Identifying Discontinuity In Piecewise Functions At X=0

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Identifying Discontinuity in Piecewise Functions at x=0

Hey math enthusiasts! Today, we're diving into the world of piecewise functions and figuring out which ones are discontinuous at the point where x equals 0. Discontinuity, in simple terms, means the function has a "break" or "jump" at a specific point. Let's break down each option and see which one fits the bill. This is a common topic in calculus and precalculus, so understanding this is super important, guys. We'll be using some basic limit concepts and function evaluation to get our answers. Ready to get started? Let's go!

Understanding Discontinuity

Before we jump into the functions, let's quickly recap what it means for a function to be discontinuous at a point. A function is discontinuous at a point, say x=c, if one or more of these conditions is not met:

  1. The function must exist at that point. Meaning, f(c) must be defined.
  2. The limit of the function must exist at that point. This implies the left-hand limit and the right-hand limit must be equal. Mathematically, lim x→c- f(x) = lim x→c+ f(x).
  3. The value of the function at that point must equal the limit at that point. In other words, f(c) = lim x→c f(x).

If any of these conditions fail, the function is discontinuous at x=c. There are various types of discontinuities like jump discontinuity, removable discontinuity, and infinite discontinuity. In the context of piecewise functions, we're mostly looking for jump discontinuities, where the function "jumps" from one value to another at the boundary point. So, the key is to check if the left and right limits are the same, and if they match the function's value at that point. Let's keep these rules in mind as we analyze each of the provided options. Understanding these principles will help us to dissect each piecewise function effectively and find the one that shows a break at x=0. The main goal here is to find where the graph “jumps” or has a hole, indicating a discontinuity.

The Importance of Limits

Understanding the concept of limits is essential when dealing with the continuity of functions, especially piecewise functions. Limits help us to analyze the behavior of a function as it approaches a certain point. The limit of a function at a point exists if the function approaches the same value from both the left and the right sides of that point. For a function to be continuous at a point, the limit at that point must exist, the function must be defined at that point, and the limit and the function's value must be the same. Piecewise functions are specifically designed with different definitions for different intervals of x-values. Therefore, limits are crucial to examine the transition points, like x=0 in our problem. If the limits from the left and right at x=0 are not equal, then we can confirm a discontinuity. Visualizing the graph can be really helpful. A continuous function will have a smooth curve without any gaps or jumps, while a discontinuous function will display one or more of these issues. Evaluating the limits helps us identify those points where a function might be behaving unexpectedly, and that’s what we are trying to do here.

Analyzing the Piecewise Functions

Now, let's carefully examine each of the provided piecewise functions. We'll check the left-hand and right-hand limits at x=0 and see if they match the function's value at that point. Keep in mind that for a function to be continuous at x=0, both the left and right-hand limits must be equal, and this value must equal the function's value at x=0. Let’s look at each option systematically. We will determine the limit from both sides (left and right) and check if the function is defined at x = 0 and what the value is at x = 0 to evaluate the continuity.

Option A: f(x) = { x^2 for x < 0; x^2 for x >= 0 }

In this function, both parts of the definition are the same: x^2. Let's find the limit from the left and the right.

  • Left-hand limit (x → 0-): As x approaches 0 from the left, f(x) = x^2. Thus, lim x→0- x^2 = 0^2 = 0.
  • Right-hand limit (x → 0+): As x approaches 0 from the right, f(x) = x^2. Thus, lim x→0+ x^2 = 0^2 = 0.
  • Value at x = 0: Since the function is defined as x^2 for x >= 0, f(0) = 0^2 = 0.

Since the left-hand limit, the right-hand limit, and the function's value at x=0 are all equal to 0, this function is continuous at x=0. So, it is not our answer!

Option B: f(x) = { x + 1 for x < 0; x + 1 for x >= 0 }

This function also has the same definition on both sides of x=0. Let's do the limits and function value check.

  • Left-hand limit (x → 0-): As x approaches 0 from the left, f(x) = x + 1. Thus, lim x→0- (x + 1) = 0 + 1 = 1.
  • Right-hand limit (x → 0+): As x approaches 0 from the right, f(x) = x + 1. Thus, lim x→0+ (x + 1) = 0 + 1 = 1.
  • Value at x = 0: Since the function is defined as x + 1 for x >= 0, f(0) = 0 + 1 = 1.

Again, the left-hand limit, the right-hand limit, and the function's value at x=0 are all equal to 1. This function is also continuous at x=0. Moving on!

Option C: f(x) = { x + 2 for x < 0; x - 2 for x >= 0 }

Here's where things get interesting, guys! This is the one we are looking for. The function changes its definition at x=0. Let's see what happens to the limits.

  • Left-hand limit (x → 0-): As x approaches 0 from the left, f(x) = x + 2. Thus, lim x→0- (x + 2) = 0 + 2 = 2.
  • Right-hand limit (x → 0+): As x approaches 0 from the right, f(x) = x - 2. Thus, lim x→0+ (x - 2) = 0 - 2 = -2.

Since the left-hand limit (2) and the right-hand limit (-2) are not equal, the limit does not exist at x=0. Therefore, this function is discontinuous at x=0. Bingo!

Conclusion: The Discontinuous Function

After analyzing all the options, we've found our answer. Option C, f(x) = { x + 2 for x < 0; x - 2 for x >= 0 }, is the piecewise function that is discontinuous at x=0. This is because the function "jumps" from a value of 2 (approached from the left) to a value of -2 (approached from the right) at x=0. The discontinuity is a jump discontinuity, as the function makes a sudden leap at that point. This confirms that at the point x=0, the function's graph has a break, satisfying our definition of discontinuity. Remember to carefully evaluate the limits from both sides of the point where the piecewise function changes definitions. That is the key to identifying discontinuities in piecewise functions.

Summary of Analysis

To recap:

  • Option A: Continuous at x=0.
  • Option B: Continuous at x=0.
  • Option C: Discontinuous at x=0 (Jump Discontinuity).

So there you have it, folks! Understanding how to analyze the continuity of piecewise functions is a super important skill in calculus. Keep practicing these types of problems, and you'll become a pro in no time! Remember to always check both the left and right limits and the function value to determine the continuity at a point. Thanks for tuning in, and happy calculating!