Problem 43 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Determine the values of \(x\), if any, at which each function is discontinuous. At each number where \(f\) is discontinuous, state the condition(s) for continuity that are violated. $$ f(x)=\left\\{\begin{array}{ll} x+5 & \text { if } x<0 \\ 2 & \text { if } x=0 \\ -x^{2}+5 & \text { if } x>0 \end{array}\right. $$

Step-by-Step Solution

Answer

The function \(f(x)\) is discontinuous at \(x = 0\). The left-hand limit as \(x\) approaches 0 is 5, the right-hand limit is 5, and the function value at \(x = 0\) is 2. The violated condition for continuity is that the left-hand limit, right-hand limit, and function value must all be equal.

Step 1: Check the continuity of each piece

Each piece of the function is a simple polynomial function, and we know that polynomial functions are continuous everywhere in their domain. Therefore, each piece of the function is continuous within its defined interval.

Step 2: Check the limit at x = 0

To determine if a function is continuous at the transition points, we need to check if the left-hand limit, right-hand limit, and the function value all exist and are equal. In this function, the transition point is at \(x = 0\). Let us find the left-hand limit, the right-hand limit, and the function value at this point: - Left-hand limit: As \(x\) approaches 0 from the left, we are in the interval \(x < 0\), so we need to use the equation \(f(x) = x + 5\). Therefore, the left-hand limit is: $$ \lim_{x \to 0^{-}}(x + 5) = 0 + 5 = 5 $$ - Right-hand limit: As \(x\) approaches 0 from the right, we're in the interval \(x > 0\), so we need to use the equation \(f(x) = -x^2 + 5\). Therefore, the right-hand limit is: $$ \lim_{x \to 0^{+}}(-x^2 + 5) = -0^2 + 5 = 5 $$ - Function value at \(x = 0\): Since \(x = 0\), we use the equation \(f(x) = 2\). Therefore, the function value at \(x = 0\) is 2. Since the left-hand limit, right-hand limit, and the function value at \(x = 0\) are not equal (5 ≠ 2 ≠ 5), the function is discontinuous at \(x = 0\).

Step 3: State the violated conditions

For a function to be continuous at a point, all of the following conditions must be met: 1. The left-hand limit exists. 2. The right-hand limit exists. 3. The function value exists. 4. The left-hand limit and right-hand limit are equal, and both are equal to the function value. In this case, the left-hand limit, right-hand limit, and function value all exist, but they are not equal, so the function is discontinuous at \(x = 0\). Thus, the function is discontinuous at \(x = 0\), and the violated condition for continuity is that the left-hand limit, right-hand limit, and function value must all be equal.

Key Concepts

Left-Hand Limit

When analyzing discontinuities in a function, understanding the concept of the left-hand limit is crucial. The left-hand limit refers to the value that a function approaches as the input approaches a certain point from the left side (i.e., from smaller values). For example, when we examine the function described in the exercise:

• As \( x \) approaches 0 from values less than 0, we use the part of the piecewise function described by \( f(x) = x + 5 \).
• This means the left-hand limit at \( x = 0 \) is calculated by evaluating \( \lim_{x \to 0^{-}}(x + 5) \).
• In this case, we find that the left-hand limit is 5.

By evaluating the left-hand limit, we can understand half of the picture needed to determine if the function is continuous at that point. The left-hand limit helps us understand the behavior of the function as it nears the point from just one side.

Right-Hand Limit

The right-hand limit is another essential concept when examining where a function might be discontinuous. This limit investigates the value that the function approaches as the variable approaches a specific point from the right (i.e., from larger values). In our function:

• The right-hand limit at \( x = 0 \) involves the part of the piecewise function defined by \( f(x) = -x^2 + 5 \) when \( x > 0 \).
• We calculate the right-hand limit using \( \lim_{x \to 0^{+}}(-x^2 + 5) \).
• For this function, the right-hand limit is also 5.

Understanding the right-hand limit highlights the behavior of the function as \( x \) approaches the point from the positive side. Both left-hand and right-hand limits are vital for comparing if a function is continuous at a transition point like \( x = 0 \).

Continuity Conditions

For a function to be continuous at a point, several continuity conditions should be met. These key conditions ensure that the function does not "jump" or leave any gap at the transition points. The conditions include:

• The left-hand limit must exist at the point.
• The right-hand limit must exist at the point.
• The function must have a defined value at that point.
• Finally, the left-hand limit, right-hand limit, and the function value at the point should all be equal.

In our exercise, while the left-hand limit at \( x = 0 \) is 5, the right-hand limit is also 5; however, the value of the function at \( x = 0 \) is 2. These values are not equal, thereby violating the last condition of continuity. This shows that at \( x = 0 \), the function exhibits a discontinuity because all conditions for continuity are not satisfied simultaneously.

Piecewise Function

A piecewise function is a type of function that has multiple expressions or "pieces" for different intervals of the input variable. These functions match different conditions over various parts of their domain. For the function given in the exercise:

• One piece defines \( f(x) = x + 5 \) for \( x < 0 \).
• Another piece specifies \( f(x) = 2 \) for \( x = 0 \).
• The last piece is \( f(x) = -x^2 + 5 \) for \( x > 0 \).

Piecewise functions are common in real-world scenarios where a rule or condition changes at certain points. It is essential when analyzing such functions to examine the limits at the transition points between pieces and test for continuity. Understanding how a piecewise function switches between its different pieces helps predict where these transitions might cause discontinuities, as seen in the function's discontinuity at \( x = 0 \).