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

(a) 5, (b) 0, (c) 6, (d) \( a^6 + a^2 \), (e) \( x^2 + |x^2| \), (f) Undefined; domain: \( y \geq 0 \).

Step 1: Understanding the Function

The function is \( f(x, y) = x^2 y + \sqrt{y} \). This function involves two operations: multiplication (\( x^2 y \)) and a square root (\( \sqrt{y} \)). We need to carefully evaluate this for different values of \(x\) and \(y\). Remember that the square root is only defined for non-negative values of \(y\).

Step 2: Evaluating f(2,1)

Substitute \( x = 2 \) and \( y = 1 \) into the function: \[ f(2, 1) = (2^2)(1) + \sqrt{1} = 4 + 1 = 5. \] Thus, \( f(2, 1) = 5. \)

Step 3: Evaluating f(3,0)

Substitute \( x = 3 \) and \( y = 0 \) into the function: \[ f(3, 0) = (3^2)(0) + \sqrt{0} = 0 + 0 = 0. \] Thus, \( f(3, 0) = 0. \)

Step 4: Evaluating f(1,4)

Substitute \( x = 1 \) and \( y = 4 \) into the function: \[ f(1, 4) = (1^2)(4) + \sqrt{4} = 4 + 2 = 6. \] Thus, \( f(1, 4) = 6. \)

Step 5: Evaluating f(a, a^4)

Substitute \( x = a \) and \( y = a^4 \) into the function: \[ f(a, a^4) = (a^2)(a^4) + \sqrt{a^4} = a^{6} + a^2. \] Thus, \( f(a, a^4) = a^6 + a^2. \)

Step 6: Evaluating f(1/x, x^4)

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

Step 7: Evaluating f(2,-4)

Substitute \( x = 2 \) and \( y = -4 \) into the function: The term \( \sqrt{-4} \) is not defined in the real number system, hence \( f(2, -4) \) is undefined. The square root function requires \( y \geq 0 \).

Step 8: Finding the Natural Domain

Since \( \sqrt{y} \) is a part of the function, \( y \) must be non-negative. Thus, the natural domain of \( f \) is all real numbers for \( x \) with \( y \geq 0 \).

Key Concepts

Natural Domain

When dealing with multivariable functions, the natural domain is crucial because it tells us what input values are valid for the function. The function \( f(x, y) = x^2 y + \sqrt{y} \) involves both polynomial and radical expressions. Particularly, the square root \( \sqrt{y} \) brings an important limitation: the expression under the square root must be zero or positive, i.e., \( y \geq 0 \). This restriction ensures that the square root is defined within the realm of real numbers.

Therefore, the natural domain is the set of all \( (x, y) \) pairs where \( x \) can be any real number but \( y \) must satisfy \( y \geq 0 \). This understanding helps us focus on only viable values when evaluating the function in different scenarios.

Function Evaluation

Evaluating a function involves substituting values into the function expression and simplifying to find the result. Let's take \( f(x, y) = x^2 y + \sqrt{y} \) as an example. When asked to evaluate \( f(2,1) \), you would substitute \( x = 2 \) and \( y = 1 \):

• Calculate the first term: \( (2^2) \cdot 1 = 4 \)
• Calculate the second term: \( \sqrt{1} = 1 \)
• Add them together: \( 4 + 1 = 5 \)

So, \( f(2, 1) = 5 \). Repeating this process for different input values will provide specific outputs. In essence, function evaluation helps us understand how the function behaves under various circumstances by replacing variables with concrete numbers.

Square Root Restrictions

In mathematics, the square root of a non-negative number \( y \) is a number \( q \) such that \( q^2 = y \). However, issues arise when \( y \) is negative because there is no real number squared that will result in a negative number. Thus, the restriction \( y \geq 0 \) is necessary for the square root function.In our function \( f(x, y) = x^2 y + \sqrt{y} \), when you evaluate expressions like \( \sqrt{y} \), you must ensure that \( y \geq 0 \). For example, trying to evaluate \( f(2,-4) \) leads to an undefined result because \( \sqrt{-4} \) does not produce a real number. Always check the value of \( y \) to ensure it meets the requirement before substituting it into the function.