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At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$h(x, y)=\frac{\sqrt{x-y}}{4}$$

Short Answer

Expert verified
Question: Identify the points in the domain \(\mathbb{R}^{2}\) where the function \(h(x, y) = \frac{\sqrt{x-y}}{4}\) is continuous. Answer: The function is continuous at all points \((x, y)\) in \(\mathbb{R}^{2}\) such that \(x \geq y\).

Step by step solution

01

Identify problematic points for continuity

To do this, we start by finding where the function is undefined. The function is only undefined when the value within the square root is negative. We have: $$x - y < 0$$
02

Solve inequality for x and y

Now, we must solve the inequality for \(x\) and \(y\), which can be rearranged as: $$x < y$$ This inequality represents that the function is undefined if \(x\) is less than \(y\). It means that the points in the \(xy\)-plane above the line \(x=y\) are the problematic points for the continuity of the given function.
03

Determine the points for which the function is continuous

For the function to be continuous in the \(xy\)-plane (in \(\mathbb{R}^2\)), it must be true that the square root is never negative. This happens if: $$x - y \geq 0$$ By rearranging the inequality, we have: $$x \geq y$$ This inequality means that the function is continuous everywhere in the \(xy\)-plane below or on the line \(x=y\) (including the points on this line).
04

Write down the answer

In conclusion, the function \(h(x, y) = \frac{\sqrt{x-y}}{4}\) is continuous at all points \((x, y)\) in \(\mathbb{R}^{2}\) such that \(x \geq y\).

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