Math Problem Statement
At what points of R Superscript 2 is the following function continuous? f left parenthesis x comma y right parenthesis equals left brace Start 2 By 2 Matrix 1st Row 1st Column StartFraction xy Over 9 x squared plus 4 y squared EndFraction 2nd Column if left parenthesis x comma y right parenthesis not equals left parenthesis 0 comma 0 right parenthesis 2nd Row 1st Column 0 2nd Column if left parenthesis x comma y right parenthesis equals left parenthesis 0 comma 0 right parenthesis EndMatrix
Question content area bottom Part 1 Select the correct choice and, if necessary, fill in the answer box to complete your choice. A. All points except the point(s) enter your response here (Type an ordered pair. Simplify your answer. Use a comma to separate answers as needed.) B. All points except the points that are above yequals enter your response here (Simplify your answer.) C. All points except the points that are on the line(s) yequals enter your response here (Simplify your answer. Use a comma to separate answers as needed.) D. All points in R Superscript 2 E. All points except the points that are below yequals enter your response here (Simplify your answer.)
Solution
We are given the following function:
\begin{cases} \frac{xy}{9x^2 + 4y^2} & \text{if } (x, y) \neq (0, 0) \\ 0 & \text{if } (x, y) = (0, 0) \end{cases}$$ ### Step 1: Understand Continuity The function is continuous at points $$(x, y) \neq (0, 0)$$ because the rational expression $$\frac{xy}{9x^2 + 4y^2}$$ is well-defined and continuous everywhere except possibly at $$(0, 0)$$. We need to determine if the function is continuous at the origin, i.e., at $$(0, 0)$$. ### Step 2: Test Continuity at $$(0, 0)$$ To check the continuity of $$f$$ at $$(0, 0)$$, we need to check if the limit of $$f(x, y)$$ as $$(x, y) \to (0, 0)$$ exists and equals $$f(0, 0)$$. We know that: $$f(0, 0) = 0$$ Thus, we want to see if: $$\lim_{(x, y) \to (0, 0)} \frac{xy}{9x^2 + 4y^2} = 0$$ We will approach the origin along different paths and see if the limit is the same. #### Path 1: Along the x-axis ($$y = 0$$) When $$y = 0$$, the function becomes: $$f(x, 0) = \frac{x \cdot 0}{9x^2 + 4(0)^2} = 0$$ So, along the x-axis, the limit is $$0$$. #### Path 2: Along the y-axis ($$x = 0$$) When $$x = 0$$, the function becomes: $$f(0, y) = \frac{0 \cdot y}{9(0)^2 + 4y^2} = 0$$ So, along the y-axis, the limit is also $$0$$. #### Path 3: Along the line $$y = x$$ When $$y = x$$, the function becomes: $$f(x, x) = \frac{x \cdot x}{9x^2 + 4x^2} = \frac{x^2}{13x^2} = \frac{1}{13}$$ Along this path, the limit is $$\frac{1}{13}$$, which is **not** equal to 0. ### Step 3: Conclusion Since the limits along different paths yield different values (0 along the axes and $$\frac{1}{13}$$ along the line $$y = x$$), the limit as $$(x, y) \to (0, 0)$$ does not exist. Therefore, the function is **not continuous** at $$(0, 0)$$. ### Step 4: Final Answer The function is continuous at all points in $$\mathbb{R}^2$$ except at the point $$(0, 0)$$. So, the correct choice is: **A. All points except the point(s) $$(0, 0)$$**. --- Would you like further clarification or additional examples related to continuity? Here are 5 questions to expand on this topic: 1. What is the definition of continuity at a point in terms of limits? 2. How can we check the continuity of a function using polar coordinates? 3. What is the difference between continuity of a function in a single variable and in two variables? 4. Can you think of another example where a function is not continuous at a single point? 5. How does the path-dependent limit show that a function is not continuous at a point? **Tip:** When checking continuity at a point for functions of two variables, always test the limit along multiple paths (e.g., lines, curves) to ensure the limit is the same from all directions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Continuity
Functions of Two Variables
Limits
Formulas
f(x, y) = (xy / (9x² + 4y²))
f(0, 0) = 0
Theorems
Limit of a function in multiple variables
Continuity of functions of two variables
Suitable Grade Level
Grades 11-12 (Advanced Calculus/Multivariable Calculus)
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