Problem 41 - Calculus: Early Transcendentals

Question

Find the numbers at which \( f \) is discontinuous. At which of these numbers is \( f \) continuous from the right, from the left, or neither? Sketch the graph of \( f \). \( f(x) = \left\\{ \begin{array}{ll} x^2 & \mbox{if \) x < -1 \(}\\\ x & \mbox{if \) -1 \le x < 1 \(} \\ 1/x & \mbox{if \) x \ge 1 \(} \end{array} \right.\)

Step-by-Step Solution

Answer

Discontinuous at \( x = -1 \); continuous at \( x = 1 \) from both sides.

Step 1: Analyze the piecewise function

The function \( f(x) \) is defined with three different expressions for different intervals of \( x \). 1. \( f(x) = x^2 \) for \( x < -1 \) 2. \( f(x) = x \) for \( -1 \le x < 1 \) 3. \( f(x) = \frac{1}{x} \) for \( x \ge 1 \) We'll check the points where the function switches from one expression to another for discontinuity.

Step 2: Check continuity at x = -1

First, analyze the point \( x = -1 \): - The left side as \( x \to -1^- \) is \( f(x) = x^2 \), so \( f(-1^-) = (-1)^2 = 1 \). - The right side as \( x \to -1^+ \) is \( f(x) = x \), so \( f(-1^+) = -1 \). - Since \( f(-1^-) eq f(-1^+) \), \( f(x) \) is discontinuous at \( x = -1 \). - It is continuous from neither right nor left at this point.

Step 3: Check continuity at x = 1

Next, analyze the point \( x = 1 \): - The left side as \( x \to 1^- \) is \( f(x) = x \), so \( f(1^-) = 1 \). - The right side as \( x \to 1^+ \) is \( f(x) = \frac{1}{x} \), so \( f(1^+) = \frac{1}{1} = 1 \). - Since \( f(1^-) = f(1^+) \), \( f(x) \) is continuous at \( x = 1 \). - Thus, \( f(x) \) is both continuous from the right and from the left at \( x = 1 \).

Step 4: Create a sketch of the graph

To sketch the graph:- For \( x < -1 \), graph \( y = x^2 \).- For \( -1 \le x < 1 \), graph \( y = x \).- For \( x \ge 1 \), graph \( y = \frac{1}{x} \).- Clearly mark the open and closed intervals to represent the discontinuities and where the graphs connect, particularly highlighting the discontinuity at \( x = -1 \) with a break in the line.

Key Concepts

Continuity from Left or Right

Continuity is a key concept in calculus and refers to the behavior of functions at a given point. In a piecewise function like the one given, it is crucial to check the transitions between different sections of the function, which usually happen at boundary points. Here, we focus on the continuity from the left and the right for the given piecewise function:

• For continuity from the left at a point, the limit of the function as it approaches that point from the left must equal the function's value at that point.
• For continuity from the right, the limit of the function as it approaches from the right must equal the function’s value at that point.

Consider the piecewise function described in the exercise. At each point where the expressions change, check the continuity by evaluating the limit of each side:- At \( x = -1 \), the left limit \( f(x) = x^2 \) provides a value of 1 and the right limit \( f(x) = x \) gives -1. Since these limits are unequal, the function is not continuous from the left or the right.- At \( x = 1 \), the left limit \( f(x) = x \) and the right limit \( f(x) = \frac{1}{x} \) both evaluate to 1, confirming continuity from both directions. This means the graph smoothly connects at this point.

Graphing Piecewise Functions

Graphing piecewise functions involves plotting different parts of a function defined by various expressions over certain intervals. Each part of the function is drawn on the same coordinate plane, taking care to adjust the graph's behavior at the boundaries that separate these expressions.For graphing the given function:

• For \( x < -1 \), use \( y = x^2 \). This produces a part of a parabola, curving upwards with a closed circle at \( x = -1 \) if included.
• For \( -1 \le x < 1 \), plot the line \( y = x \). This is a straight line through the origin with an open circle at \( x = 1 \), due to the next section taking over.
• For \( x \ge 1 \), graph \( y = \frac{1}{x} \), a hyperbolic curve approaching the coordinate axes with an open point reflecting its exclusion from \( x = 1 \).

To clearly indicate where the function transitions, use open or closed circles to mark the limits of intervals, ensuring that discontinuities at points like \( x = -1 \) are evident. This visual representation helps easily recognize how each section connects or fails to connect to another.

Limits and Continuity Analysis

The analysis of limits and continuity is pivotal in understanding how piecewise functions behave at their border points. When you calculate the limit of a function approaching a boundary from both the left and the right, it gives invaluable insight into whether a function is continuous or discontinuous at that specific point.For the given function:

• At \( x = -1 \), calculate the limits from the left \( (1) \) and right \( (-1) \). Since the limits differ, \( f(x) \) is discontinuous at \( x = -1 \).
• On the contrary, at \( x = 1 \), both the left and right limits are consistent, \( (1) \), indicating the function is continuous at this point.

This process of analyzing limits provides a clear, mathematical proof of continuity or its absence at various points within piecewise functions. It's a fundamental aspect of calculus that ensures we understand the nature of graphs in relation to real-world modeling.