R}\), because \(x + 2y\) can produce any real number
depending on the values of \(x\) and
\(y\).
Problem 9: \(f(x, y) = x -
2y\)
- Domain: The domain is \(\mathbb{R}^2\).
- Range: The range is \(\mathbb{R}\).
Problem 10: \(f(x, y) = \frac{1}{x
+ 2y}\)
- Domain: The domain excludes points where the
denominator is zero. Therefore, \(x + 2y \neq
0\).
- Range: The range is \(\mathbb{R} \setminus \{0\}\), since the
function never evaluates to zero but can approach any other real
number.
Problem 11: \(f(x, y) =
\frac{1}{x^2 + y^2 + 1}\)
- Domain: The domain is all of \(\mathbb{R}^2\), as the denominator \(x^2 + y^2 + 1\) is always positive.
- Range: The range is \((0,
1]\), as the denominator is at least 1, making the function
positive and never exceeding 1.
Problem 12: \(f(x, y) = \sin(x)
\cos(y)\)
- Domain: The domain is \(\mathbb{R}^2\).
- Range: The range is \([-1, 1]\), as the product of two functions
each ranging from -1 to 1.
Problem 13: \(f(x, y) = \sqrt{9 -
x^2 - y^2}\)
- Domain: The domain is the disk \(x^2 + y^2 \leq 9\), including all points
within and on the boundary of a circle with radius 3 centered at the
origin.
- Range: The range is \([0,
3]\), as the function represents the z-coordinate in a
hemisphere.
Problem 14: \(f(x, y) =
\frac{1}{\sqrt{x^2 + y^2 - 9}}\)
- Domain: The domain excludes the disk \(x^2 + y^2 \leq 9\) to avoid non-real and
division by zero errors, so \(x^2 + y^2 >
9\).
- Range: The range is \((0,
\infty)\), as the denominator can approach zero from the positive
side making the function increase towards infinity but never reaching
zero.