Math Problem Statement

Solution

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Problem 8:

Given the function f(x,y)f(x, y) defined as:

\frac{7x^2 + y^2}{x^2 + y^2} & \text{if } (x, y) \neq (0,0), \\ 0 & \text{if } (x, y) = (0,0), \end{cases}$$ Options: a) $$f(x, 0)$$ does not exist. b) $$f(0, y) = 7$$. c) $$f(0, 0)$$ does not exist. d) $$f(0, 0) = 0$$. To check the correctness, let's analyze the function. 1. For $$(x, y) \neq (0,0)$$: $$f(x, y) = \frac{7x^2 + y^2}{x^2 + y^2}$$ For specific values: - If $$y = 0$$: $$f(x, 0) = \frac{7x^2}{x^2} = 7$$ - If $$x = 0$$: $$f(0, y) = \frac{y^2}{y^2} = 1$$ 2. At $$(x, y) = (0,0)$$: The function is defined as: $$f(0,0) = 0$$ Now let's examine each option: - **Option a)** $$f(x, 0)$$ does not exist. - This is incorrect, $$f(x, 0) = 7$$ for all $$x \neq 0$$. - **Option b)** $$f(0, y) = 7$$. - This is incorrect, $$f(0, y) = 1$$ for all $$y \neq 0$$. - **Option c)** $$f(0, 0)$$ does not exist. - This is incorrect, $$f(0, 0) = 0$$. - **Option d)** $$f(0, 0) = 0$$. - This is correct. Therefore, the correct option for problem 8 is **d)**. ### Problem 9: Given the function $$f(x, y)$$ defined as: $$f(x, y) = \begin{cases} \frac{x^2 y - x^3 y^2}{x^2 + y^2} & \text{if } (x, y) \neq (0,0), \\ 0 & \text{if } (x, y) = (0,0), \end{cases}$$ Options: a) $$f(0,0) = 1$$ and $$f(0,0) = 2$$. b) $$f(0,0) = 0$$ and $$f(x,y) = 0$$. c) $$f(0,0) = 0$$ and $$f(x,0) = 0$$. d) $$f(0,0) = 0$$ and $$f(0,y) = 2$$. To check the correctness, let's analyze the function. 1. For $$(x, y) \neq (0,0)$$: $$f(x, y) = \frac{x^2 y - x^3 y^2}{x^2 + y^2}$$ For specific values: - If $$y = 0$$: $$f(x, 0) = \frac{x^2 \cdot 0 - x^3 \cdot 0^2}{x^2 + 0^2} = 0$$ - If $$x = 0$$: $$f(0, y) = \frac{0 \cdot y - 0 \cdot y^2}{0 + y^2} = 0$$ 2. At $$(x, y) = (0,0)$$: The function is defined as: $$f(0,0) = 0$$ Now let's examine each option: - **Option a)** $$f(0,0) = 1$$ and $$f(0,0) = 2$$. - This is incorrect, $$f(0,0) = 0$$. - **Option b)** $$f(0,0) = 0$$ and $$f(x,y) = 0$$. - This is incorrect, $$f(x,y) \neq 0$$ for all $$(x,y) \neq (0,0)$$. - **Option c)** $$f(0,0) = 0$$ and $$f(x,0) = 0$$. - This is correct, since we found $$f(x,0) = 0$$. - **Option d)** $$f(0,0) = 0$$ and $$f(0,y) = 2$$. - This is incorrect, $$f(0,y) = 0$$ for all $$y \neq 0$$. Therefore, the correct option for problem 9 is **c)**. ### Summary: - Problem 8: Correct answer is **d)**. - Problem 9: Correct answer is **c)**. Would you like more details on these solutions or have any other questions? Here are some related questions: 1. What is the limit definition of a function's continuity? 2. How can we evaluate the limit of a function with two variables? 3. What is the difference between pointwise and uniform convergence? 4. How can you check if a multivariable function is continuous at a point? 5. What are the different types of discontinuities? 6. How do we differentiate a function of several variables? 7. What are partial derivatives and how are they used? 8. How do we find the gradient of a function with several variables? **Tip**: Always check the definition of a piecewise function at the boundary points to ensure continuity.

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Math Problem Analysis

Mathematical Concepts

Piecewise Functions
Limits
Continuity

Formulas

Piecewise function definition
Limit definition

Theorems

Definition of continuity

Suitable Grade Level

High School