Continuity Of A Piecewise Function: A Detailed Solution

Oct 28, 2025 by Admin 56 views

Hey guys! Today, we're diving deep into a fascinating problem in calculus: determining the points of continuity for a piecewise function. Specifically, we'll be looking at a function defined differently for rational and irrational numbers. This type of problem really tests our understanding of continuity, limits, and the nature of real numbers. So, buckle up, and let's get started!

Understanding the Problem

Before we jump into solving, let's make sure we fully grasp the problem. We're given a piecewise function, which means it's defined by different formulas over different parts of its domain. In this case, our function, let's call it g(x), behaves one way for rational numbers and another way for irrational numbers. The big question is: where is this function continuous? To answer this, we need to recall the definition of continuity and think about how rational and irrational numbers intertwine on the number line.

Defining the Piecewise Function

Our function g(x) is defined as follows:

g(x) = 
  \begin{cases}
    2x - 2 & \text{if } x \in \mathbb{Q} \\
    x + 3 & \text{if } x \in \mathbb{R} \setminus \mathbb{Q}
  \end{cases}

Here:

${ \mathbb{Q} }$ represents the set of rational numbers (numbers that can be expressed as a fraction ${ p/q }$ , where ${ p }$ and ${ q }$ are integers and ${ q \neq 0 }$ ).
${ \mathbb{R} }$ is the set of all real numbers.
${ \mathbb{R} \setminus \mathbb{Q} }$ denotes the set of irrational numbers (real numbers that cannot be expressed as a fraction).

So, if you plug in a rational number into g(x), you use the formula ${ 2x - 2 }$ . If you plug in an irrational number, you use ${ x + 3 }$ . This split definition is what makes this problem interesting!

What is Continuity?

Now, let's quickly recap what it means for a function to be continuous at a point. A function g(x) is continuous at a point c if three conditions are met:

g(c) is defined (the function has a value at c).
The limit of g(x) as x approaches c exists ( ${ \lim_{x \to c} g(x) }$ exists).
The limit of g(x) as x approaches c is equal to g(c) ( ${ \lim_{x \to c} g(x) = g(c) }$ ).

In simpler terms, for a function to be continuous at a point, there shouldn't be any breaks, jumps, or holes at that point. The function should smoothly transition through that point. This concept is crucial for solving our problem.

Solving for Continuity Points

Okay, let's get down to the nitty-gritty and find the points where g(x) is continuous. The key to solving this problem lies in understanding how the rational and irrational numbers are distributed. They are densely packed within the real numbers. This means that between any two real numbers, you can always find both a rational number and an irrational number.

The Strategy

Given the dense nature of rational and irrational numbers, for the limit of g(x) to exist at a point c, the two pieces of the function must approach the same value as x approaches c. In other words, the limits from the rational and irrational "sides" must match. This is because no matter how close we get to c, we'll always encounter both rational and irrational numbers.

So, our strategy is to find the points c where the two pieces of our function are equal:

2c - 2 = c + 3

This equation represents the condition where the rational part of the function and the irrational part of the function "meet." Let's solve it!

Solving the Equation

Solving the equation ${ 2c - 2 = c + 3 }$ is straightforward algebra. Let's walk through it:

Subtract c from both sides:
```
2c - c - 2 = c - c + 3
c - 2 = 3
```
Add 2 to both sides:
```
c - 2 + 2 = 3 + 2
c = 5
```

So, we've found that c = 5 is the only point where the two pieces of the function have the same value. This is a strong candidate for a point of continuity.

Verifying Continuity at c = 5

Now we need to formally verify that g(x) is indeed continuous at c = 5. To do this, we need to check the three conditions of continuity we discussed earlier:

g(5) is defined:

Since 5 is a rational number, we use the rational part of the function:
```
g(5) = 2(5) - 2 = 10 - 2 = 8
```
So, g(5) = 8, and the function is defined at c = 5.
The limit of g(x) as x approaches 5 exists:

We need to show that the limit exists and is the same whether we approach 5 through rational or irrational numbers. Let's consider the limits:
- Limit through rational numbers:
```
\lim_{x \to 5, x \in \mathbb{Q}} g(x) = \lim_{x \to 5} (2x - 2) = 2(5) - 2 = 8
```
- Limit through irrational numbers:
```
\lim_{x \to 5, x \in \mathbb{R} \setminus \mathbb{Q}} g(x) = \lim_{x \to 5} (x + 3) = 5 + 3 = 8
```
Since both limits are equal to 8, the overall limit exists and is equal to 8:
```
\lim_{x \to 5} g(x) = 8
```
The limit equals the function value:

We have ${ \lim_{x \to 5} g(x) = 8 }$ and ${ g(5) = 8 }$ , so the third condition is also satisfied.

Since all three conditions for continuity are met, we can confidently conclude that g(x) is continuous at c = 5.

Discontinuity Elsewhere

But what about other points? Let's think about what happens if ${ x \neq 5 }$ . If x is not equal to 5, then ${ 2x - 2 }$ will not be equal to ${ x + 3 }$ . Because rational and irrational numbers are densely packed, we can always find sequences of rational numbers and sequences of irrational numbers that approach any given x. However, since the function values for the rational and irrational sequences will approach different limits ( ${ 2x - 2 }$ and ${ x + 3 }$ , respectively), the overall limit will not exist at any point other than x = 5. Therefore, the function is discontinuous everywhere except at x = 5.

Conclusion

So, guys, we've successfully navigated through this tricky problem! We've shown that the piecewise function g(x) is continuous only at the point x = 5. This problem highlights the importance of understanding the definition of continuity and the properties of real numbers, particularly the dense nature of rational and irrational numbers. Remember, for a piecewise function like this to be continuous, the pieces have to "meet" at a point, and the limits from both sides must agree.

This type of question is a classic in real analysis and calculus courses, so mastering this concept will definitely help you in your mathematical journey. Keep practicing, and you'll become a pro at these continuity problems in no time! If you have any questions, feel free to ask. Happy problem-solving!