Evaluating Piecewise Functions A Comprehensive Guide
Hey guys! Today, we're diving into the world of piecewise functions. These functions might look a little intimidating at first, but trust me, they're not as scary as they seem. Think of them like a set of instructions – depending on the input you give, you follow a different rule. We will solve the problem of evaluating a piecewise function. Our function today is defined as follows:
g(x) = \begin{cases}
\frac{1}{2}x + 3 & \text{if } x \leq -4 \\
-x - 1 & \text{if } -4 < x < 1 \\
x^2 - 4 & \text{if } x \geq 1
\end{cases}
We're going to break down how to evaluate this function for different values of x. So, let's get started!
Understanding Piecewise Functions
Before we jump into the calculations, let's make sure we're all on the same page about what a piecewise function actually is. A piecewise function, in simple terms, is a function that's defined by multiple sub-functions, each applying to a specific interval of the domain. It's like having different rules for different situations. The key here is understanding which "piece" of the function applies to the input value you're given.
Think of it like this: imagine you're driving and the speed limit changes depending on where you are. In a school zone, the speed limit might be lower than on a highway. A piecewise function works similarly – the "rule" (or sub-function) changes based on the input (x in our case).
In our example, we have three pieces:
- The first piece, (1/2)x + 3, applies when x is less than or equal to -4. This is like our school zone – if our input is in this range, we follow this specific rule.
- The second piece, -x - 1, applies when x is greater than -4 and less than 1. This is like driving on a regular road with a different speed limit.
- The third piece, x^2 - 4, applies when x is greater than or equal to 1. This could be like our highway section, with yet another speed limit.
To evaluate a piecewise function, the most crucial step is identifying which interval your input x falls into. Once you know the correct interval, you simply plug your x value into the corresponding sub-function. This is where careful reading and attention to detail are important. Make sure you're using the correct inequality signs (≤, <, >, ≥) to determine the appropriate piece.
For example, if we want to find g(-5), we first need to figure out which interval -5 belongs to. Since -5 is less than or equal to -4, we'll use the first piece of the function. But if we want to find g(0), we'll need to use the second piece because 0 falls between -4 and 1. And for g(2), we'll use the third piece since 2 is greater than or equal to 1. See how it works? Let's put this into practice with our specific examples!
a) Evaluating g(2)
Let's start with finding the value of g(2). Remember, the first step is to figure out which piece of our piecewise function applies when x = 2. Looking at our function definition:
g(x) = \begin{cases}
\frac{1}{2}x + 3 & \text{if } x \leq -4 \\
-x - 1 & \text{if } -4 < x < 1 \\
x^2 - 4 & \text{if } x \geq 1
\end{cases}
We need to determine which of the conditions on the right-hand side is satisfied when x = 2. The first condition, x ≤ -4, is not true because 2 is not less than or equal to -4. The second condition, -4 < x < 1, is also not true because 2 is not less than 1. However, the third condition, x ≥ 1, is true because 2 is greater than or equal to 1. So, we know that we need to use the third piece of the function to evaluate g(2).
The third piece of our function is g(x) = x² - 4. This is the rule we'll follow when x is greater than or equal to 1. Now, it's a simple matter of substitution. We replace x with 2 in this expression:
g(2) = (2)² - 4
Next, we perform the calculation. First, we square 2, which gives us 4:
g(2) = 4 - 4
Finally, we subtract 4 from 4, which gives us 0:
g(2) = 0
Therefore, the value of g(2) is 0. Guys, we've successfully evaluated the piecewise function for x = 2! Notice how crucial it was to first identify the correct piece of the function to use. This is the key to working with piecewise functions. Now, let's move on to the next part and find g(-1).
b) Evaluating g(-1)
Okay, let's tackle g(-1) now. Just like before, our first step is to figure out which piece of our piecewise function applies when x = -1. Let's revisit our function definition:
g(x) = \begin{cases}
\frac{1}{2}x + 3 & \text{if } x \leq -4 \\
-x - 1 & \text{if } -4 < x < 1 \\
x^2 - 4 & \text{if } x \geq 1
\end{cases}
We need to check each condition on the right-hand side to see which one holds true when x = -1. The first condition, x ≤ -4, is not true because -1 is not less than or equal to -4. However, the second condition, -4 < x < 1, is true because -1 is greater than -4 and less than 1. The third condition, x ≥ 1, is not true because -1 is not greater than or equal to 1. Therefore, we know that we need to use the second piece of the function to evaluate g(-1).
The second piece of our function is g(x) = -x - 1. This is the rule we follow when x is between -4 and 1 (not including -4 and 1). Now, we substitute x with -1 in this expression:
g(-1) = -(-1) - 1
Remember that subtracting a negative number is the same as adding its positive counterpart. So, -(-1) becomes +1:
g(-1) = 1 - 1
Finally, we subtract 1 from 1, which gives us 0:
g(-1) = 0
Therefore, the value of g(-1) is 0. Great job, guys! We've successfully evaluated the piecewise function for x = -1. Again, the key was to correctly identify the interval that -1 falls into and then use the corresponding sub-function. Let's move on to our final evaluation and find g(-6).
c) Evaluating g(-6)
Alright, let's find the value of g(-6). By now, you guys probably know the drill! Our first step, as always, is to figure out which piece of the piecewise function applies when x = -6. Let's take a look at the function definition one more time:
g(x) = \begin{cases}
\frac{1}{2}x + 3 & \text{if } x \leq -4 \\
-x - 1 & \text{if } -4 < x < 1 \\
x^2 - 4 & \text{if } x \geq 1
\end{cases}
We need to check each condition on the right-hand side to see which one is true when x = -6. The first condition, x ≤ -4, is true because -6 is less than or equal to -4. The second condition, -4 < x < 1, is not true because -6 is not greater than -4. The third condition, x ≥ 1, is also not true because -6 is not greater than or equal to 1. So, we know that we need to use the first piece of the function to evaluate g(-6).
The first piece of our function is g(x) = (1/2)x + 3. This is the rule we'll follow when x is less than or equal to -4. Now, we substitute x with -6 in this expression:
g(-6) = (1/2)(-6) + 3
Next, we perform the multiplication. One-half times -6 is -3:
g(-6) = -3 + 3
Finally, we add -3 and 3, which gives us 0:
g(-6) = 0
Therefore, the value of g(-6) is 0. Fantastic! We've successfully evaluated the piecewise function for x = -6. You guys are becoming pros at this! The key takeaway here is the importance of carefully checking the conditions to determine which piece of the function applies to the given input.
Conclusion
So, there you have it! We've successfully evaluated our piecewise function for three different values of x: 2, -1, and -6. In all three cases, the value of the function was 0. Remember, the most important thing when working with piecewise functions is to carefully determine which piece of the function applies to your input value. Once you've done that, it's just a matter of plugging in the value and doing the math. You've got this!
I hope this step-by-step guide has helped you understand how to evaluate piecewise functions. Keep practicing, and you'll become a master in no time! If you have any questions, feel free to ask. Happy function-evaluating, guys!
