Problem 25
Question
Give the domain and range of the functions of three variables. $$ f(x, y, z)=\sqrt{z-x^{2}+y^{2}} $$
Step-by-Step Solution
Verified Answer
Domain: \(z \geq x^2 - y^2\); Range: \([0, \infty)\).
1Step 1: Understand the Function
The function is given by \( f(x, y, z) = \sqrt{z - x^2 + y^2} \). It includes a square root, which affects the domain due to the necessity that the expression inside must be non-negative.
2Step 2: Determine the Domain Condition
For the expression under the square root to be valid, it must be non-negative. Thus, we need to find the condition where \( z - x^2 + y^2 \geq 0 \). This means \( z \geq x^2 - y^2 \).
3Step 3: Express the Domain
The domain of the function consists of all points \((x, y, z)\) such that \( z \geq x^2 - y^2 \). There are no further restrictions on \(x\) and \(y\), so they can be any real numbers. Therefore, the domain is the set of all \((x, y, z)\) where this inequality is satisfied.
4Step 4: Determine the Range of the Function
The range of a function is the set of all possible output values. Since the square root function produces non-negative values, the smallest value \(f(x, y, z)\) can take is 0 (when \(z = x^2 - y^2\)), and it increases without bound as \(z\) increases. Therefore, the range is \([0, \infty)\).
Key Concepts
DomainRangeSquare Root FunctionInequalities
Domain
When evaluating multivariable functions, the domain defines the set of all possible inputs for which the function is valid. In the given exercise, the function is \( f(x, y, z) = \sqrt{z - x^2 + y^2} \). A crucial feature of this function is the square root, which dictates certain conditions for its domain.
A square root function can only process non-negative numbers. Hence, the expression inside the square root, \( z - x^2 + y^2 \), must be greater than or equal to zero. This requirement translates to the inequality \( z \geq x^2 - y^2 \).
A square root function can only process non-negative numbers. Hence, the expression inside the square root, \( z - x^2 + y^2 \), must be greater than or equal to zero. This requirement translates to the inequality \( z \geq x^2 - y^2 \).
- The domain consists of all points \((x, y, z)\) where the inequality is satisfied.
- There are no restrictions on \(x\) and \(y\), meaning they can be any real number.
Range
The range of a function is about what outputs are possible from the inputs within the domain. For our function \( f(x, y, z) = \sqrt{z - x^2 + y^2} \), the output depends on the non-negative values created by the square root function.
As the square root of any real number is always non-negative, the smallest value our function can yield is 0. This happens when the expression inside the square root equals 0, specifically when \( z = x^2 - y^2 \).
As the square root of any real number is always non-negative, the smallest value our function can yield is 0. This happens when the expression inside the square root equals 0, specifically when \( z = x^2 - y^2 \).
- The range is all non-negative numbers, written as \([0, \infty)\).
- The function produces larger values as \(z\) grows larger than \(x^2 - y^2\).
Square Root Function
Square root functions are a foundational concept in both single-variable and multivariable calculus. They return only non-negative values (zero or positive), based on their characteristics.
For the given multivariable function \( f(x, y, z) = \sqrt{z - x^2 + y^2} \), the role of the square root is to ensure that the outputs remain non-negative if the expression inside it is valid. This makes it crucial to determine conditions like \( z \geq x^2 - y^2 \) for validity.
For the given multivariable function \( f(x, y, z) = \sqrt{z - x^2 + y^2} \), the role of the square root is to ensure that the outputs remain non-negative if the expression inside it is valid. This makes it crucial to determine conditions like \( z \geq x^2 - y^2 \) for validity.
- Square roots tackle non-negative real numbers, protecting against imaginary results.
- They impose restrictions prompting inequalities on input expressions, crucial for the domain.
Inequalities
Inequalities play a vital role in defining domains, especially in functions involving square roots. In mathematics, they encourage finding conditions under which expressions are valid. For \( f(x, y, z) = \sqrt{z - x^2 + y^2} \), we face the inequality \( z \geq x^2 - y^2 \).
Understanding and solving inequalities allows us to determine where a function makes sense. Solving \( z \geq x^2 - y^2 \) indicates that:
Understanding and solving inequalities allows us to determine where a function makes sense. Solving \( z \geq x^2 - y^2 \) indicates that:
- The values of \(z\) must not be less than \(x^2 - y^2\) for real results of the function.
- It identifies a surface of inputs dictating a threshold for valid calculations.
Other exercises in this chapter
Problem 25
In Exercises \(25-28,\) a function \(w=F(x, y, z),\) a vector \(\vec{v}\) and a point \(P\) are given. (a) Find \(\nabla F(x, y, z)\). (b) Find \(D_{\vec{u}} F\
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Find \(f_{x}, f_{y}, f_{x x}, f_{y y}, f_{x y}\) and \(f_{y x}\). $$ f(x, y)=\frac{\ln x}{4 y} $$
View solution Problem 26
In Exercises \(23-26,\) find \(\frac{d y}{d x}\) using Implicit Differentiation and Theorem \(12.5 .3 .\) $$ \ln \left(x^{2}+x y+y^{2}\right)=1 $$
View solution Problem 26
A function \(w=F(x, y, z),\) a vector \(\vec{v}\) and a point \(P\) are given. (a) Find \(\nabla F(x, y, z)\). (b) Find \(D_{\vec{u}} F\) at \(P,\) where \(\vec
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