Problem 19 - Contemporary Precalculus

Question

Determine whether or not the function is continuous at the given number. $$f(x)=\left\\{\begin{array}{cl} -2 x+4 & \text { if } x \leq 2 \\ 2 x-4 & \text { if } x>2 \end{array} \text { at } x=2\right.$$

Step-by-Step Solution

Answer

Answer: Yes, the function is continuous at x = 2.

Step 1: Find the left-hand limit at x = 2

To find the left-hand limit, we will use the function definition for x ≤ 2 which is given by: $$ f(x) = -2x + 4 $$ We need to find the limit as x approaches 2 from the left side: $$ \lim _{x \rightarrow 2^{-}} (-2x + 4) $$ By substituting x = 2, we get: $$ \lim _{x \rightarrow 2^{-}} (-2x + 4) = -2(2) + 4 = 0 $$

Step 2: Find the right-hand limit at x = 2

To find the right-hand limit, we will use the function definition for x > 2 which is given by: $$ f(x) = 2x - 4 $$ We need to find the limit as x approaches 2 from the right side: $$ \lim _{x \rightarrow 2^{+}} (2x - 4) $$ By substituting x = 2, we get: $$ \lim _{x \rightarrow 2^{+}} (2x - 4) = 2(2) - 4 = 0 $$

Step 3: Compare the left-hand limit, right-hand limit and the function's value at x = 2

In order to prove that the function is continuous at x = 2, we must check that the left-hand limit, right-hand limit and the function's value at x = 2 are equal. From Step 1, we have the left-hand limit as 0, and from Step 2, we have the right-hand limit as 0. Now, let's find the function's value at x = 2: Since x = 2 satisfies the condition x ≤ 2, we will use the function definition for x ≤ 2 to find f(2): $$ f(2) = -2(2) + 4 = 0 $$ Now, we will compare the left-hand limit, right-hand limit, and the function's value at x = 2: Left-hand limit = Right-hand limit = Function's value at x = 2 = 0 Since all three values are equal, we can conclude that the function is continuous at x = 2.

Key Concepts

Left-hand limit

The left-hand limit of a function describes the behavior of the function as the input value approaches a particular point from the left side. In the case of piecewise defined functions, such as the one described in the original exercise, it is crucial to identify which part of the function applies for the left side of the given value.
For the given function, when evaluating at \(x = 2\), the left-hand limit involves using the expression for \(x \leq 2\). Therefore, we'll consider the piece \(-2x + 4\).
To calculate the left-hand limit, substitute the value into the expression:

• \(\lim_{x \rightarrow 2^{-}}(-2x + 4) = -2(2) + 4 = 0\)

This means that as \(x\) approaches 2 from the left, the output of the function approaches 0. This is an integral part of determining the continuity of the function at this point.

Right-hand limit

The right-hand limit of a function refers to the function's behavior as the input reaches a specific point from the right side. For piecewise functions, it's important to use the appropriate expression that is applicable for inputs greater than the given point.
In our exercise, we analyze the function at \(x = 2\) using the expression that applies to \(x > 2\), which is \(2x - 4\). We substitute \(x = 2\) into this new context.
The calculation for the right-hand limit is as follows:

• \(\lim_{x \rightarrow 2^{+}}(2x - 4) = 2(2) - 4 = 0\)

This calculation tells us that as \(x\) approaches 2 from the right side, the function output approaches 0.
Finding both the left-hand and right-hand limits is crucial in determining overall limits and continuity of piecewise functions.

Evaluating limits

Evaluating limits is a fundamental concept in calculus, especially when discussing the continuity of functions. Limits provide insight into both the behavior of a function as the input approaches a specific value, and whether that function is continuous at that point.
In the context of piecewise functions, it's important to separately evaluate both the left-hand and right-hand limits.
When both limits at a point agree, it suggests that the function might be continuous there, pending verification from the actual value of the function at that point. Calculating limits involves:

• Finding the left-hand limit as the input approaches from below the point.
• Finding the right-hand limit as the input approaches from above the point.

If both limits at \(x = 2\) are equal to each other and to the function's value, \(f(x)\), then the function can be stated as continuous.
For our exercise, both limits and the function's value at x = 2 are 0, confirming the continuity.

Piecewise defined functions

Piecewise defined functions consist of different expressions applied to specific intervals of the input. Understanding which part of the function is used at different sections is key to defining and evaluating the function accurately.
In our exercise, the function is defined as being \(-2x + 4\) when \(x\) is less than or equal to 2, and \(2x - 4\) when \(x\) is greater than 2. This division impacts the calculation of limits and evaluating continuity.
To determine if the function is continuous at a join-point such as \(x = 2\), one must:

• Check the value and limits using the expressions from both intervals.
• Ensure all values and limits converge to the same point at \(x = 2\).

Piecewise functions often appear in real-world engineering and economic applications, where the situation changes based on different conditions or thresholds.
Understanding how these functions are set up allows us to capture distinct behaviors within a single unified function.