Problem 14 - Calculus Early Transcendentals

Question

Sketch the graph of an example of a function \(f\) that satisfies all of the given conditions. $$\begin{array}{l}{\lim _{x \rightarrow 0^{-}} f(x)=1, \quad \lim _{x \rightarrow 0^{+}} f(x)=-1, \quad \lim _{x \rightarrow 2^{-}} f(x)=0} \\\ {\lim _{x \rightarrow 2^{+}} f(x)=1, \quad f(2)=1, \quad f(0) \text { is undefined }}\end{array}$$

Step-by-Step Solution

Answer

The function has jump discontinuities at \(x = 0\) and \(x = 2\), with the specified limits. \(f(x)\) is undefined at \(x = 0\) and defined at \(x = 2\).

Step 1: Understanding the Problem

We need to sketch a function \(f(x)\) that satisfies the specified limits and values. It's important to consider that the function may have discontinuities at \(x = 0\) and \(x = 2\), as indicated by the different behavior of the limits from the left and right.

Step 2: Determine Left-Hand and Right-Hand Limits at Discontinuities

For \(x = 0\), the limit from the left is \(1\) \(\left(\lim_{x \to 0^-} f(x) = 1\right)\) and from the right is \(-1\) \(\left(\lim_{x \to 0^+} f(x) = -1\right)\). This suggests a jump discontinuity at \(x = 0\). Therefore, \(f(x)\) might have a step-like behavior around \(x = 0\), switching from \(1\) to \(-1\) as \(x\) crosses \(0\). At \(x = 0\), \(f(0)\) is undefined, thus there's an open circle at both \((0,1)\) and \((0,-1)\).

Step 3: Understanding the Behavior at x = 2

For \(x = 2\), the limit from the left is \(0\) \(\left(\lim_{x \to 2^-} f(x) = 0\right)\), and the limit from the right is \(1\) \(\left(\lim_{x \to 2^+} f(x) = 1\right)\). This suggests another jump discontinuity. Additionally, \(f(2) = 1\), so \(f(x)\) should have a closed circle at \((2,1)\) but approach \(0\) from the left and \(1\) from the right, indicating another step-like change.

Step 4: Sketch the Graph

Plot the function on a graph:1. For \(x < 0\), draw a line approaching \(y = 1\) as \(x\) approaches \(0\) from the left. Open circle at \((0,1)\).2. For \(x > 0\), draw a separate line approaching \(y = -1\) as \(x\) approaches \(0\) from the right. Open circle at \((0,-1)\).3. At \(x = 2\), from the left, draw a line approaching \(y = 0\) with an open circle, and from the right, draw a line approaching \(y = 1\) with an open circle. Include a closed circle at \((2,1)\) to represent \(f(2) = 1\).

Step 5: Verify the Sketch

Check that the graph includes all stipulated conditions: - The jump from 1 to -1 at \(x = 0\), confirming the limits \(\lim_{x \to 0^-} f(x) = 1\) and \(\lim_{x \to 0^+} f(x) = -1\).- The open circles at \((0,1)\) and \((0,-1)\) since \(f(0)\) is undefined.- The left-side limit of \(0\) and right-side limit of \(1\) at \(x = 2\), validating \(\lim_{x \to 2^-} f(x) = 0\) and \(\lim_{x \to 2^+} f(x) = 1\), and the closed circle at \((2,1)\) for \(f(2) = 1\).

Key Concepts

Jump Discontinuity

A jump discontinuity occurs in a graph when there is a sudden "jump" from one value to another. In simpler terms, the graph of a function doesn't follow a smooth curve through a point but instead leaps abruptly from one point to another.
For example, in our function at

• **Point 1:** The limit from the left at **\(x = 0\)** is \(1\), and the limit from the right is \(-1\). This dramatic switch, without a connecting line, is a classic jump discontinuity.
• **Point 2:** Similarly, at **\(x = 2\)**, the limit from the left reaching **\(0\)** and from the right aiming for **\(1\)**, creates another discontinuous leap.

When you graph these points, use open circles to show that the function doesn't exist between the jump coordinates.

Left-Hand Limit

The left-hand limit of a function is essentially the value that a function approaches as you get infinitely close to a certain point from the left side. You can think of it as traveling along a number line from negative infinity to a specific point.
For example, in this function:

• Approaching **\(x = 0^-\)** from the left, the function values tend towards **\(1\)**. This tells us our graph line should come close to the coordinate (0,1) but will remain unconnected.
• Similarly, as you approach **\(x = 2^-\)** from the left, the function aims towards **\(0\)**. This implies the graph should smoothly head to (2,0) from the left.

Right-Hand Limit

The right-hand limit shows the behavior of a function as it gets infinitely close to a particular point from the right side. Visualize moving along a number line from positive infinity to your desired point.
In this scenario:

• When coming from the right at **\(x = 0^+\)**, the function heads towards **\(-1\)**. Illustrate this by approaching the point (0,-1) on your graph without connecting it.
• Similarly, for **\(x = 2^+\)**, the function aligns itself closer to **\(1\)**, meaning the graph line heading to (2,1) should be drawn from the right.

These sides of the limits demonstrate how to extend the graphed lines while respecting the points of discontinuity.

Undefined Function Value

In certain points, a function might be undefined, meaning there's no specific value assigned at that coordinate. This often results in an open circle on a graph, indicating the lack of a defined point.
In this function presentation:

• **\(f(0)\) is undefined**, meaning there's no actual recorded value at **\(x = 0\)**. Hence, both points (0,1) and (0,-1) should feature open circles, representing this undefined status.
• On the other hand, where the function **relies on specific values**, such as **\(f(2) = 1\)**, a closed circle should be used to indicate a defined point.

Understanding an undefined function value helps interpret the graphical representation properly and acknowledge jumps or holes in the graph.