Problem 1 - Calculus Early Transcendentals: Pearson New International Edition

Question

Let \(f(x, y)=x^{2} y+\sqrt{y}\). Find each value. (a) \(f(2,1)\) (b) \(f(3,0)\) (c) \(f(1,4)\) (d) \(f\left(a, a^{4}\right)\) (e) \(f\left(1 / x, x^{4}\right)\) (f) \(f(2,-4)\) What is the natural domain for this function?

Step-by-Step Solution

Answer

The values are (a) 5, (b) 0, (c) 6, (d) \(a^6 + a^2\), (e) \(2x^2\), (f) undefined. The natural domain is \(y \geq 0\).

Step 1: Evaluate f(2,1)

Plug in the values 2 for \(x\) and 1 for \(y\) into the function. The function becomes: \[ f(2, 1) = (2)^2 \cdot 1 + \sqrt{1} = 4 + 1 = 5 \]

Step 2: Evaluate f(3,0)

Plug in the values 3 for \(x\) and 0 for \(y\) into the function. The function becomes: \[ f(3, 0) = (3)^2 \cdot 0 + \sqrt{0} = 0 + 0 = 0 \]

Step 3: Evaluate f(1,4)

Plug in the values 1 for \(x\) and 4 for \(y\) into the function. The function becomes: \[ f(1, 4) = (1)^2 \cdot 4 + \sqrt{4} = 4 + 2 = 6 \]

Step 4: Evaluate f(a, a^4)

Substitute \(a\) for \(x\) and \(a^4\) for \(y\). The function becomes: \[ f(a, a^4) = a^2 \cdot a^4 + \sqrt{a^4} = a^6 + a^2 \]

Step 5: Evaluate f(1/x, x^4)

Substitute \(1/x\) for \(x\) and \(x^4\) for \(y\). The function becomes: \[ f\left(\frac{1}{x}, x^4\right) = \left(\frac{1}{x}\right)^2 \cdot x^4 + \sqrt{x^4} = \frac{x^4}{x^2} + x^2 = x^2 + x^2 = 2x^2 \]

Step 6: Evaluate f(2,-4)

Plug in the values 2 for \(x\) and -4 for \(y\) into the function. The function becomes: \[ f(2, -4) = (2)^2 \cdot (-4) + \sqrt{-4} \] However, the square root of a negative number is not defined in the set of real numbers, so \(f(2, -4)\) is undefined for real numbers.

Step 7: Determine the Natural Domain

The natural domain of \(f(x, y) = x^2 y + \sqrt{y}\) is dependent on avoiding negative numbers under the square root function. Thus, \(y\) must be non-negative. The natural domain is defined as \[ \{ (x, y) \mid y \geq 0 \} \] because the square root function requires non-negative values.

Key Concepts

Function Evaluation

Function evaluation is the process of finding the value of a function for specific input values given for its variables. To evaluate a multivariable function like \( f(x, y) = x^2 y + \sqrt{y} \), you need to plug in the values for \( x \) and \( y \) into the function's formula. This involves substituting each variable with the given numbers and performing the arithmetic operations that follow.
For example, if we want to evaluate \( f(2,1) \), we plug 2 for \( x \) and 1 for \( y \) which yields:

• Substitute: \( (2)^2 \cdot 1 + \sqrt{1} = 4 + 1 = 5 \)

This step-by-step substitution allows you to compute the output of the function for any given pair of inputs \( (x, y) \). Remember that correct substitution and order of operations are crucial to accurately evaluating any function.

Natural Domain

The natural domain of a function refers to all possible input values for which the function is defined and gives real number outputs. It is important to consider any restrictions in the operation of the function. In the case of \( f(x, y) = x^2 y + \sqrt{y} \), the presence of the square root function imposes a restriction.
Square roots of negative numbers are undefined in the real number system, which means for the function to output real numbers, \( y \) must be non-negative. Thus, the natural domain of the function is:

• All pairs \( (x, y) \) such that \( y \geq 0 \)

This simply means the function can take any \( x \) but \( y \) must be zero or a positive number to satisfy the conditions of the square root operation.

Square Root Function

The square root function is one that takes a number and produces another number whose square is the original number. Mathematically, for any non-negative number \( a \), \( \sqrt{a} \) is a number such that \( b^2 = a \). The condition of non-negativity is pivotal when working with real numbers.
In our function \( f(x, y) = x^2 y + \sqrt{y} \), the term \( \sqrt{y} \) enforces \( y \geq 0 \) to ensure that the square root function remains defined within the real numbers.

• Cannot compute \( \sqrt{y} \) if \( y < 0 \) in real number arithmetic.
• The function is continuous and smooth for any \( y \geq 0 \).

Understanding the square root component is crucial in identifying the natural domain and evaluating any numeric examples involving the square root operation.