Math Problem Statement
Determine if the following piecewise function is continuous:
f(x) = { (x^2 - 9)/(x - 3), x ≠ 3;
9, x = 3 }.
Solution
The image contains a mathematical problem regarding the continuity of a piecewise function , defined as:
\frac{x^2 - 9}{x - 3}, & \text{for } x \neq 3 \\ 9, & \text{for } x = 3 \end{cases}$$ The problem asks whether this function is continuous or not. ### Step-by-step solution: 1. **Simplify the function for $$x \neq 3$$:** The expression $$\frac{x^2 - 9}{x - 3}$$ can be simplified using the difference of squares formula: $$x^2 - 9 = (x - 3)(x + 3)$$ So, $$\frac{x^2 - 9}{x - 3} = \frac{(x - 3)(x + 3)}{x - 3} = x + 3 \quad \text{for } x \neq 3.$$ Thus, for $$x \neq 3$$, $$f(x) = x + 3$$. 2. **Check the limit as $$x \to 3$$:** Since for $$x \neq 3$$, $$f(x) = x + 3$$, we can find the limit of $$f(x)$$ as $$x$$ approaches 3: $$\lim_{x \to 3} f(x) = \lim_{x \to 3} (x + 3) = 6.$$ 3. **Check the value of $$f(3)$$:** From the definition of the function, $$f(3) = 9$$. 4. **Determine continuity:** A function is continuous at $$x = 3$$ if: $$\lim_{x \to 3} f(x) = f(3).$$ In this case, the limit as $$x \to 3$$ is 6, but $$f(3) = 9$$. Since $$\lim_{x \to 3} f(x) \neq f(3)$$, the function is **not continuous** at $$x = 3$$. ### Conclusion: The function is not continuous at $$x = 3$$, so the correct answer is **False**. Let me know if you need further clarification, or if you have more questions! --- Here are five related questions for practice: 1. What does it mean for a function to be continuous at a point? 2. How do you check if a piecewise function is continuous at the boundary point? 3. Can a function be discontinuous at one point but continuous everywhere else? 4. What is the difference between a removable discontinuity and a jump discontinuity? 5. How can limits be used to determine continuity at a point? **Tip:** When simplifying rational functions, always check if the factor you are canceling makes the denominator zero, as it indicates a point of potential discontinuity.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Continuity
Limits
Algebra
Formulas
Difference of squares: a^2 - b^2 = (a - b)(a + b)
Limit formula for continuity
Theorems
Continuity of a function at a point
Limit properties
Suitable Grade Level
Grades 11-12 or Early University