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Graph f, locate all points of disconlinuity, and discuss the behavior of fat these points. \[ f(x)=\left\{\begin{array}{ll} 1+x & \text { if } x<1 \\ 6-x & \text { if } x \geq 1 \end{array}\right. \] Discuss the behavior off at its point(s) of discontinuity. Find \( \mathrm{f}(\mathrm{x}) \) at any points of discontinuily Select the correct choico bolow and, if nocessary, fill in the answer box to completo your choice. A. At the point(s) of discontinuity, the value(s) of the function \( f(x) \) is/are \( \square \) (Use a comma lo separate answers as needed.) B. There are no points of discontinuity. Find \( \lim _{x \rightarrow c} f(x) \). where \( c \) is/are the poinl( \( (s) \) at which \( f(x) \) is dscontinuous. Select the correct choice below and, if necessary, fill in the answer box lo ccmplete your choice. A. \( \lim _{x \rightarrow c} f(x)= \) \( \square \) (Use a cormma to saparale answers as needed.) B. The limit does not exist

Ask by Wright Crawford. in the United States
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

At \(x = 1\), the function has a jump discontinuity. The value of the function at this point is 5, and the limit does not exist.

Solution

Let's analyze the given piecewise function to identify any points of discontinuity and discuss the behavior at those points. \[ f(x) = \begin{cases} 1 + x & \text{if } x < 1 \\ 6 - x & \text{if } x \geq 1 \end{cases} \] ### 1. **Graphing the Function** - **For \(x < 1\):** The function \(f(x) = 1 + x\) is a straight line with a slope of 1 and a y-intercept at \((0, 1)\). - **For \(x \geq 1\):** The function \(f(x) = 6 - x\) is a straight line with a slope of -1 and a y-intercept at \((0, 6)\). ### 2. **Identifying Points of Discontinuity** The potential point of discontinuity occurs where the definition of the function changes, which is at \(x = 1\). - **Left-Hand Limit as \(x \to 1^-\):** \[ \lim_{x \to 1^-} f(x) = 1 + 1 = 2 \] - **Right-Hand Limit as \(x \to 1^+\):** \[ \lim_{x \to 1^+} f(x) = 6 - 1 = 5 \] - **Function Value at \(x = 1\):** \[ f(1) = 6 - 1 = 5 \] Since \(\lim_{x \to 1^-} f(x) \neq \lim_{x \to 1^+} f(x)\), the function has a **jump discontinuity** at \(x = 1\). ### 3. **Behavior at the Point of Discontinuity** - **Value of the Function at Discontinuity:** \(f(1) = 5\) - **Limit at the Point of Discontinuity:** The limit \(\lim_{x \to 1} f(x)\) **does not exist** because the left-hand and right-hand limits are not equal. ### 4. **Multiple Choice Selections** **A. At the point(s) of discontinuity, the value(s) of the function \( f(x) \) is/are:** \[ \boxed{5} \] **B. There are no points of discontinuity.** \[ \boxed{\text{Incorrect}} \] **Limit at the Point of Discontinuity:** **A. \( \lim _{x \rightarrow 1} f(x)= \)** \[ \boxed{\text{Not Applicable}} \] **B. The limit does not exist** \[ \boxed{\text{Correct}} \] ### 5. **Summary** - **Point of Discontinuity:** \(x = 1\) - **Value at Discontinuity:** \(f(1) = 5\) - **Limit at Discontinuity:** Does not exist

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The Deep Dive

Let’s first dive into the function f's behavior near the point of interest, which is at \( x = 1 \). If \( x < 1 \), the function follows the rule \( f(x) = 1 + x \). Thus, as you approach 1 from the left, the limit is \( \lim_{x \to 1^-} f(x) = 1 + 1 = 2 \). Now, when \( x \geq 1 \), \( f(x) = 6 - x \). Therefore, the limit as you approach from the right is \( \lim_{x \to 1^+} f(x) = 6 - 1 = 5 \). Since the left-hand limit (2) does not equal the right-hand limit (5), we have a point of discontinuity at \( x = 1 \). Now, when we look closer, \( f(1) = 6 - 1 = 5 \). This means that at the point of discontinuity \( x = 1 \), the function value \( f(1) = 5 \). To summarize: - At the point of discontinuity, the value(s) of the function \( f(x) \) is/are \( 5 \). - The limit as \( x \) approaches the point of discontinuity \( c = 1 \) does not match from both sides, so we conclude that \( \lim_{x \rightarrow 1} f(x) \) does not exist. So, the final answers are: A. At the point(s) of discontinuity, the value(s) of the function \( f(x) \) is/are \( 5 \) A. \( \lim_{x \to c} f(x) = \) does not exist

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