Math Problem Statement

Find all points of discontinuity for the following piecewise function: f(x)=⎧⎩⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪⎪−2x, if x<−2x2, if −2≤x<03x, if 0<x<3xx−4, if 3≤x≤6xx−5, if 6<x≤7x2, if 7<x

Solution

To find all points of discontinuity for the given piecewise function:

\begin{cases} -2x & \text{if } x < -2 \\ x^2 & \text{if } -2 \leq x < 0 \\ 3x & \text{if } 0 < x < 3 \\ \frac{x}{x-4} & \text{if } 3 \leq x \leq 6 \\ \frac{x}{x-5} & \text{if } 6 < x \leq 7 \\ x^2 & \text{if } x > 7 \end{cases}$$ We need to check for discontinuities at the boundaries of each interval, as well as within the intervals where the function might not be defined. ### Step 1: Check at $$x = -2$$ - Left-hand limit: $$\lim_{x \to -2^-} f(x) = \lim_{x \to -2^-} -2x = -2(-2) = 4$$ - Right-hand limit: $$\lim_{x \to -2^+} f(x) = \lim_{x \to -2^+} x^2 = (-2)^2 = 4$$ - Function value: $$f(-2) = (-2)^2 = 4$$ Since the left-hand limit, right-hand limit, and function value are all equal at $$x = -2$$, there is **no discontinuity** at this point. ### Step 2: Check at $$x = 0$$ - Left-hand limit: $$\lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x^2 = 0^2 = 0$$ - Right-hand limit: $$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} 3x = 3(0) = 0$$ Since the left-hand and right-hand limits are both 0, and the function is not defined at $$x = 0$$ (the intervals are open at $$x = 0$$), there is **no discontinuity** at this point. ### Step 3: Check at $$x = 3$$ - Left-hand limit: $$\lim_{x \to 3^-} f(x) = \lim_{x \to 3^-} 3x = 3(3) = 9$$ - Right-hand limit: $$\lim_{x \to 3^+} f(x) = \lim_{x \to 3^+} \frac{x}{x-4} = \frac{3}{3-4} = \frac{3}{-1} = -3$$ - Function value: $$f(3) = \frac{3}{3-4} = -3$$ The left-hand limit and right-hand limit are not equal at $$x = 3$$, so there is a **discontinuity** at $$x = 3$$. ### Step 4: Check at $$x = 4$$ - Within the interval $$3 \leq x \leq 6$$, the function $$f(x) = \frac{x}{x-4}$$ has a vertical asymptote at $$x = 4$$, making it undefined at this point. Therefore, there is a **discontinuity** at $$x = 4$$. ### Step 5: Check at $$x = 6$$ - Left-hand limit: $$\lim_{x \to 6^-} f(x) = \lim_{x \to 6^-} \frac{x}{x-4} = \frac{6}{6-4} = \frac{6}{2} = 3$$ - Right-hand limit: $$\lim_{x \to 6^+} f(x) = \lim_{x \to 6^+} \frac{x}{x-5} = \frac{6}{6-5} = \frac{6}{1} = 6$$ - Function value: $$f(6) = \frac{6}{6-4} = 3$$ The left-hand limit and right-hand limit are not equal at $$x = 6$$, so there is a **discontinuity** at $$x = 6$$. ### Step 6: Check at $$x = 7$$ - Left-hand limit: $$\lim_{x \to 7^-} f(x) = \lim_{x \to 7^-} \frac{x}{x-5} = \frac{7}{7-5} = \frac{7}{2} = 3.5$$ - Right-hand limit: $$\lim_{x \to 7^+} f(x) = \lim_{x \to 7^+} x^2 = 7^2 = 49$$ - Function value: $$f(7) = \frac{7}{7-5} = 3.5$$ The left-hand limit and right-hand limit are not equal at $$x = 7$$, so there is a **discontinuity** at $$x = 7$$. ### Conclusion The points of discontinuity for the function $$f(x)$$ are at $$x = 3$$, $$x = 4$$, $$x = 6$$, and $$x = 7$$. Would you like a detailed explanation of any part? Here are some related questions: 1. How do you determine the type of discontinuity (jump, infinite, or removable)? 2. What is the difference between left-hand and right-hand limits? 3. How does the definition of a piecewise function affect continuity? 4. Can a function have an infinite number of discontinuities? 5. How do you analyze the continuity of more complex piecewise functions? **Tip:** When checking continuity, always consider both limits (left-hand and right-hand) and the function's actual value at the point of interest.

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Math Problem Analysis

Mathematical Concepts

Piecewise functions
Limits
Discontinuities

Formulas

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Theorems

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Suitable Grade Level

Advanced High School