Problem 6 - The Calculus with Analytic Geometry

Question

In Exercises 3 through 11 , find the domain and range of the function \(f\) and draw a sketch showing as a shaded region in \(R^{2}\) the set of points in the domain of \(f\). $$ f(x, y)=\sqrt{\frac{x-y}{x+y}} $$

Step-by-Step Solution

Answer

Domain: \((x, y)\ \text{such that}\ x \geq y \ \text{and} \ x + y \eq 0 \). Range: \( [0, \infty) \).

Step 1: Identify the Function

The function given is \( f(x, y) = \sqrt{\frac{x-y}{x+y}} \).

Step 2: Determine the Validity for the Square Root

The expression inside the square root, \( \frac{x-y}{x+y} \), must be non-negative. Hence, \( \frac{x-y}{x+y} \geq 0 \).

Step 3: Solve the Inequality

To solve \( \frac{x-y}{x+y} \geq 0 \), consider the numerator and denominator separately: - Numerator: \(x - y \geq 0 \) implies \(x \geq y \) - Denominator: \(x + y eq 0 \) because it makes the fraction undefined.Combining these, the inequality is satisfied when \(x \geq y \) and \(x + y \eq 0 \).

Step 4: Domain of the Function

The domain of \( f \) is all pairs \((x, y)\) such that \( x \geq y \) and \( x + y \eq 0 \).

Step 5: Determine the Range

Since the square root function only produces non-negative values: The range of \( f \) is all non-negative real numbers, so \( [0, \infty) \).

Step 6: Sketch the Domain

Draw the line \(x = y\). The region to the right (including the line) is \(x \geq y\). Exclude the line \(x + y = 0\) (or \(x = -y\)). The shaded region represents the domain.

Key Concepts

domain and range

Let's dive into understanding domain and range in the context of this exercise. The function given is \(f(x, y) = \sqrt{\frac{x-y}{x+y}}\). The domain of a function consists of all possible input values (in this case, pairs of \(x\) and \(y\)) that do not cause the function to become undefined or invalid. Since we have a square root, its argument must be non-negative. Therefore, we need to ensure \(\frac{x-y}{x+y} \geq 0\).

To ensure the denominator \(x + y\) doesn't equal zero, we must exclude points where \(x + y = 0\). The domain is where \(x \geq y\) and \(x + y \eq 0\). For the range, consider that the square root function only gives non-negative outputs. Hence, the range of \(f(x, y)\) is \([0, \infty)\).

inequalities

To solve the inequality \(\frac{x-y}{x+y} \geq 0\), we need to consider both the numerator and denominator separately.

• Numerator \(x - y \geq 0\) implies \(x \geq y\).
  
• Denominator \(x + y \eq 0\), to avoid undefined values. This means \(x \eq -y\).
  

Combining these, the inequality satisfies where \(x \geq y\) and \(x + y \eq 0\). Thus, the domain must account for both conditions, ensuring the inequality holds and the function remains defined.

sketching regions

Sketching the domain of \(f(x, y)\) helps visualize where the function is defined. Begin by drawing the line \(x = y\). The area on the plane where \(x \geq y\) lies to the right of this line. However, we must also exclude the line \(x = -y\) since \(x + y = 0\) is undefined for our function.

The final shaded region will be everything to the right of \(x = y\), excluding the line \(x = -y\). This graphical approach can significantly aid in understanding where the function \(f(x, y)\) is valid.