Problem 4 - Calculus Early Transcendentals

Question

What is the domain of \(h(x, y)=\sqrt{x-y} ?\)

Step-by-Step Solution

Answer

Question: Determine the domain of the function \(h(x, y) =\sqrt{x-y}\). Answer: The domain of the function \(h(x, y)=\sqrt{x-y}\) is all points (x, y) in the real plane where x is greater than or equal to y, which can be represented as \(\text{Domain}(h) = \{(x, y) \in \mathbb{R}^2 ~|~ x \ge y\}\).

Step 1: Identify the restriction

For a square root function, the expression inside the square root must be greater than or equal to 0. For our given function, \(h(x, y) = \sqrt{x-y},\) we need to find when \(x-y \ge 0.\)

Step 2: Solve the inequality

To find when \(x-y \ge 0,\) we can simply rearrange the inequality: $$x - y \ge 0 \Rightarrow x \ge y$$

Step 3: Determine the domain

The domain of the function h(x, y) consists of all the points (x, y) for which x is greater than or equal to y. So, the domain of h(x, y) can be described as $$\text{Domain}(h) = \{(x, y) \in \mathbb{R}^2 ~|~ x \ge y\}$$ Thus, the domain of the function \(h(x, y) =\sqrt{x-y}\) is all points (x, y) where x is greater than or equal to y.