Problem 20 - Single Variable Calculus

Question

Find the numbers, if any, where the function is discontinuous. \(f(x)=\left\\{\begin{array}{ll}x+2 & \text { if } x<3 \\ \ln (x-2)+5 & \text { if } x \geq 3\end{array}\right.\)

Step-by-Step Solution

Answer

The function \(f(x)\) is continuous everywhere and has no points of discontinuity since the left and right limits are equal at the transition point x = 3.

Step 1: Find the left limit of the function at x = 3

To find the left limit, we will evaluate the limit of the function \(x + 2\) as x approaches 3 from the left (that is, \(x \rightarrow 3^-\)): \[\lim_{x \rightarrow 3^-} (x+2)\] Simply plug in x = 3 into the equation: \[\lim_{x \rightarrow 3^-} (x+2) = 3 + 2 = 5\]

Step 2: Find the right limit of the function at x = 3

To find the right limit, we will evaluate the limit of the function \(\ln (x-2) + 5\) as x approaches 3 from the right (that is, \(x \rightarrow 3^+\)): \[\lim_{x \rightarrow 3^+} (\ln (x-2) + 5)\] Plug in x = 3 into the equation: \[\lim_{x \rightarrow 3^+} (\ln (x-2) + 5) = \ln (3-2) + 5 = \ln (1) + 5\] Since the natural logarithm of 1 is 0: \[\lim_{x \rightarrow 3^+} (\ln (x-2) + 5) = 0 + 5 = 5\]

Step 3: Compare the left and right limits

Now, compare the left and right limits at x = 3: \[\lim_{x \rightarrow 3^-} (x+2) = 5\] \[\lim_{x \rightarrow 3^+} (\ln (x-2) + 5) = 5\] Since the left and right limits are equal at x = 3, the function is continuous at x = 3.

Step 4: Conclusion

Since the left and right limits are equal at x = 3, the function is continuous at the transition point. Therefore, the function \(f(x)\) is continuous everywhere and has no points of discontinuity.

Key Concepts

Understanding Piecewise Functions

Piecewise functions are functions defined by different expressions based on which part of their domain is being considered. Typically, they are written using a brace to show each condition and its corresponding expression. In the exercise, the function \[ f(x) = \begin{cases} x + 2 & \text{if } x < 3 \ \ln(x-2) + 5 & \text{if } x \geq 3 \end{cases} \] is an example of a piecewise function.

• The first piece applies when \( x < 3 \), where the expression is \( x + 2 \).
• The second piece applies when \( x \geq 3 \), with the expression \( \ln(x-2) + 5 \).

Piecewise functions allow you to capture a variety of behaviors within a single formal expression. When dealing with them, it’s crucial to understand how the function behaves at the meeting points or boundaries. These are the values of \( x \) where the expression changes, like \( x = 3 \) in our example.

Exploring Limits of Functions

Limits are a core concept in calculus that help us understand the behavior of functions as they approach a certain point. For piecewise functions like in the exercise, examining the limits at the boundary points is essential to determine continuity. When approaching the point \( x = 3 \), we looked at the limits from both the left and right:

• The left-hand limit is what the function approaches as \( x \) gets arbitrarily close to 3 from values less than 3. Here, \( \lim_{x \to 3^-} (x+2) = 5 \).
• The right-hand limit is what the function approaches as \( x \) gets close to 3 from values greater than or equal to 3. We found \( \lim_{x \to 3^+} (\ln(x-2) + 5) = 5 \).

Continuity at a point requires that both these limits exist, are equal, and that the function is defined at that point. Since both limits equal 5, continuity is confirmed at \( x = 3 \).

The Role of the Natural Logarithm

The natural logarithm function, denoted as \( \ln(x) \), is a crucial component in calculus and is defined for all positive \( x \). It is the inverse of the exponential function with base \( e \), approximately equal to 2.718. In our piecewise function, the natural logarithm appears in the expression \( \ln(x-2) + 5 \). This part of the piecewise function is only valid for \( x \geq 3 \) because:

• The argument \( x-2 \) must be greater than zero for \( \ln(x-2) \) to be defined, hence \( x > 2 \).
• When \( x = 3 \), \( \ln(3-2) \) simplifies to \( \ln(1) = 0 \), as the natural log of 1 is 0. Thus, the result is 5.

The properties of the natural logarithm, such as its domain and how it behaves near its undefined points (like \( x = 2 \) here), are essential for ensuring our function’s continuity across pieces.