Piecewise Functions: Identifying Valid Function Relations

Hey guys! Let's dive into the world of piecewise functions and figure out what makes them tick. Specifically, we're going to look at how to determine whether a piecewise relation actually defines a function. Buckle up, it's gonna be a fun ride!

Understanding Piecewise Functions

Before we jump into analyzing the given relations, let's make sure we're all on the same page about what piecewise functions are. A piecewise function is basically a function that is defined by multiple sub-functions, each applying to a certain interval of the main function's domain. Think of it like a set of different rules that apply depending on where you are on the x-axis.

Each "piece" of the function is defined by:

• An expression (like x^2 or -3x).
• A domain interval (like x < -2 or -2 ≤ x ≤ 4).

The key thing to remember is that for a piecewise relation to be a function, it must pass the vertical line test. This means that for any single value of x, there can only be one corresponding value of y. In the context of piecewise functions, this translates to ensuring that the intervals don't overlap in a way that creates multiple y values for the same x value.

Let's consider this more closely. Imagine you have a piecewise function defined as follows:

f(x) = {
  x + 1,  x < 0
  x^2,   x >= 0
}

Here, for any x less than 0, the function behaves like x + 1. For any x greater than or equal to 0, the function behaves like x^2. At x = 0, the function uses the second rule, so f(0) = 0^2 = 0. There's no ambiguity, and this is a valid function.

However, if we had something like this:

f(x) = {
  x + 1,  x <= 0
  x^2,   x >= 0
}

Now, at x = 0, we have two definitions: f(0) could be 0 + 1 = 1 (from the first rule) or 0^2 = 0 (from the second rule). This is problematic because a function must have a single, well-defined output for each input. Thus, this is NOT a valid function.

Now that we have a solid grasp of what piecewise functions are and what makes them valid, let's apply this knowledge to the given options.

Analyzing the Piecewise Relations

We need to determine which of the provided piecewise relations defines a function. Remember, the key is to make sure that there's no overlap in the x intervals that would cause a single x value to map to multiple y values.

Option A

Let's take a look at option A:

$y=\left\{\begin{array}{cl}x^2, & x<-2 \\ 0, & -2 \geq 4\end{array}\right.$

Here's how we analyze it:

• For x < -2, y = x^2.
• For -2 ≤ x ≤ 4, y = 0.
• For x ≥ 4, y = -x^2.

Notice that the intervals are nicely separated. At x = -2, the function transitions from x^2 to 0. At x = 4, it transitions from 0 to -x^2. There's no overlap, and for each x, there's only one defined y. So, option A looks like a valid function!

To be absolutely sure, let's check the transition points:

• At x = -2, the first piece isn't defined at -2 (it's for x strictly less than -2), and the second piece is defined at -2, with y = 0. There's no conflict.
• At x = 4, the second piece is defined at 4, with y = 0, and the third piece is defined at 4, with y = -4^2 = -16. Here is a problem! Since there are two y values at x=4, this is not a valid function.

Thus, option A is not a valid function.

Option B

Now, let's examine option B:

$y=\left\{\begin{aligned} x^2, & x \leq-2 \\ 4, & -2 x \geq 1\end{aligned}\right.$

Alright, let's break it down:

• For x ≤ -2, y = x^2.
• For -2 Let's look at what happens at x = -2. At x = -2, y = x^2 which gives us (-2)^2 = 4. Also, since x can be less than or equal to -2 on the first part of the function and greater than or equal to -2 on the second part, we can plug x = -2 into the second part and it yields y = 4. We get the same y-value in both cases, which is valid.
• For -2 < x < 1, y = 4.
• For x >= 1, y = 2*x - 3.

Again, let's check the transition points:

• At x = -2, we have y = 4 for both adjacent function definitions. Thus, it is a valid function.
• At x = 1, y = 4 for the function definition on the left and y = 2(1) - 3 = -1 for the function definition on the right. There are two different y-values for the same x-value, meaning it is NOT a valid function.

Thus, option B is not a valid function.

Option C

Finally, let's check out option C:

$y=\left{\begin{aligned}-3 x, & x<-2 \-8, & -2

Let's see what's going on here:

• For x < -2, y = -3x.
• For -2 x = -2, the first part isn't defined at -2 (it's for x strictly less than -2), and the second piece is defined at -2, with y = -8. There's no conflict.
• At x = 6, the second piece is defined at 6, with y = -8, and the third piece is defined at 6, with y = x - 14 = 6 - 14 = -8. The y value is the same for both function definitions, meaning it is valid.

Thus, option C is a valid function.

Conclusion

After analyzing all three options, we found that only option C defines a valid function. The key to identifying valid piecewise functions is to ensure that the intervals don't overlap in a way that creates multiple y values for the same x value. Also, we have to check the y-values at the transition points to make sure the function behaves correctly.

So, there you have it! I hope this explanation helps you better understand piecewise functions and how to determine whether they define a valid function. Keep practicing, and you'll become a piecewise pro in no time!