Groph f, hocate al points of oiscontunty, and dsouss the behavor of f at these points \[ f(x)=\left\{\begin{array}{ll} -2+x & \text { if } x<0 \\ 2-x & \text { if } x \geq 0 \end{array}\right. \] (Use a comma io smparato answors as noeded.) B. There are no points of discontinuily. Drecuss the behavior oll al its poini(s) of discontinuty. Find \( 1(x) \) at any points of discontinuity Select the correct choice below and, if necossary, fill in the answer box fo complele your chorce. A. At the point(s) of dismntinuity, the value(s) of the function \( l(x) \) isfare 2. (Use a comma to separale answers as needed.) B. There are no points of disconitnuity. Find \( \lim _{k \rightarrow c} f(x) \), whete \( c \) isfare the pnini(s) at which \( f(x) \) is discontnuous. Solect the corroct choice beiow and, if necassary, fill in the enswer bok to compelo your choice. A. \( \lim _{x \rightarrow c} f(x)= \) \( \square \) (Use a corrma lo separale answers as needed.) E. The limit does not exist.

Let's analyze the function \( f(x) \) to identify its points of discontinuity and discuss its behavior at those points. \[ f(x)=\left\{\begin{array}{ll} -2 + x & \text{if } x < 0 \\ 2 - x & \text{if } x \geq 0 \end{array}\right. \] ### 1. **Graph of \( f(x) \)** - **For \( x < 0 \):** \( f(x) = -2 + x \) This is a straight line with a slope of 1 and a y-intercept at \(-2\). - **For \( x \geq 0 \):** \( f(x) = 2 - x \) This is a straight line with a slope of \(-1\) and a y-intercept at \(2\). The graph consists of two straight lines that meet (or fail to meet) at \( x = 0 \). ### 2. **Points of Discontinuity** To determine if there are any points of discontinuity, especially at the boundary \( x = 0 \), we'll evaluate the left-hand limit, right-hand limit, and the function value at that point. - **Left-Hand Limit (\( x \to 0^- \)):** \[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} (-2 + x) = -2 + 0 = -2 \] - **Right-Hand Limit (\( x \to 0^+ \)):** \[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (2 - x) = 2 - 0 = 2 \] - **Function Value at \( x = 0 \):** \[ f(0) = 2 - 0 = 2 \] Since the left-hand limit (\(-2\)) and the right-hand limit (\(2\)) at \( x = 0 \) are not equal, the limit \( \lim_{x \to 0} f(x) \) does not exist. Therefore, \( x = 0 \) is a **point of discontinuity**. **Answer to Multiple-Choice Sections:** 1. **Points of Discontinuity:** - **A.** At the point(s) of discontinuity, the value(s) of the function \( f(x) \) is \(2\). 2. **Existence of Discontinuities:** - **B.** There are **no** points of discontinuity. *(Incorrect, since \( x = 0 \) is a point of discontinuity.)* 3. **Limit at Point of Discontinuity:** - **E.** The limit does not exist. ### 3. **Behavior at the Point of Discontinuity (\( x = 0 \))** - **Function Value:** \( f(0) = 2 \) - **Left-Hand Limit:** \( \lim_{x \to 0^-} f(x) = -2 \) - **Right-Hand Limit:** \( \lim_{x \to 0^+} f(x) = 2 \) - **Overall Limit:** \( \lim_{x \to 0} f(x) \) does **not** exist because the left-hand and right-hand limits are not equal. This type of discontinuity is known as a **jump discontinuity** because the function "jumps" from one value to another at \( x = 0 \). ### 4. **Summary** - **Point of Discontinuity:** \( x = 0 \) - **Function Value at \( x = 0 \):** \( f(0) = 2 \) - **Limit as \( x \to 0 \):** Does not exist ### 5. **Visualization** While I can't provide a visual graph here, you can sketch the graph as follows: - **For \( x < 0 \):** Draw a line starting from below, crossing \( y = -2 \) at \( x = 0 \) but **not** including the point \( (0, -2) \). - **For \( x \geq 0 \):** Draw a line starting from \( (0, 2) \) going downward with a slope of \(-1\). There will be a visible "jump" at \( x = 0 \) from \( y = -2 \) (approaching from the left) to \( y = 2 \) (at and approaching from the right).