Problem 22
Question
Find the range of each function \(f(x, y)\), when defined on the specified domain \(D\). \(f(x, y)=\frac{x}{y} ; D=\\{(x, y): 0 \leq x \leq 1,1 \leq y \leq 2\\}\)
Step-by-Step Solution
Verified Answer
The range of the function is \([0, 1]\).
1Step 1: Identify the Function and Domain
The function is given as \(f(x, y) = \frac{x}{y}\) and the domain is \(D = \{(x, y) : 0 \leq x \leq 1, 1 \leq y \leq 2\}\). This means \(x\) can vary between 0 and 1, and \(y\) can vary between 1 and 2.
2Step 2: Examine the Function Behavior
Since \(f(x, y) = \frac{x}{y}\), the function depends on how both \(x\) and \(y\) vary. The maximum value for \(\frac{x}{y}\) occurs when \(x\) is maximized and \(y\) is minimized. Conversely, the minimum value occurs when \(x\) is minimized and \(y\) is maximized.
3Step 3: Determine Maximum Value
The function \(f\) reaches its maximum when \(x\) is maximum, i.e., 1, and \(y\) is minimum, i.e., 1. Therefore, the maximum value is \(f(1, 1) = \frac{1}{1} = 1\).
4Step 4: Determine Minimum Value
The function \(f\) reaches its minimum when \(x\) is minimum, i.e., 0, and \(y\) is maximum, i.e., 2. Therefore, the minimum value is \(f(0, 2) = \frac{0}{2} = 0\).
5Step 5: Conclude the Range
Since the function increases as \(x\) increases and decreases as \(y\) increases, with the calculations above, any intermediate values of \(x\) and \(y\) will yield values of \(f(x, y)\) between 0 and 1. Thus, the range of \(f(x, y)\) is \([0, 1]\).
Key Concepts
Function RangeTwo-Variable FunctionsDomain Analysis
Function Range
Understanding the range of a function is crucial in mathematical analysis. It tells us all possible values a function can take when applied to its domain. For the function \(f(x, y) = \frac{x}{y}\), we consider its behavior over the specified domain \(D = \{(x, y) : 0 \leq x \leq 1, 1 \leq y \leq 2\}\).
This means every possible value produced by \(f(x, y)\) as \((x,y)\) varies within the domain belongs to the closed interval \([0, 1]\). Understanding this concept helps visualize the implication of the function on its domain.
- The maximum value of \(f\) occurs when \(x\) is 1 and \(y\) is 1, giving us \(f(1, 1) = 1\).
- The minimum value of \(f\) is when \(x\) is 0 and \(y\) is 2, resulting in \(f(0, 2) = 0\).
This means every possible value produced by \(f(x, y)\) as \((x,y)\) varies within the domain belongs to the closed interval \([0, 1]\). Understanding this concept helps visualize the implication of the function on its domain.
Two-Variable Functions
Two-variable functions, like \(f(x, y) = \frac{x}{y}\), depend on two independent variables: \(x\) and \(y\). Unlike single-variable functions, they map ordered pairs of numbers into a single real number.
Such functions allow us to examine interactions between two changing parameters. For example, when both \(x\) and \(y\) vary, the value of \(f(x, y)\) reveals how variations in one variable affect the entire expression. Some useful properties of two-variable functions include:
Such functions allow us to examine interactions between two changing parameters. For example, when both \(x\) and \(y\) vary, the value of \(f(x, y)\) reveals how variations in one variable affect the entire expression. Some useful properties of two-variable functions include:
- They exhibit different behaviors depending on domain restrictions - for this function, \(x\) influences the numerator while \(y\) dictates the denominator.
- Their visualization often involves surfaces in three-dimensional space, providing insights into how different pairs \((x, y)\) yield distinct outcomes.
Domain Analysis
Analyzing a function's domain is a fundamental step. It helps in understanding where the function is defined and reveals applicable constraints. The domain of a function is essentially all ordered pairs \((x, y)\) for which the function produces a valid result. For \(f(x, y) = \frac{x}{y}\), the domain ensures that we do not have division by zero, so \(y\) must never be zero. In the specified domain \(D = \{(x, y): 0 \leq x \leq 1, 1 \leq y \leq 2\}\), \(y\) comfortably stays away from zero, ensuring valid outputs. Key points about domain analysis include:
- Identify constraints - here, \(x\) and \(y\)'s ranges enforce valid expressions throughout the specified domain.
- Recognize that each valid input \((x, y)\) leads to a unique value of \(f(x, y)\), here ranging from 0 to 1.
Other exercises in this chapter
Problem 22
Find the indicated partial derivatives. \(g(v, w)=\frac{w^{2}}{v+w} ; g_{v}(1,1)\)
View solution Problem 22
Use the definition of continuity to show that $$f(x, y)=\sqrt{9+x^{2}+y^{2}}$$ is continuous at \((0,0)\).
View solution Problem 22
Find the linearization of \(f(x, y)\) at the indicated point \(\left(x_{0}, y_{0}\right) .\) \(f(x, y)=e^{9 x+2 y} ;(1,2)\)
View solution Problem 22
Find the global maxima and minima of $$f(x, y)=2 x^{2}+y^{2}-6 y+3$$ on the disk $$ D=\left\\{(x, y): x^{2}+y^{2} \leq 16\right\\} $$
View solution