Problem 39 - Calculus Early Transcendentals

Question

\(37-39\) 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. \(f(x)=\left\\{\begin{array}{ll}{x+2} & {\text { if } x<0} \\ {e^{x}} & {\text { if } 0 \leq x \leqslant 1} \\ {2-x} & {\text { if } x>1}\end{array}\right.\)

Step-by-Step Solution

Answer

Discontinuities at \( x = 0 \) (continuous from right) and \( x = 1 \) (continuous from left).

Step 1: Analyze Each Piece of the Function

The function \( f(x) \) is defined as a piecewise function. Let's examine each piece:1. **For** \( x < 0 \), \( f(x) = x + 2 \).2. **For** \( 0 \leq x \leq 1 \), \( f(x) = e^x \).3. **For** \( x > 1 \), \( f(x) = 2 - x \).We need to examine the points where the definition of the function changes: at \( x = 0 \) and \( x = 1 \). These are potential discontinuities.

Step 2: Check Continuity at x = 0

To check for continuity at \( x = 0 \), calculate the left-hand limit, right-hand limit, and the function value at \( x = 0 \):- **Left-hand limit** as \( x \to 0^- \), \( f(x) = x + 2 \). So, \( \lim_{x \to 0^-} f(x) = 0 + 2 = 2 \).- **Right-hand limit** as \( x \to 0^+ \), \( f(x) = e^x \). The limit is \( \lim_{x \to 0^+} e^x = e^0 = 1 \).- **Function value at** \( x = 0 \) is \( f(0) = e^0 = 1 \).The left-hand and right-hand limits (2 and 1, respectively) are not equal, so \( f(x) \) is discontinuous at \( x = 0 \). The function is continuous from the right at this point because the right-hand limit equals \( f(0) \).

Step 3: Check Continuity at x = 1

To check for continuity at \( x = 1 \), calculate the left-hand limit, right-hand limit, and the function value at \( x = 1 \):- **Left-hand limit** as \( x \to 1^- \), \( f(x) = e^x \). Here, \( \lim_{x \to 1^-} e^x = e^1 = e \).- **Right-hand limit** as \( x \to 1^+ \), \( f(x) = 2 - x \). The limit is \( \lim_{x \to 1^+} (2 - x) = 2 - 1 = 1 \).- **Function value at** \( x = 1 \) is \( f(1) = e^1 = e \).The left-hand limit and the function value are equal but differ from the right-hand limit, so \( f(x) \) is discontinuous at \( x = 1 \). The function is continuous from the left.

Step 4: Sketch the Graph

With the analysis from steps 2 and 3, we can sketch the graph:1. **For** \( x < 0 \): A line with an equation \( y = x + 2 \) extends from top left towards up to \( x = 0 \) with an open circle at \( (0, 2) \).2. **For** \( 0 \leq x \leq 1 \): The exponential curve \( y = e^x \) starts at \( (0, 1) \) and ends at \( (1, e) \) with a closed circle.3. **For** \( x > 1 \): A line with an equation \( y = 2 - x \) starts with an open circle at \( (1, 1) \) and moves downwards.

Key Concepts

Continuity

Continuity in the context of piecewise functions means that the graph of the function has no breaks, holes, or jumps at a given point. For a function to be continuous at a certain point, three conditions must be met:

• The function is defined at that point.
• The left-hand and right-hand limits at that point exist.
• The left-hand and right-hand limits are equal to each other and to the function's value at that point.

Let's take a closer look at the piecewise function provided in the exercise. This function switches definitions at certain points: specifically, at \(x = 0\) and \(x = 1\). To determine if the function is continuous at these points, we examine the limits approaching each from both sides, as well as checking the function's actual value at those points.

Discontinuity

Discontinuity occurs at a point if the graph of a function has a break, hole, or jump at that point. This means the function is not smooth at that particular value of \(x\).In the analyzed piecewise function, we identified potential points of discontinuity at \(x = 0\) and \(x = 1\) because these are the locations where the definition of the function changes.

• At \(x = 0\), the function is discontinuous because the left-hand limit \((2)\) and right-hand limit \((1)\) are not equal. The function has a jump discontinuity here.
• At \(x = 1\), the function is also discontinuous because the left-hand limit \((e)\) and right-hand limit \((1)\) are not equal. There is also a jump discontinuity at this point.

These discontinuities are key to understanding the behavior of the piecewise function across its range.

Limits

Limits are a fundamental concept in calculus that describe the value that a function approaches as the input approaches a certain point. They are crucial in analyzing the behavior of piecewise functions at the points where their definitions change.To determine if our piecewise function is continuous at \(x = 0\) and \(x = 1\), we calculate the limits from both sides of each point. If the limits from the left and the right are equal, and they match the function value at that point, the function is continuous there.

• If the limits are not equal or don't match the function value, this indicates a discontinuity.

The process of calculating these limits provides insight into the behavior of the function as it transitions between pieces, as we will explore further in the left-hand and right-hand limit sections.

Left-hand Limit

The left-hand limit of a function at a particular point is the value that the function approaches as the input \(x\) approaches the point from the left side, denoted by \(x \to a^-\).In our exercise, for the piecewise function, we looked at:

• At \(x = 0\): The left-hand limit is calculated as \(\lim_{x \to 0^-} (x + 2) = 2\).
• At \(x = 1\): The left-hand limit is calculated as \(\lim_{x \to 1^-} e^x = e\).

These calculations show what the function approaches from the left side at these points. When analyzing continuity, the left-hand limit is compared with the right-hand limit and the actual function value to determine if there is continuity or a discontinuity at that point.

Right-hand Limit

The right-hand limit is similar to the left-hand limit but taken as the input \(x\) approaches a given point from the right, denoted by \(x \to a^+\).In the exercise, we computed the right-hand limits of the piecewise function:

• At \(x = 0\): The right-hand limit is \(\lim_{x \to 0^+} e^x = 1\).
• At \(x = 1\): The right-hand limit is \(\lim_{x \to 1^+} (2 - x) = 1\).

The right-hand and left-hand limits help determine the nature of any discontinuity. If they are equal and match the function's value at that point, the function is continuous. However, in both cases here, the right and left limits differ, contributing to the discontinuities identified at \(x = 0\) and \(x = 1\).