Graph 1, locato all points of disconlinuity, and discuss the behavicr of lal these points. \[ f(x)=\left\{\begin{array}{c} -2+x \text { if } x<0 \\ 2-x \text { it } x \geq 0 \end{array}\right. \] Find \( { }^{\prime}(x) \) et any points \( n \) idscontiruily Select the conrect choice below and, il necessary, fill in the answer box to comptele your chrict. A. Al the poink(s) of discontinuity, the value(s) of the function \( \mathrm{f}(\mathrm{x}) \) isfare 2. (Use a cornma to sepaiats enswars as needed) B. There are no points of discontinuity Find \( \lim _{x \rightarrow c} f(x) \), where \( c \) is/are the ponils) al which \( f(x) \) is ciscontinuous. Select the correct chorce Detcw and, if necessary, fil in liw answer box to completo your choico. A. \( \lim _{x \rightarrow c} 1(x)=\square \) \( \square \) (Use a cullura to separalo answers as neoded) B. The limil does nol exist.

Let's analyze the given piecewise function to identify points of discontinuity, discuss their behavior, and determine the limits where the function is continuous. ## Given Function \[ f(x) = \begin{cases} -2 + x & \text{if } x < 0 \\ 2 - x & \text{if } x \geq 0 \end{cases} \] ## 1. Points of Discontinuity ### Identifying Discontinuities A function can be discontinuous at points where the definition changes or where it is not defined. In this case, the function definition changes at \( x = 0 \). We'll check for continuity at this point. ### Checking Continuity at \( x = 0 \) 1. **Left-Hand Limit (\( x \to 0^- \))**: \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-2 + x) = -2 + 0 = -2 \] 2. **Right-Hand Limit (\( x \to 0^+ \))**: \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2 - x) = 2 - 0 = 2 \] 3. **Function Value at \( x = 0 \)**: \[ f(0) = 2 - 0 = 2 \] Since the left-hand limit (\(-2\)) does not equal the right-hand limit (\(2\)), the limit \( \lim_{x \to 0} f(x) \) does not exist. Therefore, \( f(x) \) is **discontinuous** at \( x = 0 \). ### Conclusion on Discontinuity - **Point of Discontinuity**: \( x = 0 \) - **Function Value at Discontinuity**: \( f(0) = 2 \) **Correct Choice**: **A.** All the point(s) of discontinuity, the value(s) of the function \( f(x) \) is 2. *Explanation*: There is only one point of discontinuity at \( x = 0 \), and at this point, the function value is \( 2 \). ## 2. Limits at Points of Continuity For points where \( f(x) \) is continuous (i.e., \( x \neq 0 \)), the limit as \( x \) approaches \( c \) is simply \( f(c) \) because the function is linear (hence continuous) in those intervals. ### Expression of Limits - **For \( c < 0 \)**: \[ \lim_{x \to c} f(x) = -2 + c \] - **For \( c > 0 \)**: \[ \lim_{x \to c} f(x) = 2 - c \] ### Conclusion on Limits **Correct Choice**: **A.** \[ \lim_{x \to c} f(x) = \begin{cases} -2 + c & \text{if } c < 0 \\ 2 - c & \text{if } c > 0 \end{cases} \] *Explanation*: For all points \( c \) where \( f(x) \) is continuous (i.e., \( c \neq 0 \)), the limit exists and is equal to the function value at that point. --- ## 3. Graph of \( f(x) \) To visualize the function and its discontinuity: 1. **For \( x < 0 \)**: \( f(x) = x - 2 \) This is a straight line with a slope of \( 1 \) and a y-intercept at \( (0, -2) \). 2. **For \( x \geq 0 \)**: \( f(x) = 2 - x \) This is a straight line with a slope of \( -1 \) and a y-intercept at \( (0, 2) \). 3. **At \( x = 0 \)**: There is a **jump discontinuity** because the left-hand limit (\(-2\)) does not equal the right-hand limit (\(2\)), and the function value is \( 2 \). ### Sketch of the Graph ``` f(x) | 3 | * 2 |------------* 1 | / 0 |------/ -1| / -2|--* |------------------- -2 -1 0 1 2 x ``` - The asterisk (\(*\)) at \( x = 0 \) on the lower part indicates the limit from the left (\(-2\)), and the solid dot (\(*\)) on the upper part indicates the function value at \( x = 0 \) (\(2\)). ## Summary - **Discontinuity**: \( x = 0 \), where \( f(0) = 2 \). - **Limits at Points of Continuity**: For all \( c \neq 0 \), \[ \lim_{x \to c} f(x) = f(c) \] - **Graph**: Consists of two linear pieces with a jump discontinuity at \( x = 0 \).