Problem 2
Question
Let \(f(x, y)=y / x+x y\). Find each value. (a) \(f(1,2)\) (b) \(f\left(\frac{1}{4}, 4\right)\) (c) \(f\left(4, \frac{1}{4}\right)\) (d) \(f(a, a)\) (e) \(f\left(1 / x, x^{2}\right)\) (f) \(f(0,0)\) What is the natural domain for this function?
Step-by-Step Solution
Verified Answer
(a) 4; (b) 17; (c) 17/16; (d) 1 + a²; (e) x³ + x; (f) Undefined; domain excludes x = 0.
1Step 1: Evaluate f(1,2)
To find \( f(1, 2) \), substitute \( x = 1 \) and \( y = 2 \) into the function: \[ f(1, 2) = \frac{2}{1} + 1 \cdot 2. \] Simplifying, we get \( f(1, 2) = 2 + 2 = 4 \).
2Step 2: Evaluate f(1/4, 4)
Substitute \( x = 1/4 \) and \( y = 4 \) into the function: \[ f\left(\frac{1}{4}, 4\right) = \frac{4}{1/4} + \frac{1}{4} \cdot 4. \] Simplify the expression: \( \frac{4}{1/4} = 16 \) and \( \frac{1}{4} \cdot 4 = 1 \). Thus, \( f\left(\frac{1}{4}, 4\right) = 16 + 1 = 17 \).
3Step 3: Evaluate f(4, 1/4)
Substitute \( x = 4 \) and \( y = 1/4 \) into the function: \[ f\left(4, \frac{1}{4}\right) = \frac{1/4}{4} + 4 \cdot \frac{1}{4}. \] Simplify the expression: \( \frac{1}{4} \cdot \frac{1}{4} = \frac{1}{16} \) and \( 4 \cdot \frac{1}{4} = 1 \). Thus, \( f\left(4, \frac{1}{4}\right) = \frac{1}{16} + 1 = \frac{17}{16} \).
4Step 4: Evaluate f(a, a)
Substitute \( x = a \) and \( y = a \) into the function: \[ f(a, a) = \frac{a}{a} + a \cdot a. \] Simplify: \( \frac{a}{a} = 1 \) (provided \( a eq 0 \)) and \( a \cdot a = a^2 \). Thus, \( f(a, a) = 1 + a^2 \).
5Step 5: Evaluate f(1/x, x²)
Substitute \( x = 1/x \) and \( y = x^2 \) into the function: \[ f\left(\frac{1}{x}, x^2\right) = \frac{x^2}{1/x} + \frac{1}{x} \cdot x^2. \] Simplify: \( \frac{x^2}{1/x} = x^3 \) and \( \frac{1}{x} \cdot x^2 = x \). Thus, \( f\left(\frac{1}{x}, x^2\right) = x^3 + x \).
6Step 6: Evaluate f(0,0) and domain determination
Attempt to substitute \( x = 0 \) and \( y = 0 \) into the function: \[ f(0, 0) = \frac{0}{0} + 0 \cdot 0. \] The term \( \frac{0}{0} \) is undefined, so \( f(0, 0) \) is undefined. The natural domain of \( f(x, y) \), considering \( \frac{y}{x} \) must be defined, excludes \( x = 0 \). Therefore, the natural domain is all \( (x, y) \in \mathbb{R}^2 \) except \( x = 0 \).
Key Concepts
Natural DomainUndefined ExpressionSimplificationSubstitution in Functions
Natural Domain
In mathematics, the concept of a natural domain refers to the set of all input values (in this case, (x, y) pairs) for which the function is properly defined and yields a real number. For the function given, \( f(x, y) = \frac{y}{x} + x \, y \), we must be cautious about where each component of the expression can exist.
Importantly, division by zero is undefined, which directly affects our domain. This means we cannot allow \( x \) to be zero, since \( \frac{y}{x} \) would become \( \frac{y}{0} \), an undefined expression.
Thus, the natural domain includes all real numbers for \( (x, y) \) except where \( x = 0 \). In formal terms, this can be expressed as "all \( \mathbb{R}^2 \) except \( x = 0 \)." Understanding the natural domain is crucial, as it guides us to valid inputs that ensure the mathematical operations within the function remain defined and computable.
Importantly, division by zero is undefined, which directly affects our domain. This means we cannot allow \( x \) to be zero, since \( \frac{y}{x} \) would become \( \frac{y}{0} \), an undefined expression.
Thus, the natural domain includes all real numbers for \( (x, y) \) except where \( x = 0 \). In formal terms, this can be expressed as "all \( \mathbb{R}^2 \) except \( x = 0 \)." Understanding the natural domain is crucial, as it guides us to valid inputs that ensure the mathematical operations within the function remain defined and computable.
Undefined Expression
An undefined expression typically arises in mathematics when a situation violates one of the fundamental rules of arithmetic or algebra, such as division by zero mentioned above.
When evaluating the function at any given point, it’s essential to check if any terms lead to undefined situations. For instance, in this exercise scenario, when \( f(0,0) \) was evaluated, we encountered the term \( \frac{0}{0} \).
Mathematically, \( \frac{0}{0} \) is undefined because it does not yield a consistent value and represents an indeterminate form. In practical terms, avoiding such undefined expressions is vital, as they can lead to errors in calculations and interpretations of mathematical results.
By systematically checking for these conditions, the integrity of evaluating a function anywhere within its natural domain can be maintained.
When evaluating the function at any given point, it’s essential to check if any terms lead to undefined situations. For instance, in this exercise scenario, when \( f(0,0) \) was evaluated, we encountered the term \( \frac{0}{0} \).
Mathematically, \( \frac{0}{0} \) is undefined because it does not yield a consistent value and represents an indeterminate form. In practical terms, avoiding such undefined expressions is vital, as they can lead to errors in calculations and interpretations of mathematical results.
By systematically checking for these conditions, the integrity of evaluating a function anywhere within its natural domain can be maintained.
Simplification
Simplification in mathematics is the process of reducing an expression to its most compact and understandable form.
For a function like \( f(x, y) = \frac{y}{x} + x \, y \), simplification involves substituting specific values for \( x \) and \( y \) and conducting operations to combine terms and reduce fractions where possible.
Through the given solutions, the process of simplification was demonstrated for various function evaluations:
By mastering simplification, you gain the ability to tackle even more complex expressions with confidence, ensuring they are both manipulated accurately and interpreted clearly.
For a function like \( f(x, y) = \frac{y}{x} + x \, y \), simplification involves substituting specific values for \( x \) and \( y \) and conducting operations to combine terms and reduce fractions where possible.
Through the given solutions, the process of simplification was demonstrated for various function evaluations:
- For instance, \( f(1, 2) \) was simplified from \( \frac{2}{1} + 1 \cdot 2 \) to \( 4 \).
- Each time, fractions are resolved first, followed by multiplication. Finally, all components are added together.
By mastering simplification, you gain the ability to tackle even more complex expressions with confidence, ensuring they are both manipulated accurately and interpreted clearly.
Substitution in Functions
Substitution is a straightforward yet critical concept when evaluating functions. It is the process of replacing variables with given numbers or expressions. For the function \( f(x, y) \), substitution helps you compute specific values of the function by replacing \( x \) and \( y \) with actual numbers.
For example, to evaluate \( f(a, a) \), we replace both \( x \) and \( y \) with \( a \), giving us \( \frac{a}{a} + a^2 \). This yields the simplified result of \( 1 + a^2 \).
In the context of this function, it's evident that every substitution requires care, especially if substitutions result in undefined forms such as division by zero.
It's also interesting to notice how substitution in different function configurations such as \( f(1/x, x^2) \) impacts the resulting expression.
Practicing substitution not only helps simplify evaluations but also increases understanding of how varying inputs influence a function.
For example, to evaluate \( f(a, a) \), we replace both \( x \) and \( y \) with \( a \), giving us \( \frac{a}{a} + a^2 \). This yields the simplified result of \( 1 + a^2 \).
In the context of this function, it's evident that every substitution requires care, especially if substitutions result in undefined forms such as division by zero.
It's also interesting to notice how substitution in different function configurations such as \( f(1/x, x^2) \) impacts the resulting expression.
Practicing substitution not only helps simplify evaluations but also increases understanding of how varying inputs influence a function.
Other exercises in this chapter
Problem 2
In Problems 1-8, find the directional derivative of \(f\) at the point \(\mathbf{p}\) in the direction of \(\mathbf{a}\). \(f(x, y)=y^{2} \ln x ; \mathbf{p}=(1,
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In Problems 1-6, find dw/dt by using the Chain Rule. Express your final answer in terms of \(t\). $$ w=x^{2} y-y^{2} x ; x=\cos t, y=\sin t $$
View solution Problem 2
\(\lim _{(x, y) \rightarrow(-2,1)}\left(x y^{3}-x y+3 y^{2}\right)\)
View solution Problem 2
In Problems 1-16, find all first partial derivatives of each function. \(f(x, y)=\left(4 x-y^{2}\right)^{3 / 2}\)
View solution