Piecewise Functions: Solving For Specific X Values

Hey guys! Today, we're diving deep into the fascinating world of piecewise functions. These are pretty neat because they're like a mathematical mashup, defined by different rules over different intervals of their domain. Think of it as a function that wears different hats depending on where you 'are' on the x-axis. We're going to tackle a specific problem: evaluating a piecewise function, $g(x)$, for several given values of $x$. This is a super important skill in understanding how functions behave, and once you get the hang of it, you'll be a piecewise pro! So, let's get ready to put on our analytical thinking caps and break down this function step-by-step. We've got a set of x-values to plug in: $g(-4)$, $g(5)$, $g(-2)$, $g(0)$, and $g(6)$. Each of these requires us to first figure out which piece of the function definition applies, and then perform the calculation. It's like a little detective game, but with numbers! By the end of this, you'll be able to confidently evaluate any piecewise function thrown your way. Let's jump right in and see how it's done!

Understanding Piecewise Functions

Alright, let's get to the core of what we're dealing with: piecewise functions. So, what exactly is a piecewise function? Imagine you have a single function, $g(x)$ in our case, but it's not described by just one single formula. Instead, it's defined by multiple smaller formulas, or 'pieces', each of which is valid only over a specific interval of the input values (the x-values). Our particular function, $g(x)$, is defined as follows:

$$g(x) = \begin{cases} -5|x|+7 & \text { for } x \leq-2 \\ 3 x+2 & \text { for } -2 < x < 5 \\ 6 & \text { for } x \geq 5 \end{cases}$$

See those conditions on the right? '$x \leq -2$', '$-2 < x < 5$', and '$x \geq 5$'? These are the crucial 'rules' that tell us which formula to use for any given input $x$. If your $x$ value is less than or equal to -2, you use the first formula, $-5|x|+7$. If your $x$ value is strictly between -2 and 5 (meaning it's greater than -2 BUT less than 5), you use the second formula, $3x+2$. And finally, if your $x$ value is greater than or equal to 5, you use the third 'formula', which is just the constant value 6. It's super important to pay close attention to these interval boundaries (like -2 and 5) and whether they are included (using $\leq$ or $\geq$) or excluded (using $<$ or $>$). These boundaries are where the function 'switches' from one rule to another. Understanding these intervals is the key to correctly evaluating any piecewise function. It's the foundation upon which all our calculations will be built, so let's make sure we've got this down pat. Don't worry if it seems a bit complex at first; with a bit of practice, it'll feel second nature, guys!

Evaluating $g(-4)$

Okay, team, let's kick things off by evaluating $g(-4)$. The very first step, and I can't stress this enough, is to determine which interval the input value $x = -4$ falls into. We need to look at our conditions: $x \leq -2$, $-2 < x < 5$, and $x \geq 5$. Let's check them one by one for $x = -4$:

1. Is $-4 \leq -2$? Yes, it is! -4 is indeed less than or equal to -2. This means the first rule of our piecewise function applies.

2. We don't even need to check the other intervals since we've found the correct one. But just for completeness, $-2 < -4 < 5$ is false, and $-4 \geq 5$ is also false.

Since $x = -4$ satisfies the condition $x \leq -2$, we must use the corresponding formula: $g(x) = -5|x|+7$. Now, we substitute $x = -4$ into this formula:

$g(-4) = -5|-4|+7$

Remember the absolute value! $|-4|$ is simply 4 because the absolute value of a number is its distance from zero, which is always positive. So, we have:

$g(-4) = -5(4)+7$

Now, we just perform the arithmetic. First, multiplication:

$g(-4) = -20+7$

And then, addition:

$g(-4) = -13$

So, there you have it! For an input of $x = -4$, the output of our function $g(x)$ is $-13$. See? Not too bad! It's all about identifying the correct piece of the function. Keep this process in mind as we move on to the next values. You guys are doing great!

Evaluating $g(5)$

Moving right along, let's tackle $g(5)$. Again, the critical first step is to figure out which interval $x = 5$ belongs to. Let's revisit our conditions:

• $x \leq -2$: Is $5 \leq -2$? No, this is false.
• $-2 < x < 5$: Is $-2 < 5 < 5$? No, this is false because 5 is not strictly less than 5.
• $x \geq 5$: Is $5 \geq 5$? Yes, this is true! 5 is equal to 5, so it satisfies the 'greater than or equal to' condition.

Since $x = 5$ satisfies the condition $x \geq 5$, we use the third rule defined for our function $g(x)$. This rule states that for any $x$ in this interval, $g(x) = 6$. It's a constant! This means no matter what value of $x$ we plug in as long as it's 5 or greater, the output will always be 6.

So, for $x = 5$:

$g(5) = 6$

And that's it! It's that simple when you hit a constant piece of a piecewise function. The value is just the constant itself. This highlights how different parts of the domain can lead to very different behaviors for the function. Pretty cool, right? Keep that focus, and let's nail the next one!

Evaluating $g(-2)$

Alright folks, let's evaluate $g(-2)$. This one is particularly interesting because $x = -2$ is one of our boundary points. We need to be super careful about which interval definition includes this boundary. Let's check our conditions again:

• $x \leq -2$: Is $-2 \leq -2$? Yes, this is true! Since -2 is equal to -2, it satisfies the 'less than or equal to' condition.

• $-2 < x < 5$: Is $-2 < -2 < 5$? No, this is false because -2 is not strictly greater than -2.

• $x \geq 5$: Is $-2 \geq 5$? No, this is false.

So, $x = -2$ falls into the first interval, where $x \leq -2$. The formula for this interval is $g(x) = -5|x|+7$. Now we substitute $x = -2$ into this formula:

$g(-2) = -5|-2|+7$

First, let's handle the absolute value: $|-2|$ is 2.

$g(-2) = -5(2)+7$

Next, the multiplication:

$g(-2) = -10+7$

Finally, the addition:

$g(-2) = -3$

So, $g(-2) = -3$. It's crucial to get these boundary points right. If the definition for the middle piece had been $-2 \leq x < 5$ (instead of $-2 < x < 5$), then $x=-2$ would have belonged to that second piece. Always double-check those inequality signs, guys! This carefulness is what separates a good evaluation from a tricky mistake.

Evaluating $g(0)$

Now, let's find the value of $g(0)$. We need to determine which interval $x = 0$ belongs to. Let's check:

• $x \leq -2$: Is $0 \leq -2$? No, this is false.
• $-2 < x < 5$: Is $-2 < 0 < 5$? Yes, this is true! 0 is greater than -2 AND less than 5.
• $x \geq 5$: Is $0 \geq 5$? No, this is false.

Since $x = 0$ fits the condition $-2 < x < 5$, we use the second formula for $g(x)$, which is $g(x) = 3x+2$. Now, we substitute $x=0$ into this formula:

$g(0) = 3(0)+2$

Perform the multiplication:

$g(0) = 0+2$

And the addition:

$g(0) = 2$

Therefore, $g(0) = 2$. This is a straightforward evaluation once we've identified the correct piece of the function. It really shows how the middle 'linear' piece of the function behaves for values between -2 and 5. Keep up the awesome work, everyone!

Evaluating $g(6)$

Last but certainly not least, let's evaluate $g(6)$. We need to find the right interval for $x = 6$.

• $x \leq -2$: Is $6 \leq -2$? No, false.
• $-2 < x < 5$: Is $-2 < 6 < 5$? No, false (6 is not less than 5).
• $x \geq 5$: Is $6 \geq 5$? Yes, this is true! 6 is greater than 5.

Since $x = 6$ satisfies the condition $x \geq 5$, we use the third rule for $g(x)$, which is the constant function $g(x) = 6$.

So, for $x = 6$:

$g(6) = 6$

Just like $g(5)$, this value is 6 because it falls into the interval where the function is defined as a constant 6. It's a good reminder that piecewise functions can have constant segments, which are super easy to evaluate once you've identified them. And that concludes our evaluations for all the given x-values!

Conclusion

And there you have it, guys! We've successfully navigated through evaluating our piecewise function $g(x)$ for several different values of $x$. Remember the key steps: 1. Identify the input value ($x$). 2. Determine which interval the input value falls into by checking the conditions for each piece. 3. Use the corresponding formula for that interval to calculate the output. We saw how different pieces of the function yield different results, from linear calculations like $3x+2$ to absolute value expressions like $-5|x|+7$, and even constant values like 6. We also paid special attention to boundary points, like $x=-2$, ensuring we used the correct inequality. Mastering piecewise functions is a fundamental skill in mathematics, and by working through these examples, you've taken a significant step forward. Keep practicing, keep questioning, and you'll find these problems become second nature. Happy calculating!