Question

In Exercises 23 and \(24,\) sketch a graph of a function \(f\) that satisfies the given values. (There are many correct answers.) \(f(0)\) is undefined. \(\lim _{x \rightarrow 0} f(x)=4\) \(f(2)=6\) \(\lim _{x \rightarrow 2} f(x)=3\)

Short answer

The graph of the function \(f\) should show an open circle at (0,4) because the function is undefined at \(x=0\) but approaches 4 as \(x\) nears 0. There should also be an open circle at (2,3) because \(f(x)\) approaches 3 as \(x\) nears 2, but the function at \(x=2\) is 6, not 3. So, there should be a distinct point at (2,6). In essence, the graph exhibits the given properties of the function.

Step-by-step solution

Understanding the Limits and Function Values

Start by analyzing the given limit values and function values. The limit as \(x\) approaches 0 for \(f(x)\) is 4, meaning the value of \(f(x)\) approaches 4 as \(x\) gets closer to 0. Yet, \(f(0)\) is undefined, so at \(x=0\), there will be no point on the graph. Similarly, as \(x\) approaches 2, \(f(x)\) approaches 3, but \(f(2)\) is not 3 but 6. This information will help to guide the sketch of the function graph.

Sketching the Graph Based on Limits

Next, sketch a preliminary graph based on the provided limits. On this initial sketch, place an open circle at \(x=0\) and \(x=2\) to indicate that the function is undefined at those points. The \(y\) values of these open circles should be at the limit values. For \(x=0\), place the open circle at \(f(x)=4\), and for \(x=2\), place the open circle at \(f(x)=3\). After this step, the preliminary sketch should show that as \(x\) approaches 0, \(f(x)\) approaches 4, and as \(x\) approaches 2, \(f(x)\) approaches 3, but it is undefined at these particular \(x\) values.

Adding the Function Points

Following the preliminary sketch, add the function values where \(f(x)\) is defined. At \(x=2\), \(f(2)\) is equal to 6, so add a point on the graph at \(x=2\) and \(f(x)=6\). After this step, the graph should reflect the properties of the function \(f\) as given in the exercise: \(f(x)\) is undefined at \(x=0\), it approaches 4 as \(x\) approaches 0, it equals 6 at \(x=2\), and it approaches 3 as \(x\) approaches 2.