Math Problem Statement
Answer step by step
Determine if the function f(x) = { 3x + 2 for x ≠ 1, -17 for x = 1 } is continuous at x = 1. Is f(x) = { x + 1 if x ≤ 1, 2 if 1 < x ≤ 2, 1/(x - 3) if x > 2 } continuous at x = 1, x = 2, x = 3?
Solution
Let's solve the two parts of the problem one by one:
Problem 1:
Determine if the function is continuous at , where the function is given as:
\begin{cases} 3x + 2 & \text{for } x \neq 1 \\ -17 & \text{for } x = 1 \end{cases}$$ #### Step-by-step approach: 1. **Check the value of the function at $$x = 1$$:** - From the definition, we know that $$f(1) = -17$$. 2. **Find the limit of $$f(x)$$ as $$x \to 1$$:** - The function is defined as $$3x + 2$$ for all values except $$x = 1$$. Thus, the limit as $$x \to 1$$ is: $$\lim_{x \to 1} f(x) = \lim_{x \to 1} (3x + 2) = 3(1) + 2 = 5$$ 3. **Check the condition for continuity:** - A function is continuous at $$x = 1$$ if: $$\lim_{x \to 1} f(x) = f(1)$$ - In this case, the limit is $$5$$, but $$f(1) = -17$$. - Since these two values are not equal, **the function is not continuous at $$x = 1$$**. ### Problem 2: Determine if the piecewise function is continuous at $$x = 1$$, $$x = 2$$, and $$x = 3$$. The function is given as: $$f(x) = \begin{cases} x + 1 & \text{if } x \leq 1 \\ 2 & \text{if } 1 < x \leq 2 \\ \frac{1}{x-3} & \text{if } x > 2 \end{cases}$$ #### Step-by-step approach: 1. **Continuity at $$x = 1$$:** - To be continuous at $$x = 1$$, the following must be true: $$\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = f(1)$$ - The value of the function at $$x = 1$$ is $$f(1) = 1 + 1 = 2$$ (from the first piece). - The left-hand limit as $$x \to 1^-$$ is $$\lim_{x \to 1^-} (x + 1) = 1 + 1 = 2$$. - The right-hand limit as $$x \to 1^+$$ is $$\lim_{x \to 1^+} 2 = 2$$. - Since the limits match and equal $$f(1)$$, **the function is continuous at $$x = 1$$**. 2. **Continuity at $$x = 2$$:** - To be continuous at $$x = 2$$, the following must be true: $$\lim_{x \to 2^-} f(x) = \lim_{x \to 2^+} f(x) = f(2)$$ - The value of the function at $$x = 2$$ is $$f(2) = 2$$ (from the second piece). - The left-hand limit as $$x \to 2^-$$ is $$\lim_{x \to 2^-} 2 = 2$$. - The right-hand limit as $$x \to 2^+$$ is $$\lim_{x \to 2^+} \frac{1}{x-3} = \frac{1}{2-3} = -1$$. - Since the right-hand limit and $$f(2)$$ do not match, **the function is not continuous at $$x = 2$$**. 3. **Continuity at $$x = 3$$:** - To be continuous at $$x = 3$$, the function must be defined at $$x = 3$$, and the limits from both sides must exist and match. - The function is not defined at $$x = 3$$, since the piece $$\frac{1}{x - 3}$$ is undefined at $$x = 3$$. - Therefore, **the function is not continuous at $$x = 3$$**. ### Final Answers: - For problem 1, the function is **not continuous at $$x = 1$$**. - For problem 2: - The function is **continuous at $$x = 1$$**. - The function is **not continuous at $$x = 2$$**. - The function is **not continuous at $$x = 3$$**. Would you like further details on any part of the solution? Here are some related questions: 1. What are the different types of discontinuities in functions? 2. How do you test for continuity at a point for piecewise functions? 3. What happens if the limit from the left does not equal the limit from the right? 4. Can you describe a function that is continuous at a point but has a jump discontinuity elsewhere? 5. How can we find the limits of rational functions as $$x$$ approaches a value? **Tip**: To test for continuity at a point, always ensure the limit from both sides matches the function value at that point!Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Continuity
Piecewise Functions
Limits
Formulas
Limit of a function: lim_{x→a} f(x)
Continuity condition: lim_{x→a} f(x) = f(a)
Theorems
A function is continuous at a point if the left-hand limit, right-hand limit, and the function value at that point are equal.
Suitable Grade Level
Grades 11-12