Math Problem Statement
Complete the following for f(x) = 3x if 0 <= x < 2 and f(x) = x + 3 if 2 <= x <= 4. (a) Determine the domain of f. (b) Evaluate f(-1), f(1), and f(5). (c) Graph f. (d) Is f continuous on its domain?
Solution
The question you're working on deals with a piecewise function defined as follows:
\begin{cases} 3x & \text{if } 0 \leq x < 2, \\ x + 3 & \text{if } 2 \leq x \leq 4. \end{cases}$$ Let's go through each part of the question. ### (a) **Determine the domain of $$f$$:** The function is defined for $$0 \leq x \leq 4$$, covering both intervals. Therefore, the domain of $$f$$ is: $$\boxed{[0, 4]}.$$ ### (b) **Evaluate \( f(-1), f(1), f(5):** - **$$f(-1)$$**: Since $$x = -1$$ is outside the domain $$[0, 4]$$, $$f(-1)$$ is **undefined**. - **$$f(1)$$**: For $$0 \leq 1 < 2$$, the rule is $$f(x) = 3x$$. So, $$f(1) = 3 \times 1 = 3$$. - **$$f(5)$$**: Since $$x = 5$$ is outside the domain $$[0, 4]$$, $$f(5)$$ is **undefined**. ### (c) **Graph $$f$$:** The correct graph (labeled **D**) consists of two line segments: - From $$(0, 0)$$ to $$(2, 6)$$, using the rule $$f(x) = 3x$$. - From $$(2, 5)$$ to $$(4, 7)$$, using the rule $$f(x) = x + 3$$. ### (d) **Is $$f$$ continuous on its domain?** $$f$$ is **not continuous** at $$x = 2$$, because: - The left-hand limit at $$x = 2$$ (from the first interval) is $$6$$ (since $$3 \times 2 = 6$$). - The right-hand limit at $$x = 2$$ (from the second interval) is $$5$$ (since $$2 + 3 = 5$$). Since these limits do not match, there is a jump discontinuity at $$x = 2$$. Do you need further details on any part? --- ### 5 Related Questions: 1. What is a piecewise function and how is it graphed? 2. How can you determine if a function is continuous at a point? 3. What are the conditions for a function to be continuous on an interval? 4. How do you evaluate limits for piecewise functions? 5. What are the different types of discontinuities in functions? ### Tip: When working with piecewise functions, always check the boundaries of each interval carefully to ensure correct evaluation and continuity.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Domain and Range
Continuity
Graphing Functions
Formulas
f(x) = 3x for 0 <= x < 2
f(x) = x + 3 for 2 <= x <= 4
Theorems
Continuity Theorem
Piecewise Function Evaluation
Suitable Grade Level
Grades 9-11
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