Problem 20
Question
Describe the largest set \(S\) on which it is correct to say that \(f\) is continuous. \(f(x, y)=\frac{1}{\sqrt{1+x+y}}\)
Step-by-Step Solution
Verified Answer
The largest set \(S\) is \(\{(x, y) \mid x + y > -1\}\).
1Step 1: Determine Domain of the Square Root
Since the function involves a square root in the denominator, we need to ensure that the expression under the square root is greater than zero to avoid division by zero and complex numbers. Thus, we need the inequality \(1 + x + y > 0\).
2Step 2: Solve the Inequality for \(x\) and \(y\)
The inequality \(1 + x + y > 0\) can be rewritten as \(x + y > -1\). This describes the half-plane in the Cartesian coordinate system above the line \(x + y = -1\).
3Step 3: Identify the Continuity Requirement
For \(f(x, y)\) to be continuous, it must be defined over its entire domain. The expression within the square root does not allow equality \(=0\) as it would make the denominator zero, thus, \(x + y\) must be strictly greater than \(-1\).
4Step 4: Describe the Largest Set \(S\)
The largest set \(S\) on which the function is continuous is all points \((x, y)\) in the plane such that \(x + y > -1\). This does not include the boundary line \(x + y = -1\).
Key Concepts
Domain of Multivariable FunctionsInequalities in CalculusContinuity on Open Sets
Domain of Multivariable Functions
When working with multivariable functions, determining the domain is a key step. The domain refers to all the possible input values for which the function is defined. In our example, the function is given by \( f(x, y) = \frac{1}{\sqrt{1 + x + y}} \). This involves a square root in the denominator, which brings special conditions: the argument inside the square root must be positive to ensure that the function outputs real numbers and to avoid division by zero.
To find the domain, we consider the inequality \(1 + x + y > 0\). By solving this inequality, we can describe the set of all \((x, y)\) pairs that make the function valid and continuous. In this instance, the solution to \(1 + x + y > 0\) is \(x + y > -1\).
To find the domain, we consider the inequality \(1 + x + y > 0\). By solving this inequality, we can describe the set of all \((x, y)\) pairs that make the function valid and continuous. In this instance, the solution to \(1 + x + y > 0\) is \(x + y > -1\).
- This inequality reveals a half-plane in the coordinate system.
- The set does not include the line where \(x + y = -1\).
Inequalities in Calculus
Inequalities help in understanding the regions where a function holds certain properties. In calculus, they are used to identify specific sets where functions are defined or continuous. For the function \( f(x, y) = \frac{1}{\sqrt{1 + x + y}} \), the inequality \(1 + x + y > 0\) determines where \(f\) is real-valued and defined.
Solving inequalities involves finding sets of points on a plane or space that satisfy the given conditions. Here, the inequality translates into the requirement \(x + y > -1\), excluding equal points like \(x + y = -1\).
Solving inequalities involves finding sets of points on a plane or space that satisfy the given conditions. Here, the inequality translates into the requirement \(x + y > -1\), excluding equal points like \(x + y = -1\).
- This inequality marks a boundary, a line on the plane where the function would not exist.
- Understanding these lines can help in graphing functions and predicting their behavior.
Continuity on Open Sets
The concept of continuity in calculus is foundational for understanding how functions behave across different intervals. A function is continuous if you can draw it without lifting your pencil. For multivariable functions like \( f(x, y) = \frac{1}{\sqrt{1 + x + y}} \), continuity requires the function to be defined over an open set in its domain.
In this scenario, the open set includes all points \((x, y)\) where \(x + y > -1\). It’s crucial to notice that this does not include the line \(x + y = -1\), as continuity cannot be maintained if the function approaches an undefined region (like where division by zero occurs).
In this scenario, the open set includes all points \((x, y)\) where \(x + y > -1\). It’s crucial to notice that this does not include the line \(x + y = -1\), as continuity cannot be maintained if the function approaches an undefined region (like where division by zero occurs).
- Open sets exclude boundary points, making functions smooth over these ranges.
- Continuity ensures the function doesn't jump or experience interruptions when moving across its domain.
Other exercises in this chapter
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