Continuity In Piecewise Functions Finding K For Continuity

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In the fascinating world of calculus, continuity plays a pivotal role in understanding the behavior of functions. A function is said to be continuous if its graph can be drawn without lifting the pen, meaning there are no abrupt breaks, jumps, or holes. This concept is particularly intriguing when dealing with piecewise functions, which are defined by different expressions over different intervals of their domain. In this article, we delve into the intricacies of determining the continuity of piecewise functions and calculating the values of parameters that ensure this continuity. Specifically, we will tackle two problems involving piecewise functions, each requiring us to find the value of a constant that makes the function continuous at a given point. Understanding these concepts is crucial for students and enthusiasts of mathematics, providing a solid foundation for more advanced topics in calculus and analysis.

Continuity: The Backbone of Calculus

Before diving into the problems, let's first understand the core concept of continuity. In simple terms, a function f(x) is continuous at a point x = a if three conditions are met:

  1. f(a) is defined (the function exists at that point).
  2. The limit of f(x) as x approaches a exists.
  3. The limit of f(x) as x approaches a is equal to f(a).

For piecewise functions, continuity is often a concern at the points where the function's definition changes. To ensure continuity at such points, the left-hand limit (the limit as x approaches from the left) and the right-hand limit (the limit as x approaches from the right) must be equal, and both must be equal to the function's value at that point.

The Piecewise Function

Our first problem presents us with the following piecewise function:

f(x) = egin{cases}
    kx + 1 & 	ext{for } x < 2 \ 
    2x - 3 & 	ext{for } x 	ext{ ≥ } 2

We are tasked with finding the value of k that makes this function continuous at x = 2. This is a classic problem that elegantly demonstrates the application of continuity principles to piecewise functions.

The Approach

To solve this, we need to ensure that the function's left-hand limit, right-hand limit, and value at x = 2 all coincide. Let's break down the steps:

  1. Evaluate the left-hand limit: This means we consider the limit as x approaches 2 from values less than 2. In this interval, f(x) = kx + 1. So, we need to calculate lim{x→2−} (kx + 1).
  2. Evaluate the right-hand limit: Here, we look at the limit as x approaches 2 from values greater than or equal to 2. The relevant expression is f(x) = 2x - 3. We need to find lim{x→2+} (2x - 3).
  3. Evaluate the function at x = 2: According to the function's definition, f(2) = 2(2) - 3.
  4. Equate the limits and the function value: For continuity, the left-hand limit, right-hand limit, and f(2) must all be equal. This will give us an equation in terms of k, which we can solve.

Step-by-Step Solution

Let's follow the steps outlined above to find the value of k.

1. Left-Hand Limit

The left-hand limit is given by:

lim{x→2−} (kx + 1)

Since kx + 1 is a polynomial, we can evaluate the limit by direct substitution:

lim{x→2−} (kx + 1) = k(2) + 1 = 2k + 1

2. Right-Hand Limit

The right-hand limit is given by:

lim{x→2+} (2x - 3)

Again, we can use direct substitution:

lim{x→2+} (2x - 3) = 2(2) - 3 = 4 - 3 = 1

3. Function Value at x = 2

The function's value at x = 2 is:

f(2) = 2(2) - 3 = 1

4. Equating Limits and Function Value

For continuity at x = 2, we must have:

lim{x→2−} f(x) = lim{x→2+} f(x) = f(2)

Substituting the values we found:

2k + 1 = 1 = 1

From the equation 2k + 1 = 1, we can solve for k:

2k = 1 - 1
2k = 0
k = 0

Therefore, the value of k that makes the function continuous at x = 2 is 0. This means that when k is zero, the two pieces of the function seamlessly connect at x = 2, ensuring the function's continuity.

The New Challenge

Let's consider another piecewise function, this time with a slight twist. We are given:

f(x) = egin{cases}
    5x + k & 	ext{for } x 	ext{ ≤ } 2 \ 
    6x & 	ext{for } x > 2

Our goal remains the same: to find the value of k that ensures continuity at the point where the function definition changes, which is x = 2.

The Strategy

The approach to this problem mirrors the one we used for the first function. We need to evaluate the left-hand limit, the right-hand limit, and the function's value at x = 2, and then equate them. This will lead us to an equation that we can solve for k. The key difference here lies in the specific expressions defining the function in each interval, but the underlying principle of continuity remains unchanged.

Detailed Solution

Let's proceed with the steps to find k.

1. Left-Hand Limit

To determine the left-hand limit, we consider the function's behavior as x approaches 2 from the left (values less than or equal to 2). In this case, f(x) = 5x + k. Therefore, we need to compute lim{x→2−} (5x + k).

Since 5x + k is a linear function, we can directly substitute x = 2 to find the limit:

lim{x→2−} (5x + k) = 5(2) + k = 10 + k

The left-hand limit is 10 + k. This value depends on the constant k, which is what we are trying to determine.

2. Right-Hand Limit

Now, let's evaluate the right-hand limit. This involves examining the function as x approaches 2 from the right (values greater than 2). Here, f(x) = 6x. So, we need to calculate lim{x→2+} (6x).

Again, we can use direct substitution to find the limit:

lim{x→2+} (6x) = 6(2) = 12

The right-hand limit is 12. This value is constant and does not depend on k.

3. Function Value at x = 2

To find the function's value at x = 2, we refer to the function definition for x ≤ 2, which is f(x) = 5x + k. Substituting x = 2, we get:

f(2) = 5(2) + k = 10 + k

Notice that the function's value at x = 2 is the same as the left-hand limit, which is 10 + k.

4. Equating Limits and Function Value

For the function to be continuous at x = 2, the left-hand limit, the right-hand limit, and the function's value at x = 2 must all be equal. This gives us the following equation:

lim{x→2−} f(x) = lim{x→2+} f(x) = f(2)

Substituting the values we calculated:

10 + k = 12 = 10 + k

We have the equation 10 + k = 12. Solving for k:

k = 12 - 10
k = 2

Therefore, the value of k that makes the function continuous at x = 2 is 2. This ensures that the two pieces of the function seamlessly connect at x = 2, creating a continuous graph.

In this article, we've explored the concept of continuity in the context of piecewise functions. We tackled two problems, each requiring us to find the value of a constant that ensures the function's continuity at a specific point. The key takeaway is that for a piecewise function to be continuous at a point where its definition changes, the left-hand limit, the right-hand limit, and the function's value at that point must all be equal. By applying this principle, we were able to successfully determine the values of k that make the given functions continuous.

These exercises not only reinforce our understanding of continuity but also highlight the importance of limits in calculus. The ability to manipulate and solve limits is fundamental to many advanced concepts in mathematics, and these types of problems serve as excellent practice for building those skills. As we continue our journey into the world of calculus, the principles of continuity and limits will remain essential tools in our mathematical toolkit. The exploration of continuity in piecewise functions provides a solid stepping stone towards understanding more complex functions and their behavior. The skills developed here will be invaluable in tackling more challenging problems and applications of calculus in various fields of science and engineering.