Math Problem Statement
show clear step by step with canceling,facto-zing,simplifying.
Solution
Let's go through the problem step by step with clear simplifications, factoring, and canceling as needed.
Problem c)
The problem now involves a piecewise function defined as:
\begin{cases} \sqrt{x} & \text{if } x < 0, \\ 3 - x & \text{if } 0 \leq x < 3, \\ 3x - 2 & \text{if } x \geq 3. \end{cases}$$ The subproblems are: 1. **Find $$\lim_{x \to 3} f(x)$$ if it exists.** 2. **Show that $$f(x)$$ is continuous at $$x = 3$$.** 3. **Sketch the graph of $$f(x)$$.** #### 1. Find $$\lim_{x \to 3} f(x)$$ To determine whether $$\lim_{x \to 3} f(x)$$ exists, we need to check the left-hand limit ($$\lim_{x \to 3^-} f(x)$$) and the right-hand limit ($$\lim_{x \to 3^+} f(x)$$). - **Left-hand limit ($$x \to 3^-$$)**: In this region, $$f(x) = 3 - x$$. So, $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} (3 - x) = 3 - 3 = 0.$$ - **Right-hand limit ($$x \to 3^+$$)**: In this region, $$f(x) = 3x - 2$$. So, $$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} (3x - 2) = 3(3) - 2 = 9 - 2 = 7.$$ Since the left-hand limit (0) and right-hand limit (7) are not equal, $$\lim_{x \to 3} f(x)$$ **does not exist**. #### 2. Show that $$f(x)$$ is continuous at $$x = 3$$ For $$f(x)$$ to be continuous at $$x = 3$$, the following must be true: - $$f(3)$$ is defined, - $$\lim_{x \to 3^-} f(x)$$ exists, - $$\lim_{x \to 3^+} f(x)$$ exists, - $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) = f(3)$$. Let's check these conditions: - **$$f(3)$$ is defined**: From the piecewise function, $$f(3) = 3(3) - 2 = 9 - 2 = 7$$. - **Left-hand limit**: $$\lim_{x \to 3^-} f(x) = 0$$. - **Right-hand limit**: $$\lim_{x \to 3^+} f(x) = 7$$. Since $$\lim_{x \to 3^-} f(x)$$ and $$\lim_{x \to 3^+} f(x)$$ are not equal, $$f(x)$$ **is not continuous** at $$x = 3$$. #### 3. Sketch the graph of $$f(x)$$ For sketching the graph, break it down according to the piecewise function: 1. **For $$x < 0$$**: The function $$f(x) = \sqrt{x}$$ is **not defined** for real numbers since the square root of a negative number is not a real number. 2. **For $$0 \leq x < 3$$**: The function is $$f(x) = 3 - x$$, a linear function with a negative slope of -1. This line decreases as $$x$$ increases, starting from $$(0, 3)$$ and ending just before $$(3, 0)$$. 3. **For $$x \geq 3$$**: The function is $$f(x) = 3x - 2$$, another linear function but with a positive slope of 3. This line starts at $$(3, 7)$$ and continues upward. **Key Points:** - There is a **jump discontinuity** at $$x = 3$$. - The graph is not defined for $$x < 0$$. - For $$0 \leq x < 3$$, the graph is a downward sloping line. - For $$x \geq 3$$, the graph is an upward sloping line starting at $$(3, 7)$$. --- Would you like a more detailed explanation on any of these steps, or perhaps the sketch of the graph? Here are some related questions you might consider: ### Related Questions: 1. What are the conditions necessary for a limit to exist at a point? 2. How can you determine the continuity of a piecewise function? 3. What are the differences between jump discontinuity and removable discontinuity? 4. How does the behavior of limits affect the sketching of a piecewise function? 5. What role does the slope play in determining the direction of a line? ### Tip: When evaluating limits of piecewise functions, always consider the left-hand and right-hand limits separately before concluding whether the overall limit exists at a particular point.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Limits
Continuity
Graph Sketching
Formulas
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Theorems
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Suitable Grade Level
Grades 11-12