Let's break down the requirements to construct the graph step by step: ### Step 1: **Intervals of Continuity and Discontinuity** - Function is **continuous** on \([-2, 0)\), \((0, 1)\), and \((1, \infty)\). - There are **discontinuities** at \(x = 0\) and \(x = 1\). ### Step 2: **Asymptotes** - **Horizontal asymptote** at \(y = 3\): As \(x \to \pm\infty\), \(f(x) \to 3\). - **Vertical asymptote** at \(x = 0\): As \(x \to 0^-\) or \(x \to 0^+\), \(f(x) \to \pm\infty\). ### Step 3: **Limits and Point Values** - \(\lim_{x \to 1^-} f(x) = 1\) - \(\lim_{x \to 1^+} f(x) = 2\) - \(f(1) = 3\) ### Step 4: **Construct the Function** Let’s piece together a possible sketch based on these conditions. #### **On \([-2, 0)\):** - The function could approach the vertical asymptote at \(x = 0\) as \(x\) increases. - The function should also approach the horizontal asymptote \(y = 3\) as \(x\) decreases. - A possible function here is \(f(x) = 3 - \frac{1}{x}\), which approaches \(3\) as \(x \to -\infty\) and \(-\infty\) as \(x \to 0^-\). #### **On \((0, 1)\):** - The function could be \(f(x) = 3 - \frac{1}{x}\), which approaches \(+\infty\) as \(x \to 0^+\). - As \(x \to 1^-\), \(f(x)\) should approach 1. So, let's adjust the function to meet this limit. Suppose we use a linear function: \(f(x) = m(x-1) + 1\) on \((0, 1)\), and set \(f(x) \to 1\) as \(x \to 1^-\). #### **On \((1, \infty)\):** - \(\lim_{x \to 1^+} f(x) = 2\) - \(f(1) = 3\) (so there is a jump at \(x = 1\)). - Function approaches 3 as \(x \to \infty\). Suppose we use a function like \(f(x) = 3 - \frac{1}{x}\) for \(x > 1\). As \(x \to 1^+\), this approaches \(2\), and as \(x \to \infty\), it goes to \(3\). #### **Putting it All Together** Let’s define: \[ f(x) = \begin{cases} 3 - \frac{1}{x}, & x \in [-2, 0) \\ 1 + (x-1), & x \in (0, 1) \\ 3, & x = 1 \\ 3 - \frac{1}{x}, & x \in (1, \infty) \end{cases} \] **Check the conditions:** - As \(x \to 0^-\), \(f(x) \to -\infty\) - As \(x \to 0^+\), \(f(x) \to 0\) - As \(x \to 1^-\), \(f(x) \to 1\) - As \(x \to 1^+\), \(f(x) \to 2\) - \(f(1) = 3\) - As \(x \to \pm \infty\), \(f(x) \to 3\) ### Step 5: **Sketch the Graph** Here's a sketch based on above definition: 1. **For \(x \in [-2, 0)\):** Draw the curve \(3 - \frac{1}{x}\), approaching \(y = 3\) as \(x \to -\infty\), and going to \(-\infty\) as \(x \to 0^-\). 2. **At \(x = 0\):** Vertical asymptote. 3. **For \(x \in (0, 1)\):** Draw the line \(f(x) = x\), which moves from \(0\) up to \(1\) as \(x\) goes from \(0^+\) to \(1^-\). 4. **At \(x = 1\):** Plot a point at \((1, 3)\), with open circles at \(y=1\) and \(y=2\). 5. **For \(x \in (1, \infty)\):** Draw the curve \(3 - \frac{1}{x}\), starting just above \(y=2\) at \(x=1^+\), and approaching \(y=3\) as \(x \to \infty\). --- ### **Final Answer:** #### **Piecewise function:** \[ f(x) = \begin{cases} 3 - \frac{1}{x}, & x \in [-2, 0) \\ x, & x \in (0, 1) \\ 3, & x = 1 \\ 3 - \frac{1}{x}, & x \in (1, \infty) \end{cases} \] #### **Graph:** Below is a hand-drawn sketch for your reference: ``` 4 | 3 |----.-------------------.---------- (asymptote y = 3) 2 | \ / 1 | \ / 0 |------------------o----------------- -1 | \ / -2 | \ / -3 | \/ -4 | --------------------------------------- -9 -8 ... 0 1 2 ... 9 (Vertical asymptote at x = 0) (Jump at x = 1: open circles at (1,1) and (1,2), filled at (1,3)) ``` --- If you need the graph plotted, let me know and I can provide a precise plot!